Kirchof Laws

8/3/2019 Kirchof Laws

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KIRCHHOFF'S LAWS

1.

Many circuits are too complex to be solved using the rules for series or parallel circuits or thetechniques for conversion to simpler circuits described in previous chapters. For these circuits we

need more general solution methods. The most general method is given by Kirchhoffs laws,which permit the calculation of all circuit voltages and currents of circuits by a solution of a

system of linear equations.

There are two Kirchhoff laws, the voltage law and the current law. These two laws can be used

to determine all voltages and currents of circuits.

Kirchhoffs voltage law (KVL) states that the algebraic sum of the voltage rises and voltagedrops around a loop must be zero.

A loop in the above definition means a closed path in the circuit; that is, a path that leaves a node

in one direction and returns to that same node from another direction.

In our examples, we will use clockwise direction for loops; however, the same results will beobtained if the counterclockwise direction is used.

In order to apply KVL without error, we have to define the so called reference direction. Thereference direction of the unknown voltages points from the + to the sign of the assumed

voltages. Imagine using a voltmeter. You would place the voltmeter positive probe (usually red)at the components reference + terminal. If the real voltage is positive, it is in the same direction

as we assumed, and both our solution and the voltmeter will show a positive value.

When deriving the algebraic sum of the voltages, we must assign a plus sign to those voltageswhere the reference direction agrees with the direction of the loop, and negative signs in the

opposite case.

Another way to state Kirchhoffs voltage law is: the applied voltage of a series circuit equals the

sum of the voltage drops across the series elements.

The following short example shows the use of Kirchhoffs voltage law.

Find the voltage across resistor R2, given that the source voltage, VS = 100 V and that thevoltage across resistor R1 is V1 = 40 V.


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The figure below can be created with TINA Pro Version 6 and above, in which drawing tools areavailable in the schematic editor.

The solution using Kirchhoffs voltage law: -VS + V1 + V2 =0, or VS = V1 + V2

hence: V2 = VS V1 = 100-40 = 60V

Note that normally we dont know the voltages of the resistors (unless we measurethem), and we need to use both Kirchhoffs laws for the solution.

Kirchhoffs current law (KCL) states that the algebraic sum of all the currents enteringand leaving any node in a circuit is zero.

In the following, we give a + sign to currents leaving a node and a - sign to currents entering anode.

Heres a basic example demonstrating Kirchhoffs current law.


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Find the current I2 if the source

current IS = 12 A, and I1 = 8 A.

Using Kirchhoffs currentlaw at the circled node: -IS + I1 + I2 = 0 , hence: I2= IS I1 = 12 8 = 4 A, as you can checkusing TINA (next figure).

In the next example, we will use both Kirchhoffs laws plus Ohms law to calculate the current

and the voltage across the resistors.

In the figure below, you will note the Voltage Arrowabove resistors. This is a new component

available in Version 6 of TINA and works like a voltmeter. If you connect it across acomponent, the arrow determines the reference direction (to compare to a voltmeter, imagine

placing the red probe at the tail of the arrow and the black probe at the tip). When you run DCanalysis, the actual voltage on the component will be displayed on the arrow.


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Click here to loador save this circuit

To begin using Kirchhoffs current law, we see that the currents through all the components are

the same, so lets denote that current by I.

According to Kirchhoffs voltage law: VS = V1+V2+V3

Now using Ohms law: VS=I*R1+ I*R2+I *R3

And from here the current of the circuit:

I=VS /(R1+R2+R3)= 120/(10+20+30) = 2 A

Finally the voltages of the resistors:

V1=I*R1 = 2*10 = 20 V; V2 = I*R2 = 2*20 = 40 V; V3 = I*R3 =2*30 = 60 V

The same results will be seen on the Voltage Arrows by simply running TINAs interactive DCanalysis.

In this next, more complex circuit, we also use both Kirchhoffs laws and Ohms law, but wefind that we most solve a linear system of equations.


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The total number of independent applications of Kirchhoffs laws in a circuit is the number ofcircuit branches, while the total number of unknowns (the current and voltage of each branch) is

twice that . However, by also using Ohms law at each resistor and the simple equationsdefining the applied voltages and currents, we get a system of equation where the number of

unknowns is the same as the number of equations.

Find the branch currents I1, I2, I3 in the circuit below.

Clickheretoloador savethis circuitThe set of equations follows:

The nodal equation for the circled node:

- I1 - I2 - I3 = 0

or multiplying by -1

I1 + I2 + I3 = 0

The loop equations (using the clockwise direction) for the loop L1, containing V1, R1 andR3

-V1+I1*R1-I3*R3 = 0

and for the loop L2, containing V2, R2 and R3

I3*R3 I2*R2 +V2 = 0

Substituting the component values:

I1+ I2+ I3 = 0 -8+40*I1 40*I3 = 0 40*I3 20*I2 +16 = 0


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Express I1 using the nodal equation: I1 = -I2 I3

then substitute it into the second equation:

-V1 (I2 + I3)*R1 I3*R3 = 0 or 8- (I2 + I3)*40 I3*40 = 0

Express I2 and substitute it into the third equation, from which you can already calculate I3:

I2 = - (V1 + I3*(R1+R3))/R1 or I2 = -(8 + I3*80)/40

I3*R3 + R2*(V1 + I3*(R1+R3))/R1 +V2 = 0 or I3*40 + 20*(8 + I3*80)/40 + 16 = 0

And: I3 = - (V2 + V1*R2/R1)/(R3+(R1+R3)*R2/R1) or I3 = -(16+8*20/40)/(40 + 80*20/40)

Therefore I3 = - 0.25 A; I2 = -(8-0.25*80)/40 = 0.3 A and I1 = - (0.3-0.25) = - 0.05 A

Or: I1 = -50 mA; I2 = 300 mA; I3 = -250 mA.

Now lets solve the same equations with TINAs interpreter:

{Solution by TINA's Interpreter}

Sys I1,I2,I3

I1+I2+I3=0

-V1+I1*R1-I3*R3=0

I3*R3-I2*R2+V2=0

end;

I1=[-50m]

I2=[300m]

I3=[-250m]

Finally lets check the results using TINA:

Next, lets analyze the following even more complex circuit and determine its branch currents

and voltages.


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Clickheretoloador savethis circuit

Lets denote the unknown voltages and currents by adding voltage and current arrows to

components, and also show the loops (L1,L2, L3) and the nodes (N1,N2) where we will use theKirchhoffs equations.

Click here to loador save this circuitHere is the set of Kirchhoff equations for the loops (using the clockwise direction) and thenodes.

-IL + IR1 - Is = 0 (for N1)

- IR1 + IR2 + Is3 = 0 (for N2)

-Vs1 - VR3 + VIs + VL = 0 (for L1)

-VIs + Vs2 +VR2 +VR1 = 0 (for L2)

-VR2 - Vs2 + Vs3 = 0 (for L3)

Applying Ohms law:


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VL = IL*RL

VR1 =IR1*R1

VR2 = IR2*R2

VR3 = - IL*R3

This is 9 unknowns and 9 equations. The easiest way to solve this is to use TINAs

interpreter. However, if we are pressed to use hand calculations, we note that this set of

equations can be easily reduced to a system of 5 unknowns by substituting the last 4 equationsinto the L1, L2, L3 loop equations. Also, by adding equations (L1) and (L2), we can eliminate

VIs , reducing the problem to a system of 4 equations for 4 unknowns (IL, IR1 IR2, Is3). When wehave found these currents, we can easily determine VL , VR1, VR2, and VR3 using the last four

equations (Ohms law).

Substituting VL ,VR1,VR2 ,VR3 :

-IL + IR1 - Is = 0 (for N1)

- IR1 + IR2 + Is3 = 0 (for N2)

-Vs1 + IL*R3 + VIs + IL*RL = 0 (for L1)

-VIs + Vs2 + IR2*R2 + IR1*R1 = 0 (for L2)

- IR2*R2 - Vs2 + Vs3 = 0 (for L3)

Adding (L1) and (L2) we get

-IL + IR1 - Is = 0 (for N1)

- IR1 + IR2 + Is3 = 0 (for N2)

-Vs1 + IL*R3 + IL*RL + Vs2 + IR2*R2 + IR1*R1 = 0 (L1) + (L2)

- IR2*R2 - Vs2 + Vs3 = 0 (for L3)

After substituting the component values, the solution to these equations comes readily.

-IL+IR1 2 = 0

(for N1)

-IR1 + IR2 + IS3 = 0(for N2)


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-120 - + IL*90 + IL*20 + 60 + IR2*40 + IR1*30 = 0 (L1) + (L2)

-IR2*40 60 + 270 = 0(for L3)

from L3 IR2 = 210/40 = 5.25 A (I)

from N2 IS3 IR1 = - 5.25 (II)

from L1+L2 110 IL + 30 IR1 = -150 (III)

and for N1 IR1 IL = 2 (IV)

Multiply (IV) by 30 and add to (III) 140 IL = -210 hence IL = - 1.5 A

Substitute IL into (IV) IR1 = 2 + (-1.5) = 0.5 A

and IR1 into (II) IS3 = -5.25 + IR1 = -4,75 A

And the voltages: VR1 = IR1*R1 = 15 V; VR2 = IR2*R2 = 210 V;

VR3 = - IL*R3= 135 V; VL = IL*RL = - 30 V; VIs = VS1+VR3-VL = 285 V

{Solution of the original equations by TINA's Interpreter}

Sys IL,IR1,IR2,Is3,VIs,VL,VR1,VR3,VR2

-IL-Is+IR1=0

-IR1+IR2+Is3=0

-Vs1+VR3+Vis-VL=0

-Vis+VR1+VR2+Vs2=0

-Vs3+VR2+Vs2=0

VR1=IR1*R1

VR2=IR2*R2

VR3=-IL*R3

VL=IL*RL

end;IL=[-1.5]

IR1=[500m]

IR2=[5.25]

Is3=[-4.75]

VIs=[285]

VL=[-30]

VR1=[15]

VR2=[210]

VR3=[135]


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Solution of the reduced set of equations using the interpreter:

{Solution of the reduced set of equations by TINA's Interpreter}

Sys Il,Ir1,Ir2,Is3

-Il+Ir1-2=0

-Ir1+Ir2+Is3=0

-120+110*Il+60+40*Ir2+30*Ir1=0-40*Ir2+210=0

end;

Il=[-1.5]

Ir1=[500m]

Ir2=[5.25]

Is3=[-4.75]

We can also enter expressions for the voltages and have TINAs Interpreter calculate them:

Il:=-1.5;

Ir1:=0.5;

Ir2:=5.25;Is3:=-4.75;

Vl:=Il*RL;

Vr1:=Ir1*R1

Vr2:=Ir2*R2;

Vr3:=-Il*R3;

VIs:=Vs1-Vl+Vr3;

Vl=[-30]

Vr1=[15]

Vr2=[210]

Vr3=[135]

VIs=[285]

We can check

the result with TINA by simply turning on TINAs DC interactive mode or using Analysis /

DC Analysis / Nodal Voltages


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2.

The complete set of Kirchhoffs equations can be significantly simplified by the nodepotential method described in this chapter. Using this method, Kirchhoffs voltage law issatisfied automatically, and we need only write node equations to satisfy Kirchhoffs

current law, too. Satisfying Kirchhoffs voltage law is achieved by using node potentials(also called node or nodal voltages) with respect to a particular node called thereference node. In other words, all the voltages in the circuit are relative to the referencenode, which is normally considered to have 0 potential. It is easy to see that with thesevoltage definitions Kirchhoffs voltage law is satisfied automatically, since writing loopequations with these potentials leads to identity. Note that for a circuit having N nodesyou should write only N - 1 equations. Normally, the node equation for the referencenode is left out.

The sum of all currents in the circuit is zero since each current is flowing in and out of a node.Therefore, the Nth node equation is not independent from the previous N-1 equations. If we

included all the N equations, we would have an unsolvable system of equations.

The node potential method (also called nodal analysis) is the method best suited to computer

applications. Most circuit analysis programs--including TINA--are based on this method.

The steps of the nodal analysis:

1. Pick a reference node with 0 node potential and label each remaining node with V1, V2 or

N1,N2 and so on.

2. Apply Kirchhoffs current law at each node except the reference node. Use Ohms law to

express unknown currents from node potentials and voltage source voltages when necessary.For all unknown currents, assume the same reference direction (e.g. pointing out of the node)for each application of Kirchhoffs current law.

3. Solve the resulting node equations for the node voltages.

4. Determine any requested current or voltage in the circuit using the node voltages.

Let us illustrate step 2 by writing the node equation for node V 1 of the following circuitfragment:


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First, find the current from node V1 to node V2

.We will use Ohms Law at R1. The voltage across R1 is

V1 - V2 - VS1

And the current through R1 (and from node V1 to node V2) is

Note that this current has a reference direction pointing out of the V1 node. Using the convention

for currents pointing out of a node, it should be taken into account in the node equation with apositive sign.

The current expression of the branch between V1 and V3 will be similar, but since VS2 is in theopposite direction from VS1 (which means the potential of the node between VS2 and R2 is V3-

VS2), the current is

Finally, because of the indicated reference direction, IS2 should have a positive sign and IS1 anegative sign in the node equation.

The node equation:


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Now lets see a complete example to demonstrate the use of the node potential method.

Find the voltage V and the currents through the resistors in the circuit below


Since we have only two nodes in this circuit, we can reduce the solution to the determination ofone unknown quantity. By choosing the lower node as a reference node, the unknown node

voltage is the voltage were solving for, V.

Click

here

toload

or s

ave

this

circuit

The nodal equation for the upper node:

Numerically:


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Multiply by 30: 7.5+3V 30 + 1.5 V + 7.5.+ V 40 = 0 5.5 V 55 = 0

Hence: V = 10 V


Sys V

I+(V-Vs1)/R1+(V+Vs2)/R2+(V-Vs3)/R3=0

end;

V=[10]

Now lets determine the currents through the resistors. This is easy, since the same currents are

used in the nodal equation above.


{Use node potential method !}Sys V

I+(V-Vs1)/R1+(V+Vs2)/R2+(V-Vs3)/R3=0

end;

V=[10]

{The currents of the resistors}

IR1:=(V-Vs1)/R1;

IR2:=(V+Vs2)/R2;

IR3:=(V-Vs3)/R3;

IR1=[0]

IR2=[750.0001m]

IR3=[-1000m]

We can check the result with TINA by simply turning on TINAs DC interactive mode or using

the Analysis / DC Analysis / Nodal Voltages command.


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Next, lets solve the problem which was already used as the last example of the Kirchhoffs laws

chapter-


Find the voltages and currents of each element of the circuit.

Choosing the lower node as a reference node of 0 potential, the nodal voltage of N2 will be equal

to VS3, : N2 = therefore we have only one unknown nodal voltage. You may remember that

previously, using the full set of Kirchhoffs equations, even after some simplifications, we had alinear system of equations of 4 unknowns.

Writing the node equations for node N1, let us denote the nodal voltage of N1 by N1

The simple equation to solve is:


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Numerically:

Multiply by 330, we get:

3N1-360 660 + 11N1 2970 = 0 p N1= 285 V

After calculating N1, it is easy to calculate the other quantities in the circuit.

The currents:

IS3 = IR1 IR2 = 0.5 5.25 = - 4.75 A

And the voltages:

VIs = N1 = 285 V

VR1= (N1 VS3) = 285 270 = 15 V

VR2 = (VS3 VS2) = 270 60 =210 V

VL = -(N1-VS1-VR3) = -285 +120 +135 = - 30 V


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You may note that with the node potential method you still need some extra calculation todetermine the currents and voltages of the circuit. However these calculations are very simple,

much simpler than solving linear equations systems for all circuit quantities simultaneously.

We can check the result with TINA by simply turning on TINAs DC interactive mode or using

Analysis / DC Analysis / Nodal Voltages command.


Lets see further examples.

Example 1

Find the current I.


In this circuit there are four nodes, but since we have an ideal voltage source that determines the

node voltage at its positive pole, we should chose its negative pole as the reference node.

Therefore, we really have only two unknown node potentials: N1 and N2 .


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The equations for the nodes of potentials N1 and N2:

Numerically:

so the system of linear equations is:

To solve this, multiply the first equation by 3 and the second by 2, then add the two equations:

11N1 =220

and hence N1 = 20V, N2 =(50 + 5N1) / 6= 25 V

Finally the unknown current:


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The solution of a system of linear equations can be also calculated using Cramersrule.

Lets illustrate the use of Cramers rule by solving the system above again..

1. Fill in the matrix of the coefficients of unknowns:

2. Calculate the value of the determinant of the D matrix.

`D` = 7*6 (-5)*(-4) = 22

3. Place the values of the right hand side in the column of the coefficients of the

unknown variable then calculate the value of the determinant:

4. Divide the newly found determinants by the original determinant, to find the followingratios:

Hence N1 = 20 V and N2 = 25 V

To check the result with TINA, simply turn on TINAs DC interactive mode or use Analysis /DC Analysis / Nodal Voltages command. Note that using the Voltage Pincomponent of TINA,

you can directly show the node potentials assuming that the Groundcomponent is connected tothe reference node.


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Sys fi1,fi2

(fi1-fi2)/R2+(fi1-VS1)/R3+fi1/R4=0

(fi2-fi1)/R2+(fi2-VS1)/R1-Is=0

end;fi1=[20]

fi2=[25]

I:=(fi2-VS1)/R1;

I=[500m]

Example 2.Find the voltage of the resistor R4.

R1 = R3 = 100 ohm, R2 = R4 = 50 ohm, R5 = 20 ohm, R6 = 40 ohm, R7 = 75 ohm


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Clickhere to load or save this circuit

In this case, it is practical to choose the negative pole of the voltage source VS2 as the referencenode because then the positive pole of the VS2 voltage source will have VS2 = 150 node potential.

Because of this choice, however, the required V voltage is opposite to the node voltage of thenode N4; therefore V4 = - V.

The equations:

We do not present the hand calculations here, since the equations can be easily solved by TINAsinterpreter.


{Use node potential method !}

Sys V,V1,V2,V3

V1/R2+(V1-Vs2)/R1-Is=0

(V2+V)/R6+(V2-V3+Vs1)/R5+Is=0


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(V3+V)/R7+(V3-Vs2)/R3+(V3-Vs1-V2)/R5=0

(-V-V2)/R6-V/R4+(-V-V3)/R7=0

end;

V1=[116.6667]

V2=[-91.8182]

V3=[19.697]V=[34.8485]

To check the result with, TINA simply turn on TINAs DC interactive mode or use Analysis /

DC Analysis / Nodal Voltages command. Note that we have to place a few voltage pins on the

nodes to show the node voltages.


3.

Another way of simplifying the complete set of Kirchhoffs equations is the mesh or loop current

method. Using this method, Kirchhoffs current law is satisfied automatically, and the loop

equations that we write also satisfy Kirchhoffs voltage law. Satisfying Kirchhoffs current law is

achieved by assigning closed current loops called mesh or loop currents to each independent

loop of the circuit and using these currents to express all the other quantities of the circuit. Since

the loop currents are closed, the current that flows into a node must also flow out of the node;

so writing node equations with these currents leads to identity.


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Let us first consider the method of mesh currents.

We first note that the mesh current method is only applicable for planar circuits. Planar circuits

have no crossing wires when drawn on a plane. Often, by redrawing a circuit which appears to

be non-planar, you can determine that it is, in fact, planar. For non-planar circuits, use the loop

current methoddescribed later in this chapter.

To explain the idea of mesh currents, imagine the branches of the circuit as fishing net and

assign a mesh current to each mesh of the net. (Sometimes it is also said that a closed current

loop is assigned in each window of the circuit.)

Theschematic diagramThefishing net or the graph of the circuit

The technique of representing the circuit by a simple drawing, called a graph, is quite powerful.

Since Kirchhoffs laws do not depend on the nature of the components, you can disregard the

concrete components and substitute for them simple line segments, called the branches of the

graph. Representing circuits by graphs allows us to use the techniques of mathematical graph

theory. This helps us explore the topological nature of a circuit and determine the independent

loops. Come back later to this site to read more about this topic.

The steps of mesh current analysis:

1. Assign a mesh current to each mesh. Although the direction is arbitrary, it is customaryto use the clockwise direction.

2. Apply Kirchhoffs voltage law (KVL) around each mesh, in the same direction as themesh currents. If a resistor has two or more mesh currents through it, the total currentthrough the resistor is calculated as the algebraic sum of the mesh currents. In otherwords, if a current flowing through the resistor has the same direction as the meshcurrent of the loop, it has a positive sign, otherwise a negative sign in the sum. Voltagesources are taken into account as usual, If their direction is the same as the meshcurrent, their voltage is taken to be positive, otherwise negative, in the KVL equations.

Usually, for current sources, only one mesh current flows through the source, and thatcurrent has the same direction as the current of the source. If this is not the case, usethe more general loop current method, described later in this paragraph. There is noneed to write KVL equations for loops containing mesh currents assigned to currentsources.

3. Solve the resulting loop equations for the mesh currents.


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4. Determine any requested current or voltage in the circuit using the mesh currents.

Let us illustrate the method by the following example:Find the current I in the circuit below.

Click here to load or save this circuit

We see that there are two meshes (or a left and right window) in this circuit. Lets assign the

clockwise mesh currents J1 and J2 to the meshes. Then we write the KVL equations, expressing

the voltages across the resistors by Ohms law:

-V1 + J1*(Ri1+R1) J2*R1 = 0

V2 J1*R1 + J2*(R + R1) = 0

Numerically:

-12 + J1*17 J2*2 = 0

6 J1*2 +J2*14 = 0

Express J1 from the first equation: J1 = and then substitute into the second

equation: 6 2*

+14*J2 = 0

multiply by 17: 102 - 24 + 4*J2 + 238*J2 = 0 hence J2 =

and J1 =


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Finally, the required current:

{Solution using TINA's Interpreter}

{Mesh current method}

Sys J1,J2

J1*(Ri1+R1)-J2*R1-V1=0

J1*R1+J2*(R1+R)+V2=0

end;

J1=[666.6667m]

J2=[-333.3333m]

I:=J1-J2;

I=[1]

Lets check the results with TINA:

Click here to load or save this circuit

Next, lets solve the previous example again, but with the more general method of loop

currents. Using this method, the closed current loops, called loop currents, are assigned not

necessarily to the meshes of the circuit, but to arbitrary independent loops. You can ensure

that the loops are independent by having at least one component in each loop that is not

contained in any other loop. For planar circuits, the number of the independent loops is the

same as the number of meshes, which is easy to see.


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A more precise way of determining the number of independent loops is as follows.

Given a circuit with b branches and Nnodes. The number of the independent loops l is:

l= b - N+ 1

This follows from fact that the number of independent Kirchhoffs equations must be equal to the

branches in the circuit, and we already know that there are only N-1 independent node

equations. Therefore the total number of the Kirchhoffs equations is

b = N-1 + l and hence l= b - N+ 1

This equation also follows from the fundamental theorem of graph theory which will be

described later at this site.

Now lets solve the previous example again, but more simply, by using the loop current method.

With this method we are free to use loops in meshes or any other loops, but lets keep the loop

with J1 in the left mesh of the circuit. However, for the second loop we choose the loop with J2,

as shown in the figure below. The advantage of this choice is that J1 will be equal to the

requested current I, since it is the only loop current passing through R1. This means that we

dont need to calculate J2at all.Note that, unlike real currents, the physical meaning of loop

currents is dependent upon how we assign them to the circuit.


The KVL equations:

J1*(R1+ Ri1) + J2*R i1 V1 = 0


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-V1+J1*Ri1 +J2*(R + Ri) + V2 = 0

and the required current: I = J1

Numerically: J1*(15+2)+J2*15-12 = 0

-12 + J1*15 + J2*(15+12) + 6 = 0

Express J2 from the second equation:

Substitute into the first equation:

Hence: J1 = I = 1 A

Further examples.

Example 1Find the current I in the circuit below.


In this circuit, we use the method of loop currents. In the left window of the circuit we take a loop

current which we denote with I since it is equal to the requested current. The other loop current

is equal to the Is1 source current, so we denote it directly as IS1.

Note that that the direction of this loop current is notclockwise since its direction is determined

by the current source. However, since this loop current is already known, there is no need to

write the KVL equation for the loop where IS1 is taken.

Therefore the only equation to solve is:

-V1 + I*R2 + R1 *(I - IS1) = 0


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hence

I= (V1 + R1 *IS1)/( R1 + R2)

NumericallyI=(10+20*4)/(20+10)=3 A

You can also generate this result calling TINAs symbolic analysis from the Analysis/Symbolic

Analysis/DC Result menu:

Or you can solve the KVL equation by the interpreter:

{Solution by TINAs Interpreter}

{Use mesh current method}

Sys I

-V1 + I*R2 + R1 *(I - IS1) = 0

end;

I=[3]

The following example has 3 current sources and is very easy to solve by the method of loop

currents.

Example 2Find the voltage V.


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In this example, we can choose three loop currents so that each passes through only one

current source. Therefore, all the three loop currents are known, and we only need to express

the unknown voltage, V, using them.

Making the algebraic sum of the currents through R3:

V= (IS3 - IS2)*R3=(10-5)*30 = 150 V. You can verify this with TINA:.


Next, lets tackle again a problem that we have already solved in the Kirchhoffs laws and Node

potentialmethodchapters.

Example 3

Find the voltage V of the resistor R4.


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R1 = R3 = 100 ohm, R2 = R4 = 50 ohm, R5 = 20 ohm,R6 = 40 ohm, R7 = 75 ohm.

This problem needed at least 4 equations to solve in the previous chapters.

Solving this problem with the method of loop currents, we have four independent loops, but with

the proper choice of loop currents, one of the loop currents will be equal to the source current Is.


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Based on the loop currents shown in the figure above, the loop equations are:

VS1+I4*(R5+R6+R7) IS*R6I3*(R5 + R6) = 0

VS2- I3*(R1+R2) IS*R2 + I2*(R1 + R2)= 0

-VS1 + I3*(R1 + R2 + R3 + R4 + R5 + R6) + IS*(R2 +R4 + R6) I4*(R5 + R6) - I2*(R1 + R2) = 0

The unknown voltageVcan be expressed by the loop currents:

V = R4 * (I2 + I3)

Numerically:

100+I4*135-2*40-I3*60 = 0

150+I2*150-2*50-I3*150 = 0

100+I3*360+2*140-I4*60-I2*150 = 0

V = 50*(2+I3)

We can use Cramers rule to solve this system of equations:

I4 = D3/D

where D is the determinant of the system. D4, the determinant for I4, is formed by substituting

the right hand side of the system is placed for the column of I4s coefficients.

The system of equations in ordered form:

- 60* I3 +135*I4= -20

150*I2-150*I3 = - 50

-150*I2+360*I3 - 60*I4= - 180

So the determinant D:


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The solution of this system of equations is:

V = R4*(2+I3) = 34.8485 V

You can confirm the answer via the result calculated by TINA.


{Solution using TINA's Interpreter}

Sys I2,I3,I4

Vs2+I2*(R1+R2)-R2*Is-I3*(R1+R2)=0-Vs1+I3*(R1+R2+R3+R4+R5+R6)+Is*(R2+R4+R6)-I2*(R1+R2)-I4*(R5+R6)=0

Vs1+I4*(R5+R6+R7)-Is*R6-I3*(R5+R6)=0

end;

I2=[-1.6364]

I3=[-1.303]

I4=[-727.2727m]

V:=R4*(Is+I3);


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V=[34.8485]

In this example, each unknown loop current is a branch current (I1, I3 and I4); so it is easy to

check the result by comparison with the DC analysis results ofTINA.

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Kirchof Laws