UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS · unit 4 dc equivalent circuit and network theorems 1.0 Kirchoff’s Law Kirchoff’s Current Law (KCL) states at any junction in

UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS

MARLIANA/JKE/POLISAS/ET101-UNIT4 1

UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS 1.0 Kirchoff’s Law

Kirchoff’s Current Law (KCL) states at any junction in an electric circuit the total current flowing

towards that junction is equal to the total current flowing away from the junction, i.e. I = 0 Thus , referring to figure 1:

I2

I3

I1I4

I5

Figure 1

Kirchoff’s Voltage Law (KVL) states in any closed loop in a network, the algebraic sum Figure 2 of the voltage drops (i.e. products of current and resistance) taken around the loop is equal to the resultant e.m.f. acting in that loop.

V

R1

R2

Figure 2

1.1 Mesh analysis Analysis using KVL to solve for the currents around each closed loop of the network and hence determine the currents through and voltages across each elements of the network.

Mesh analysis procedure: 1. Assign a distinct current to each closed loop of the network. 2. Apply KVL around each closed loop of the network. 3. Solve the resulting simultaneous linear equation for the loop currents.

current towards = current flowing away I1 + I2+ I3 = I4 + I5

I1 + I2 + (- I3 ) + (- I4 ) + (-I5 ) = 0

I = 0

E = IR1 + IR2 E = I(R1 + R2 ) E + (- IR1 ) + (- IR2) = 0



Example 1 Find the current flow through each resistor using mesh analysis for the circuit below.

R1 R2

R3V1 V2

10Ω 20Ω

40Ω

10V 20V

Figure 3

Solution: Step 1: Assign a distinct current to each closed loop of the network.

R1 R2

R3V1 V2

10Ω 20Ω

40Ω

10V 20VI1 I2

I1 I2

I13

Figure 4

Step 2: Apply KVL around each closed loop of the network. Loop 1:

------------ equation 1

Loop 2:

--------------- equation 2 Step 3: Solve the resulting simultaneous linear equation for the loop currents. Solve equation 1 and 2 using matrix

Matrix form:

From KCL :




R1 R2

R3V1 V2

5kΩ 3kΩ

6kΩ

40V 55V

Figure 5

Solution: Step 1: Assign a distinct current to each closed loop of the network.

R1 R2

R3V1 V2

5kΩ 3kΩ

6kΩ

40V 55VI2I1

I1 I2

I3

Figure 6

Step 2: Apply KVL around each closed loop of the network. Loop 1:

------------ equation 1

Loop 2:

--------------- equation 2 Step 3: Solve the resulting simultaneous linear equation for the loop currents. Solve equation 1 and 2 using matrix

Matrix form:

From KCL :



1.2 Nodes analysis Analysis using KCL to solve for voltages at each common node of the network and hence determines the currents through and voltages across each elements of the network. Nodal analysis procedure: 1. Determine the number of common nodes and reference node within the network. 2. Assign current and its direction to each distinct branch of the nodes in the network. 3. Apply KCL at each of the common nodes in the network 4. Solve the resulting simultaneous linear equation for the nodal voltages. 5. Determine the currents through and voltages across each the elements in the

network.


R1 R2

R3V1 V2

10Ω 20Ω

40Ω

10V 20V

Figure 7

Solution: Step 1: Determine the number of common nodes and reference node within the network (Figure 8). 1 common node (Va) , reference node C Step 2: Assign current and its direction to each distinct branch of the nodes in the network (Figure 8).

R1 R2

R3V1 V2

10Ω 20Ω

40Ω

10V 20V

I1 I2

I13

Va

C

Figure 8

Step 3: Apply KCL at each of the common nodes in the network

KCL:



Step 4: Solve the resulting simultaneous linear equation for the nodal voltages.

Step 5: Determine the currents through each elements


R1 R2

R3V1 V2

5kΩ 3kΩ

6kΩ

40V 55V

Figure 9

Solution: Step 1: Determine the number of common nodes and reference node within the network (Figure 10). 1 common node (Va) , reference node C Step 2: Assign current and its direction to each distinct branch of the nodes in the network (Figure 10).

R1 R2

R3V1 V2

5kΩ 3kΩ

6kΩ

40V 55V

Va

I3

I2I1

C

Figure 10



Step 3: Apply KCL at each of the common nodes in the network

KCL: Step 4: Solve the resulting simultaneous linear equation for the nodal voltages.

Step 5: Determine the currents through each elements



TUTORIAL 1 Find the current through each resistor for the networking below using Mesh Analysis and Nodal Analysis. a)

R1 R2

R3V1 V2

4Ω 2Ω

8Ω

4V 6V

b)

R1 R2

R3V1 V2

20Ω 10Ω

15Ω

10V 15V

c)

R1

R2

R3

V1

V2

4kΩ

2kΩ

3kΩ

30V

25V

d)

R1

R2

R3

V1

V2

4Ω

12Ω

3Ω

10V

12V

e)

R1 R2

R3

V1

V2

5.6kΩ 3.3kΩ2.2kΩ

10V

20V

30V

V3



2.0 Thevenin’ s Theorem Thevenins Theorem states: "Any linear circuit containing several energy source and resistances can be replaced by just a Single Voltage in series with a Single Resistor". Thevenins equivalent circuit.

A Linear

Network

Containing

Several Energy

Source and

Resistance

RL

A

B

VTH

RTH

RL

Thevenin’s Equivalent Circuit

IL

Figure 11

Thevenin’s theorem procedure: 1. Open circuit RL and find Thevenin’s voltage (VTH). 2. Find Thevenin’s resistance (RTH) when voltage source is short circuit or current source is

open circuit and RL is open circuit. 3. Draw the Thevenin’s equivalent circuit such as in figure 11 with the value of VTH and RTH.

Find the IL which current flow through the RL. Example 5 Find the current flow through RL equal to 30Ω for the circuit in Figure 12.

R1 R2

R3V1

10Ω 20Ω

40Ω

10VRL 30Ω

Figure 12



Solution: Step 1: Open circuit RL and find Thevenin’s voltage (VTH).

R2R1

R3V1

10Ω 20Ω

40Ω

10V

VTH

Figure 13

Using VDR find VTH

Step 2: Find Thevenin’s resistance (RTH) when voltage source is short circuit

R2R1

R3

10Ω 20Ω

40Ω RTH

Figure 14

Step 3: Draw the Thevenin’s equivalent circuit with the value of VTH and RTH

RTH

RLVTH

28Ω

30Ω

8V

IL

Figure 15



Example 6 Find current flow through R4.

60Ω

30Ω 90Ω 25Ω300mA

IsR1

R2

R3 R4

Figure 16

Solution : Step 1 : Open circuit RL and find Thevenin’s voltage (VTH).

60Ω

30Ω 90Ω300mA

IsR1

R2

R3 VTH

I1 I2

Figure 17

Using CDR, find I2

Step 2: Find Thevenin’s resistance (RTH) when current source,IS is open circuit.

60Ω

30Ω 90Ω

R1

R2

R3 RTH

Figure 18

Step 3: Draw the Thevenin’s equivalent circuit with the value of VTH and RTH

RTH

RLVTH

4.5V

45Ω

25Ω

IL

Figure 19

TUTORIAL 2 1. Refer to figure 1, find the current flow

through resistor 12Ω using Thevenin’s Theorem.

36V 6Ω 12Ω

3Ω 4Ω

R3R1

R2 R4

Figure 1 2. Find the current flow through resistor 15Ω

for the circuit in figure 2 using Thevenin’s Theorem.

15V

15Ω 5Ω

3Ω 4Ω

2Ω 6Ω

Figure 2

3. Count value stream IL by using Thevenin’s Theorem.

20V3kΩ5kΩ

4kΩ 2kΩ

1kΩ

IL

Figure 3 4. Use Thevenin’s Theorem to find the current

flowing in 5Ω resistor shown in figure 4.

15V

6Ω 5Ω

8Ω4Ω

2Ω

Figure 4

5. Calculate the current flow in 30Ω resistor for the circuit in figure 5 using Thevenin’s Theorem.

R2

20Ω

10Ω 30Ω 40Ω2AIs R1 R3 R4

Figure 5 6. Refer to figure 6, find the current flow

through 50Ω using Thevenin’s Theorem.

10Ω

30Ω 40Ω 50Ω200mA

Is

Figure 6 7. Use Thevenin’s Theorem, find the current

flow through resistor R=10Ω.

R1 R2

R3V1 V2

8Ω 15Ω

10Ω

6V 10V

Figure 7 8. Use Thevenin’s Theorem, find the current

flow through resistor R=10Ω. R1 R2

R3V1 V2

8Ω 15Ω

10Ω

6V 10V

Figure 8

3.0 Norton’s Theorem Nortons Theorem states: "Any linear circuit containing several energy sources and resistances can be replaced by a single Constant Current generator in parallel with a Single Resistor".

A Linear

Network

Containing

Several Energy

Source and

Resistance

RL

A

B

RL

Norton Equivalent Circuit

RNIN

IL

Figure 20 Norton’s theorem procedure: 1. Remove RL from the circuit. Find IN by shorting links output terminal. 2. Find RN by short-circuit voltage source or open-circuit current source. 3. Draw the Norton’s equivalent circuit such as in figure 20 with the value of IN and RN. Find

the IL which current flow through the RL. Example 7 Find the current flow through RL equal to 30Ω for the circuit in Figure 21.

R1 R2

R3V1

10Ω 20Ω

40Ω

10VRL 30Ω

Figure 21

Step 1: Remove RL from the circuit. Find IN by shorting links output terminal.

R1 R2

R3V1

10Ω 20Ω

40Ω

10V

RL

30ΩIN

I1IT

Figure 22



Step 2: Find RN by short-circuit voltage source.

R1 R2

R3

10Ω 20Ω

40Ω RN

Figure 23

Step3: Draw the Norton’s equivalent circuit with the value of IN and RN. Find the IL which current

flow through the RL.

IN RN RL

IL

0.286A28Ω 30Ω

Figure 24

Using CDR, find IL

Example 6 Find current flow through R4.

60Ω

30Ω 90Ω 25Ω300mA

IsR1

R2

R3 R4

Figure 25



Solution: Step 1: Remove RL from the circuit. Find IN by shorting links output terminal.

60Ω

30Ω 90Ω 25Ω300mA

IsR1

R2

R3 R4IN

Figure 26

Current flow at 90Ω is 0A, so .

Step 2: Find RN by open-circuit current source.

60Ω

30Ω 90Ω

R1

R2

R3 RN

Figure 27

Step3: Draw the Norton’s equivalent circuit with the value of IN and RN. Find the IL which current flow through the RL.

IN RN RL

IL

100mA45Ω 25Ω

Figure 28

Using CDR, find IL



TUTORIAL 3 1. Refer to figure 1, find the current flow

through resistor 12Ω using Norton’s Theorem.

36V 6Ω 12Ω

3Ω 4Ω

R3R1

R2 R4

Figure 1 2. Find the current flow through resistor 15Ω

for the circuit in figure 2 using Norton Theorem.

15V

15Ω 5Ω

3Ω 4Ω

2Ω 6Ω

Figure 2

3. Count value stream IL by using Norton Theorem.

20V3kΩ5kΩ

4kΩ 2kΩ

1kΩ

IL

Figure 3 4. Use Norton Theorem to find the current

flowing in 5Ω resistor shown in figure 4.

15V

6Ω 5Ω

8Ω4Ω

2Ω

Figure 4

5. Calculate the current flow in 30Ω resistor for the circuit in figure 5 using Norton Theorem.

R2

20Ω

10Ω 30Ω 40Ω2AIs R1 R3 R4

Figure 5 6. Refer to figure 6, find the current flow

through 50Ω using Norton Theorem.

10Ω

30Ω 40Ω 50Ω200mA

Is

Figure 6 7. Use Norton Theorem, find the current flow

through resistor R=10Ω.

R1 R2

R3V1 V2

8Ω 15Ω

10Ω

6V 10V

Figure 7 8. Use Norton Theorem, find the current flow

through resistor R=10Ω. R1 R2

R3V1 V2

8Ω 15Ω

10Ω

6V 10V

Figure 8

4.0 Maximum Power Transfer theorem The maximum power transfer theorem states: ‘A load will receive maximum power from a linear bilateral dc network when its total resistive value equal to the Thevenin’s or Norton resistance of the network as seen by the load.’

RN RL

IL

IN

Norton Equivalent Circuit

RTH

RLVTH

IL

Thevenin Equivalent Circuit

Figure 29 For the Thevenin equivalent circuit above, maximum power will be delivered to the load when:

For the Norton equivalent circuit above, maximum power will be delivered to the load when:

There are four conditions occur when maximum power transfer took place in a circuit: 1. Value of RL equal to RTH (RL=RTH). 2. Value of current is half of the current when RL is short circuited. 3. Value of load voltage is half the Thevenin’s voltage (VL = ½VTH).

4. Percentage of efficiency,% = 50%. Where:

Example 7

Refer to figure 30, determine the load power for each of the following value of the variable load resistance and sketch the graph load power versus load resistance. a) 25Ω b) 50Ω c) 75Ω d) 100Ω e) 125Ω

RTH

RLVTH

IL

10V

75Ω

Figure 30



Solution:

a)

b)

c)

d)

e)

RTH RL I VTH VL=IRL % PL

75Ω 0 0.133A 10V 0V 0% 0W

75Ω 25Ω 0.1A 10V 2.5V 25% 0.25W

75Ω 50Ω 0.08A 10V 4V 40% 0.32W

75Ω 75Ω 0.067A 10V 5.0V 50% 0.336W

75Ω 100Ω 0.057A 10V 5.7V 57% 0.325W

75Ω 125Ω 0.05A 10V 6.5V 65% 0.312W

Figure 31

0, 0

25, 0.25

50, 0.32 75, 0.336 100, 0.325 125, 0.312

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 20 40 60 80 100 120 140

Load

Po

we

r (W

)

Load Resistance (Ω)

Load Power (PL) vs Load Resistance(RL)



Example 8

For the network in figure 32, determine the value of R for maximum power transfer to R and hence calculate the maximum power using Thevenin’s equivalent circuit.

12V 3Ω

6Ω 8Ω

R3R1

R2R

Figure 32 Solution: Open circuit R and find Thevenin’s voltage (VTH).

6Ω 8Ω

3Ω

12V

VTH

Figure 33

Using VDR find VTH

Find Thevenin’s resistance (RTH) when voltage source is short circuit

6Ω 8Ω

3Ω RTH

Figure 34

Draw the Thevenin’s equivalent circuit with the value of VTH and RTH

RTH

RVTH

10Ω

4V

IL

Figure 35

Maximum power transfer occur when R=RTH. So, the value of R is 10Ω.

Maximum power transfer,

5.0 Superposition Theorem The superposition theorem states: ‘In any network made up of linear resistances and containing more than one source of e.m.f, the resultant current flowing in any branch is the algebraic sum of the currents that would flow in that branch if each source was considered separately, all other sources being replaced at that time by their respective internal resistances.’

Removing the effect of voltage and current source

Short circuit Open circuit

Voltage source Current source

Example 9 Determine the current through resistor R2=5Ω for the network in figure 36 using superposition

theorem.

R1

R2

10Ω

5Ω I

9AV

15V

Figure 36

Solution: Step 1: V active , I inactive. So current source is open circuit.

R1

R2

10Ω

5ΩV

15V

Ia

Figure 37



Step 2: V inactive, I active. So voltage source is short circuit.

R1

R2

10Ω

5ΩI

9A

Ib

Figure 38

Using CDR

Step 3: Total current through R2=5Ω. Ia 1A Ib 6A

Example 10 Find the current flow through each resistor for the network in figure 39.

R1 R2

R3V1 V2

10Ω 20Ω

40Ω

10V 20V

Figure 39

Solution: Step 1: V1 active, V2 inactive

R1 R2

R3V1

10Ω 20Ω

40Ω

10V

I1'

I3'

I2'

Figure 40

Step 2: V1 inactive, V2 active

R1 R2

R3V2

10Ω 20Ω

40Ω20V

I1'’ I2'’

I3'’

Figure 41

Step 3: Total current flow through each resistor

IR1 I1’=0.429A I1’’=0.571A So

IR2 I2’=0.286A I2’’=0.714A So

IR3 I3’=0.143A I3’’=0.143A So



TUTORIAL 4 Find the current through each resistor for the networking below using Superposition Theorem. b)

R1 R2

R3V1 V2

4Ω 2Ω

8Ω

4V 6V

b)

R1 R2

R3V1 V2

20Ω 10Ω

15Ω

10V 15V

c)

R1

R2

R3

V1

V2

4kΩ

2kΩ

3kΩ

30V

25V

d)

R1

R2

R3

V1

V2

4Ω

12Ω

3Ω

10V

12V

e)

R1 R2

R3

V1

V2

5.6kΩ 3.3kΩ2.2kΩ

10V

20V

30V

V3

Documents

UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS · unit 4 dc equivalent circuit and network theorems 1.0 Kirchoff’s Law Kirchoff’s Current Law (KCL) states at any junction in