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Systematic Circuit Analysis
Nodal Analysis
Chapter 4 Section 1
3 Fundamental Principles
bull Ohmrsquos Law V = I R where I enters at the higher
voltage side
bull Kirchoffrsquos Voltage Law Algebraic sum of voltages around a loop
equal zero
bull Kirchoffrsquos Current Law Algebraic sum of currents entering and
leaving a node equal zero
Circuit Simplificationbull Series Resistors can be combined into a
single equivalent resistancebull Parallel Resistors can be combined into a
single equivalent resistancebull Current sources in parallel add
algebraicallybull Voltage sources in series add algebraicallybull A voltage source is series with a resistor
can be replaced by a current source in parallel or vice versa
Shortcuts
bull Voltage Divider ndash a voltage dividing across a series combination of resistorsndash Largest voltage is across the largest resistor
bull Current Divider ndash a current dividing among a parallel combination of resistorsndash Largest current is through the smallest
resistor
Nodal Analysis
bull Employs the 3 fundamental laws
bull May not require any circuit simplification
bull Consists of a straight-forward step-by-step procedure
bull Involves the solution of a set of linear equations simultaneously
bull Used by most circuit analysis computer programs like PSpice
Example
bull Find the voltages and currents
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node Vo=0 volts
bull Label the other nodes with V1 V2 etc
Voltage Notation
bull Remember that voltage is always measured with respect to some other point
bull So V1 the voltage at node 1 minus the voltage at node 0 or V1-Vo = V10 = V1
bull V12 would be the voltage at node 1 minus the voltage at node 2 or V12 = V1-V2
Voltage Notation
bull V12 could also be obtained by going through any path say node 3
V12 = V132 where V132 = V13 + V32
But V13 + V32 = (V1-V3) + (V3-V2)
And (V1-V3) + (V3-V2) = V1-V2 = V12
Circuit After Step OneV1
Vo = 0 volts
Step Two
bull Choose currents for every circuit branch that is connected to any of the unknown voltage nodesndash In this circuit choose currents for the
branches connected to V1
ndash You can also label voltages at other nodes in the circuit (nodes that have only two components) Va and Vb
Circuit after Step Two
I1
+ - I3
+ -
+I2
-
V1
Vo = 0 volts
VaVb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etc
bull The idea is to get all the currents to be expressed in terms of the node voltages so that a KCL equation can be written at each unknown node giving us as many equations as unknowns
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
3 Fundamental Principles
bull Ohmrsquos Law V = I R where I enters at the higher
voltage side
bull Kirchoffrsquos Voltage Law Algebraic sum of voltages around a loop
equal zero
bull Kirchoffrsquos Current Law Algebraic sum of currents entering and
leaving a node equal zero
Circuit Simplificationbull Series Resistors can be combined into a
single equivalent resistancebull Parallel Resistors can be combined into a
single equivalent resistancebull Current sources in parallel add
algebraicallybull Voltage sources in series add algebraicallybull A voltage source is series with a resistor
can be replaced by a current source in parallel or vice versa
Shortcuts
bull Voltage Divider ndash a voltage dividing across a series combination of resistorsndash Largest voltage is across the largest resistor
bull Current Divider ndash a current dividing among a parallel combination of resistorsndash Largest current is through the smallest
resistor
Nodal Analysis
bull Employs the 3 fundamental laws
bull May not require any circuit simplification
bull Consists of a straight-forward step-by-step procedure
bull Involves the solution of a set of linear equations simultaneously
bull Used by most circuit analysis computer programs like PSpice
Example
bull Find the voltages and currents
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node Vo=0 volts
bull Label the other nodes with V1 V2 etc
Voltage Notation
bull Remember that voltage is always measured with respect to some other point
bull So V1 the voltage at node 1 minus the voltage at node 0 or V1-Vo = V10 = V1
bull V12 would be the voltage at node 1 minus the voltage at node 2 or V12 = V1-V2
Voltage Notation
bull V12 could also be obtained by going through any path say node 3
V12 = V132 where V132 = V13 + V32
But V13 + V32 = (V1-V3) + (V3-V2)
And (V1-V3) + (V3-V2) = V1-V2 = V12
Circuit After Step OneV1
Vo = 0 volts
Step Two
bull Choose currents for every circuit branch that is connected to any of the unknown voltage nodesndash In this circuit choose currents for the
branches connected to V1
ndash You can also label voltages at other nodes in the circuit (nodes that have only two components) Va and Vb
Circuit after Step Two
I1
+ - I3
+ -
+I2
-
V1
Vo = 0 volts
VaVb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etc
bull The idea is to get all the currents to be expressed in terms of the node voltages so that a KCL equation can be written at each unknown node giving us as many equations as unknowns
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Circuit Simplificationbull Series Resistors can be combined into a
single equivalent resistancebull Parallel Resistors can be combined into a
single equivalent resistancebull Current sources in parallel add
algebraicallybull Voltage sources in series add algebraicallybull A voltage source is series with a resistor
can be replaced by a current source in parallel or vice versa
Shortcuts
bull Voltage Divider ndash a voltage dividing across a series combination of resistorsndash Largest voltage is across the largest resistor
bull Current Divider ndash a current dividing among a parallel combination of resistorsndash Largest current is through the smallest
resistor
Nodal Analysis
bull Employs the 3 fundamental laws
bull May not require any circuit simplification
bull Consists of a straight-forward step-by-step procedure
bull Involves the solution of a set of linear equations simultaneously
bull Used by most circuit analysis computer programs like PSpice
Example
bull Find the voltages and currents
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node Vo=0 volts
bull Label the other nodes with V1 V2 etc
Voltage Notation
bull Remember that voltage is always measured with respect to some other point
bull So V1 the voltage at node 1 minus the voltage at node 0 or V1-Vo = V10 = V1
bull V12 would be the voltage at node 1 minus the voltage at node 2 or V12 = V1-V2
Voltage Notation
bull V12 could also be obtained by going through any path say node 3
V12 = V132 where V132 = V13 + V32
But V13 + V32 = (V1-V3) + (V3-V2)
And (V1-V3) + (V3-V2) = V1-V2 = V12
Circuit After Step OneV1
Vo = 0 volts
Step Two
bull Choose currents for every circuit branch that is connected to any of the unknown voltage nodesndash In this circuit choose currents for the
branches connected to V1
ndash You can also label voltages at other nodes in the circuit (nodes that have only two components) Va and Vb
Circuit after Step Two
I1
+ - I3
+ -
+I2
-
V1
Vo = 0 volts
VaVb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etc
bull The idea is to get all the currents to be expressed in terms of the node voltages so that a KCL equation can be written at each unknown node giving us as many equations as unknowns
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Shortcuts
bull Voltage Divider ndash a voltage dividing across a series combination of resistorsndash Largest voltage is across the largest resistor
bull Current Divider ndash a current dividing among a parallel combination of resistorsndash Largest current is through the smallest
resistor
Nodal Analysis
bull Employs the 3 fundamental laws
bull May not require any circuit simplification
bull Consists of a straight-forward step-by-step procedure
bull Involves the solution of a set of linear equations simultaneously
bull Used by most circuit analysis computer programs like PSpice
Example
bull Find the voltages and currents
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node Vo=0 volts
bull Label the other nodes with V1 V2 etc
Voltage Notation
bull Remember that voltage is always measured with respect to some other point
bull So V1 the voltage at node 1 minus the voltage at node 0 or V1-Vo = V10 = V1
bull V12 would be the voltage at node 1 minus the voltage at node 2 or V12 = V1-V2
Voltage Notation
bull V12 could also be obtained by going through any path say node 3
V12 = V132 where V132 = V13 + V32
But V13 + V32 = (V1-V3) + (V3-V2)
And (V1-V3) + (V3-V2) = V1-V2 = V12
Circuit After Step OneV1
Vo = 0 volts
Step Two
bull Choose currents for every circuit branch that is connected to any of the unknown voltage nodesndash In this circuit choose currents for the
branches connected to V1
ndash You can also label voltages at other nodes in the circuit (nodes that have only two components) Va and Vb
Circuit after Step Two
I1
+ - I3
+ -
+I2
-
V1
Vo = 0 volts
VaVb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etc
bull The idea is to get all the currents to be expressed in terms of the node voltages so that a KCL equation can be written at each unknown node giving us as many equations as unknowns
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Nodal Analysis
bull Employs the 3 fundamental laws
bull May not require any circuit simplification
bull Consists of a straight-forward step-by-step procedure
bull Involves the solution of a set of linear equations simultaneously
bull Used by most circuit analysis computer programs like PSpice
Example
bull Find the voltages and currents
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node Vo=0 volts
bull Label the other nodes with V1 V2 etc
Voltage Notation
bull Remember that voltage is always measured with respect to some other point
bull So V1 the voltage at node 1 minus the voltage at node 0 or V1-Vo = V10 = V1
bull V12 would be the voltage at node 1 minus the voltage at node 2 or V12 = V1-V2
Voltage Notation
bull V12 could also be obtained by going through any path say node 3
V12 = V132 where V132 = V13 + V32
But V13 + V32 = (V1-V3) + (V3-V2)
And (V1-V3) + (V3-V2) = V1-V2 = V12
Circuit After Step OneV1
Vo = 0 volts
Step Two
bull Choose currents for every circuit branch that is connected to any of the unknown voltage nodesndash In this circuit choose currents for the
branches connected to V1
ndash You can also label voltages at other nodes in the circuit (nodes that have only two components) Va and Vb
Circuit after Step Two
I1
+ - I3
+ -
+I2
-
V1
Vo = 0 volts
VaVb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etc
bull The idea is to get all the currents to be expressed in terms of the node voltages so that a KCL equation can be written at each unknown node giving us as many equations as unknowns
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Example
bull Find the voltages and currents
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node Vo=0 volts
bull Label the other nodes with V1 V2 etc
Voltage Notation
bull Remember that voltage is always measured with respect to some other point
bull So V1 the voltage at node 1 minus the voltage at node 0 or V1-Vo = V10 = V1
bull V12 would be the voltage at node 1 minus the voltage at node 2 or V12 = V1-V2
Voltage Notation
bull V12 could also be obtained by going through any path say node 3
V12 = V132 where V132 = V13 + V32
But V13 + V32 = (V1-V3) + (V3-V2)
And (V1-V3) + (V3-V2) = V1-V2 = V12
Circuit After Step OneV1
Vo = 0 volts
Step Two
bull Choose currents for every circuit branch that is connected to any of the unknown voltage nodesndash In this circuit choose currents for the
branches connected to V1
ndash You can also label voltages at other nodes in the circuit (nodes that have only two components) Va and Vb
Circuit after Step Two
I1
+ - I3
+ -
+I2
-
V1
Vo = 0 volts
VaVb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etc
bull The idea is to get all the currents to be expressed in terms of the node voltages so that a KCL equation can be written at each unknown node giving us as many equations as unknowns
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node Vo=0 volts
bull Label the other nodes with V1 V2 etc
Voltage Notation
bull Remember that voltage is always measured with respect to some other point
bull So V1 the voltage at node 1 minus the voltage at node 0 or V1-Vo = V10 = V1
bull V12 would be the voltage at node 1 minus the voltage at node 2 or V12 = V1-V2
Voltage Notation
bull V12 could also be obtained by going through any path say node 3
V12 = V132 where V132 = V13 + V32
But V13 + V32 = (V1-V3) + (V3-V2)
And (V1-V3) + (V3-V2) = V1-V2 = V12
Circuit After Step OneV1
Vo = 0 volts
Step Two
bull Choose currents for every circuit branch that is connected to any of the unknown voltage nodesndash In this circuit choose currents for the
branches connected to V1
ndash You can also label voltages at other nodes in the circuit (nodes that have only two components) Va and Vb
Circuit after Step Two
I1
+ - I3
+ -
+I2
-
V1
Vo = 0 volts
VaVb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etc
bull The idea is to get all the currents to be expressed in terms of the node voltages so that a KCL equation can be written at each unknown node giving us as many equations as unknowns
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Voltage Notation
bull Remember that voltage is always measured with respect to some other point
bull So V1 the voltage at node 1 minus the voltage at node 0 or V1-Vo = V10 = V1
bull V12 would be the voltage at node 1 minus the voltage at node 2 or V12 = V1-V2
Voltage Notation
bull V12 could also be obtained by going through any path say node 3
V12 = V132 where V132 = V13 + V32
But V13 + V32 = (V1-V3) + (V3-V2)
And (V1-V3) + (V3-V2) = V1-V2 = V12
Circuit After Step OneV1
Vo = 0 volts
Step Two
bull Choose currents for every circuit branch that is connected to any of the unknown voltage nodesndash In this circuit choose currents for the
branches connected to V1
ndash You can also label voltages at other nodes in the circuit (nodes that have only two components) Va and Vb
Circuit after Step Two
I1
+ - I3
+ -
+I2
-
V1
Vo = 0 volts
VaVb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etc
bull The idea is to get all the currents to be expressed in terms of the node voltages so that a KCL equation can be written at each unknown node giving us as many equations as unknowns
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Voltage Notation
bull V12 could also be obtained by going through any path say node 3
V12 = V132 where V132 = V13 + V32
But V13 + V32 = (V1-V3) + (V3-V2)
And (V1-V3) + (V3-V2) = V1-V2 = V12
Circuit After Step OneV1
Vo = 0 volts
Step Two
bull Choose currents for every circuit branch that is connected to any of the unknown voltage nodesndash In this circuit choose currents for the
branches connected to V1
ndash You can also label voltages at other nodes in the circuit (nodes that have only two components) Va and Vb
Circuit after Step Two
I1
+ - I3
+ -
+I2
-
V1
Vo = 0 volts
VaVb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etc
bull The idea is to get all the currents to be expressed in terms of the node voltages so that a KCL equation can be written at each unknown node giving us as many equations as unknowns
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Circuit After Step OneV1
Vo = 0 volts
Step Two
bull Choose currents for every circuit branch that is connected to any of the unknown voltage nodesndash In this circuit choose currents for the
branches connected to V1
ndash You can also label voltages at other nodes in the circuit (nodes that have only two components) Va and Vb
Circuit after Step Two
I1
+ - I3
+ -
+I2
-
V1
Vo = 0 volts
VaVb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etc
bull The idea is to get all the currents to be expressed in terms of the node voltages so that a KCL equation can be written at each unknown node giving us as many equations as unknowns
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step Two
bull Choose currents for every circuit branch that is connected to any of the unknown voltage nodesndash In this circuit choose currents for the
branches connected to V1
ndash You can also label voltages at other nodes in the circuit (nodes that have only two components) Va and Vb
Circuit after Step Two
I1
+ - I3
+ -
+I2
-
V1
Vo = 0 volts
VaVb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etc
bull The idea is to get all the currents to be expressed in terms of the node voltages so that a KCL equation can be written at each unknown node giving us as many equations as unknowns
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Circuit after Step Two
I1
+ - I3
+ -
+I2
-
V1
Vo = 0 volts
VaVb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etc
bull The idea is to get all the currents to be expressed in terms of the node voltages so that a KCL equation can be written at each unknown node giving us as many equations as unknowns
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etc
bull The idea is to get all the currents to be expressed in terms of the node voltages so that a KCL equation can be written at each unknown node giving us as many equations as unknowns
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step Three
bull If you have a resistor between two nodes then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
bull For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)I2 = (V1 ndash 0) 10Ω = V1 10Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step Three Cont
bull If you have a resistor and a series voltage source between two key nodes determine the voltage at the node between the components and then use Ohmrsquos law
bull There is a 6v source and 14Ω resistor in series between node 1 and groundVa = Vo+6v = 6vI1 = (Va - V1) 14Ω = (6v - V1) 14Ω
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step Three Cont
bull There is also 5v source and 10Ω resistor in series between node 1 and ground
bull Determine the voltage at the intermediate node Vb = Vo+5v = 5v
bull Use Ohmrsquos law to determine the current I3 = (V1-Vb) 14Ω = (V1-5v) 10Ω
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Circuit after Step Three
I1=(6-V1)14Ω+ -
I3=(V1-5)10Ω + -
+ I2 =V110Ω
-
V1
Vo = 0 volts
Va=6v Vb=5v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI1 = I2 + I3
bull Substitute each value or expression into each KCL equation(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step 5 Solve
bull Solve the set of equations for each unknown voltage(6v-V1)14Ω = V110Ω + (V1-5v)10Ω
bull Multiply both sides to clear fractions70Ω[(6v-V1)14Ω = V110Ω + (V1-5v)10Ω]
30v - 5V1= 7V1 + 7V1 - 35v
bull Group like terms and solve65v = 19V1 or V1 = 342v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step 5 Solve Cont
bull Since V1 = 342v
bull Find each current by substituting the voltage values found into each current equation as appropriateI1 = (6v-V1)14Ω = (6-342)14 = 184 A
I2 = V110Ω = 34210 = 342 A
I3 = (V1-5v)10Ω = (342-5)10 = -158 A
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step 6 ndash Reality Check
I1=184A+ -
I3= -158A + -
+ I2=342A
-
V1=342v
Vo = 0 volts
Va=6v Vb=5v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Example with a Dependent Source
bull Find the voltages and currents
I1
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Handling the Dependent Source
bull The dependent source will be dependent on some voltage or current in the circuit
bull You will need to express that voltage or current in terms of the unknown node voltages
bull The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step One
bull Identify all nodes that have at least 3 components connected to them
bull Choose one of the above nodes as the ground or reference node at zero volts
bull Label the other nodes with V1 V2 etc
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Circuit after Step 1
V1
Vo = 0 volts
I1
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step Two
bull Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Circuit after Step 2
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step Three
bull Find the value of each current or find an expression for each current in terms of the unknown voltages V1 V2 etcIf you have a resistor between two nodes
then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohmrsquos Law)
I2 = (V1 ndash 0) 4Ω = V1 4Ω
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step Three Cont
bull If you have a resistor and a series voltage source between two nodes determine the voltage at the node between the components and then use Ohmrsquos lawVb = Vo+3v = 3vI1 = (V1-Vb) 2Ω = (V1-3v) 2Ω
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step Three Cont
bull If you have multiple resistors andor voltage sources between two key nodes combine them together to get one resistor and one sourceIf there is a dependent part of the source
express it in terms of the unknown node voltages
Determine the voltage at the node between the equivalent resistor and source
Use Ohmrsquos law
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Circuit for Step 3
V1
Vo = 0 volts - I3 +
+I2
-
I1
+ - Vb=3v
5v + 4ΩI1
VaVb
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step Three Cont
bull The combined dependent source is 5v+4ΩI1
So Va = V1 ndash (5v+4ΩI1)
But I1 in the dependent source was determined to be (V1-3v)2Ω
So Va = V1ndash(5v + 4Ω(V1-3v)2Ω) = -V1+1
bull I3 = (Vo-Va) 4Ω = (0-(-V1+1)) 4Ω I3 = (V1 -1v) 4Ω
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Circuit for Step 4
V1
Vo = 0 volts - +I3=(V1-1)4Ω
+ I2=V12Ω -
I1=(V1-3)2Ω+ - Vb=3v
5v + 4ΩI1
Va=1v-V1
Vb
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step 4
bull Write Kirchoffrsquos Current Law at each node with an unknown voltageI3 = I2 + I1
bull Substitute each value or expression into each KCL equation(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step 5 Solve
bull Solve the set of equations for each unknown voltage(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω
bull Multiply both sides to clear fractions4Ω[(V1 -1v) 4Ω = V1 4Ω + (V1-3v) 2Ω]
V1 - 1v = V1 + 2(V1-3v)
bull Group like terms and solve5v = 2V1 or V1 = 25v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step 5 Solving for Currents
bull Finding each current
bull I3 = (V1 -1v) 4Ω = (25v -1v) 4Ω = 375 A
bull I2 = V1 4Ω = 25v 4Ω = 625 A
bull I1 = (V1-3v) 2Ω = (25v-3v) 2Ω = -25 A
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step 6 - Checking
V1=25v
Vo = 0 volts - + I3=375A
+ I2=625A -
I1= -25A+ - Vb=3v
5v+4ΩI1=4v
Va=1v-V1=-15v
Vb
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Class Activity
bull Find the current equations at node 1
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Class Activitybull Find I2 in terms of V1
For a resistor between nodes use Ohmrsquos lawAdd polarities if not shown (+ at V1 - at V0)
I2 = V160Ω
+
_
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Class Activitybull Find I3 in terms of its value
For a current source between nodes the current is fixed by the source
I3 = 2 A
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Class Activitybull Find I4 in terms of V1
If you have series resistors combine them and then use Ohmrsquos law
I4 = V1(30+70)Ω = V1(100Ω)
+_+_
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Class Activitybull Find I1 in terms of V1
Combine series resistors even if not next to each other then find the voltage between the source and equivalent resistance
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Finding I1
100+80=180Ω
Va=100v
So I1 = (Va-V1)180Ω = (100-V1)180Ω
Vo=0v Vo=0v
V1 V1+ - + -
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Class Activitybull Find I5 in terms of V1
The voltage on the right side of the 50Ω resistor is known because of the 150v source
I5 = (V1-150v) 50Ω
+ -
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Class Activitybull Write KCL equation at node 1
I1 + I3 = I2 + I4 + I5
Substitute all the current equations(100-V1)180+2A=V160+V1100+(V1-150v)50
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Class Activity(100-V1)180+2A=V160+V1100+(V1-150v)50
Multiply both sides by 9000Ω
5000-50V1+18000=150V1+90V1+180V1-27000
Group 50000 = 470V1 Solve V1= 1064 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Class ActivityChecking
I1=-036A I2=1773A I3=2A I4=1064A I5=-872A
I1+I3= -036+2 =1964 A
I2+I4+I5 = 1773+1064-872 = 1965 A ndash Yes
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Example with 3 Unknown Nodes
bull Find Voltages and Currents
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Steps 1 amp 2
bull Find key nodes assign one to ground choose currents in branches
Vo=0 volts
V1 V2V3
Va
I1
I5
I3I2
I4
+ -
+ -
+ -
+
-
+
-
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step 3 ndash Current Equations
bull Currents for resistors between nodesI1=(V1-V3)4Ω
I3=(V2-V3)7Ω
I4=V21Ω
I5= V35Ω
bull Resistor and source between nodesVa = V1+ 9v
I2 = (Va-V2)3Ω = (V1+9v-V2)3Ω
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Step 4 ndash KCL Equations
bull KCL Equation at each key nodeAt node 1 0 = I1 + I2 + 8A
At node 2 I2 = I3 + I4
At node 3 25A + I1 + I3 = I5
bull Substituting for each currentAt 1 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
At 2 (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
At 3 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
At node 1
bull 0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A
bull Multiply both sides by 12Ω to clear fractions12Ω[0 =(V1-V3)4Ω +(V1+9v-V2)3Ω +8A]
Or 0 =3V1-3V3 +4V1+36v-4V2 +96v
bull Combining terms7V1 -4V2 -3V3 = -132v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
At node 2
bull (V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω
bull Multiply both sides by 21Ω to clear fractions21Ω[(V1+9v-V2)3Ω =(V2-V3)7Ω + V21Ω ]
Or 7V1+63v - 7V2 = 3V2 - 3V3 +21V2
bull Combining terms7V1 -31V2 +3V3 = -63v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
At node 3
bull 25A+(V1-V3)4Ω +(V2-V3)7Ω = V35Ω
bull Multiply both sides by 140Ω to clear fractions140Ω[25A+(V1-V3)4Ω +(V2-V3)7Ω =V35Ω]
Or 3500v+35V1-35V3+20V2-20V3=28V3
bull Combining terms35V1 +20V2 -83V3 = -3500v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
Set of 3 Simultaneous Eqs
bull 7V1 -4V2 -3V3 = -132v
bull 7V1 -31V2 +3V3 = -63v
bull 35V1 +20V2 -83V3 = -3500v
bull Solve by hand calculator or computerV1 = 5414 v
V2 = 7737 v
V3 = 46316 v
