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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
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
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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 :
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
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Example 2 Find the current flow through each resistor using mesh analysis for the circuit below.
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 :
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
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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.
Example 3 Find the current flow through each resistor using mesh analysis for the circuit below.
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:
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
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Step 4: Solve the resulting simultaneous linear equation for the nodal voltages.
Step 5: Determine the currents through each elements
Example 4 Find the current flow through each resistor using mesh analysis for the circuit below.
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
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
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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
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
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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
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
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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
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
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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
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
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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
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
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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
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
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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
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
MARLIANA/JKE/POLISAS/ET101-UNIT4 15
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
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
MARLIANA/JKE/POLISAS/ET101-UNIT4 17
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)
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
MARLIANA/JKE/POLISAS/ET101-UNIT4 18
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
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
MARLIANA/JKE/POLISAS/ET101-UNIT4 20
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
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
MARLIANA/JKE/POLISAS/ET101-UNIT4 22
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