1) Thevenin’s Theorem
1 Find the value of current flowing through the 10 Ω resistor
Solution
Step I: Calculation of VTH
Removing the 10 Ω resistor from the network,
For Mesh 1,
I1 = 10
Applying KVL to Mesh 2,
100 – 30I2 – 20I3 = 0
I2 = 2 A
Writing VTH equation,
5I1 – VTH – 20 I2 = 0
VTH = 5 I1 – 20 I2
= 5(10) – 20(2)
= 10 V
Step II: Calculation of RTH
Replacing the current source by an open circuit and the voltage source by a short
circuit,
RTH = 5 + (20 | | 30) = 17 Ω
Step III: Calculation of IL
20
IL = =0.37 A
5+1.67
2 Find the current through the 60 Ω resistance by using Thevenin’s theorem.
Solution
Step I: Calculation of VTH
Writing KVL equation in matrix form,
[ 600 0
120 ] [ II 12] = [ 800]
I1 = 2.67 A
I2 = 1.33 A
Writing VTH equation,
80 – 10 (I1 – I2) – VTH – 10I2 = 0
VTH = 80 – 10(I1–I2) – 10I2
= 80 – 10(2.67 – 1.33) – 10 (1.33)
= 53.3 V
Step II Calculation of RTH
Replacing voltage source by short circuit,
RTH = 16.66 W
Step III Calculation of IL
53.3
IL = =0.7 A
16.66+60
3 Find current through10 Ω resistor using Thevenin’s theorem.
Solution
Step I: Calculation of VTH
Removing the 10 Ω resistor from the network,
Applying KVL to Mesh 1,
10 – 6 I1 – 1(I1–I2) = 0
7 I1 – I2 = 10 ………….(1)
Applying KVL to Mesh 2,
– 1(I2 – I1) – 2 I2 – 3 I2 = 0
I1 – 6 I2 = 0 ………….(2)
Solving Eqs (1) and (2),
I2 = 0.24 A
Writing VTH equation,
3 I2 – VTH – 20 = 0
VTH = 3 I2 – 20
= 3 (0.24) – 20
= – 19.28 V
= 19.28 V (terminal B is positive w.r.t A)
Step II: Calculation of RTH
Replacing voltage sources by short circuits,
RTH = [(6|| 1) + 2] || 3 = 1.47 Ω
Step III: Calculation of IL
10
IL =60x =1.68 A
10+2
4 Find current through20Ω resistor using Thevenin’s theorem.
Solution
Step I: Calculation of VTH
Removing the 20 Ω resistor from the network,
From Fig.
VTH = 100 V
Step II: Calculation of RTH
Replacing the voltage source by a short circuit and the current source by an open
circuit,
RTH = 0
Step III: Calculation of IL
100
IL = = 5A
20
5 Find the value of current flowing through the 3Ω resistor.
Solution
Step I: Calculation of VTH
Removing the 3 Ω resistor from the network,
By source transformation,
Applying KVL to the mesh,
10 – 2I – 2I – 20 = 0
4I = –10
I = –2.5 A
Writing VTH equation,
10 – 2I – VTH = 0
VTH = 10 – 2I
= 10 – 2(–2.5)
= 15 V
Step II: Calculation of RTH
Replacing voltage source by a short circuit and current source by an open circuit,
RTH = (2 || 2) + 1 = 1 + 1 = 2 Ω
Step III: Calculation of IL
15
IL = = 3A
2+ 3
6 Find the value of current flowing through the 2 Ω resistor
ANS: IL=4.38
7 Find the value of current flowing through the 40 Ω resistor .
ANS: IL=0.09A
8 Determine the value of current flowing through the 24 Ω resistor.
ANS: IL=2A
9 Find the value of current flowing through the 20 Ω resistor.
ANS: IL=2.55A
10 Find the value of current flowing through the 1 Ω resistor.
2 Norton
1 For the given circuit find the Norton equivalent between points A and B.
Solution
Step I: Calculation of IN
Replacing 2 Ω resistor by short circuit,
Applying KVL to Mesh 1,
10 – 1I1 – 1(I1 – I2) = 0
2I1 = I2 = 10 …………….(1)
Applying KVL to Mesh 2,
–1(I2 – I1) – 1I2 = 0
–I1 + 2I2 = 0 ………….....(2)
Solving Eqs (1) and (2),
I1 = 6.67 A
I2 = IN = 3.33 A
Step II: Calculation of RN
Replacing voltage source by short circuit,
RN = 1.5 Ω
Step III: Norton’s equivalent network
2 Using Norton’s theorem, calculate the current flowing through the 15 Ω load resistor
in the given circuit.
Solution
Step I: Calculation of IN
Writing the current equation for the supermesh,
I1 – I2 = 4………………………….. (1)
Writing the voltage equation for the supermesh,
30 – 4 I1 – 6 I2 = 0
4 I1 + 6 I2 = 30……………………… (2)
Solving Eqs (1) and (2),
I1 = 5.4 A
I2 = 1.4 A
IN = I2 = 1.4 A
Step II: Calculation of RN
Step III: Calculation of IL
10
IL =1.4x =0.56 A
10+15
3 Find the value of current flowing in the 10 Ω resistor
Solution
Step I: Calculation of IN
Replacing the 10 Ω resistor by a short circuit,
The resistance of 40 Ω becomes redundant as it is connected across the 50 V
source
Applying KVL to Mesh 1,
50 – 50 I1 – 20 (I1 – I2) – 10 = 0
70 I1 – 20 I2 = 40……………. (1)
Applying KVL to Mesh 2,
10 – 20 (I2 – I1) = 0
–20 I1 + 20 I2 = 10…………… (2)
Solving Eqs. (1) and (2),
I1 = 1 A
I2 = 1.5 A
IN = I2 = 1.5 A
Step II: Calculation of RN
Replacing voltage sources by short circuits, resistor of 40 Ω gets shorted.
RN = 50||20 = 14.28 Ω
Step III: Calculation of IL
14.28
IL =1.5x =0.88 A
14.28+10
4 Find the value of current flowing through the 8 Ω resistor
Solution
Step I: Calculation of IN
Replacing the 8 Ω resistor by a short circuit,
The resistor of the 4 Ω gets shorted as it is in parallel with the short circuit.
Simplifying
the network by source transformation,
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I2 – I1 = 2 …………………..(1)
Applying KVL to the supermesh,
60 – 12I1 – 5 = 0
12I1 = 55 …………………...(2)
Solving Eqs (1) and (2),
I1 = 4.58 A
I2 = 6.58 A
IN = I2 = 6.58 A
Step II: Calculation of RN
Replacing the voltage source by a short circuit and the current source by an open
circuit,
RN = 12 || 4 = 3 Ω
Step III: Calculation of IL
15
IL =6.58x =1.79 A
2+ 3
5 Find value of current flowing through the 1 Ω resistor.
Solution
Step I: Calculation of IN
Replacing the 1 Ω resistor by a short circuit,
By source transformation,
Applying KVL to Mesh 1,
–3 – 3I1 – 2(I1 – I3) + 1 = 0
5I1 – 2I3 = – 2 ……………..(1)
Applying KVL to Mesh 2,
–1 – 2 (I2 – I3) – 2 I2 = 0
4I2 – 2I3 = – 1 ……………(2)
Applying KVL to Mesh 3,
–2 (I3 –I1) – 2 (I3 –I2) = 0
–2I1 – 2I2 + 4I3 = 0 ………..(3)
Solving Eqs. (1), (2) and (3),
I1 = – 0.64 A
I2 = – 0.55 A
I3 = – 0.59 A
IN = I3 = – 0.59 A
Step II: Calculation of RN
Replacing the voltage source by a short circuit and the current source by an open
circuit,
RN = 2.2 Ω
Step III: Calculation of IL
2.2
IL =0.59x =0.41 A
2.2+ 1
6
3 Maximum Power Transfer Theorem
1 For the given circuit find the value of R Lfor maximum power transfer and calculate
themaximum power absorbed by R L.
Solution
Step I: Calculation of VTH
Removing the variable resistor RL from the network,
Applying KVL to Mesh 1,
10 – 2I1 – 1(I1 – I2) – 8 (I1 – I2) = 0
11I1 – 9 I2 = 10 ……………………….(1)
Applying KVL to Mesh 2
–4 I2 – 5 I2 – 8 ( I2 –I1) – 1 ( I2 –I1) = 0
–9I1 + 18 I2 = 0 ………………………..(2)
Solving Eqs (1) and (2),
I1 = 1.54 A
I2 = 0.77 A
Writing VTH equation,
– 1 (I2 – I1) – 4 I2 – VTH = 0
VTH = –1(I2 – I1) – 4 I2
= –1(0.77 – 1.54) – 4(0.77)
= –2.31 V
= 2.31 V (the terminal B is positive w.r.t. A)
Step II: Calculation of RTH
Replacing the voltage source by a short circuit,
Redrawing the network,
Converting the upper delta into equivalent star network,
4X2
R1 = =0.73Ω
4+ 2+ 5
4 X5
R2 = =1.82Ω
4+ 2+ 5
5X2
R3 = =0.91Ω
4+ 2+ 5
Simplifying the network,
RTH = 1.82 + (1.73 || 8.91) = 3.27 W
Step III: Value of RL
For maximum power transfer
RL = RTH = 3.27 Ω
Step IV: Calculation of PMAX
V Th 2 (2.31)2
PMAX= = =0.41W
4 × R Th 4 X 3.27
2 For the given circuit, find the value of ‘ R L’ so that maximum power is dissipated in
it.Also, find Pmax .
Solution
Step I: Calculation of VTH
Removing the resistor R L from the network,
Applying KVL to Mesh 1,
8 – 2I1 – 1I1 – 2(I1– I2) = 0
5 I1 – 2I2 = 8……………….. (1)
Applying KVL to Mesh 2,
–2(I2– I1) – 3I2 – 8I2 + 12 = 0
–2 I1 + 13I2= 12…………… (2)
Solving Eqs. (1) and (2),
I1 = 2.1 A
I2 = 1.25 A
Writing VTH equation,
1I1 + 10 – VTH + 3I2 = 0
VTH = 1I1 + 10 + 3I2
= 1(2.1) + 10 + 3(1.25)
= 15.85 V
Step II: Calculation of RTH
Replacing the voltage sources by short circuits,
Converting the delta network formed by resistors of 2 Ω, 1 Ω and 2 Ω into
equivalent star
network,
2X1
R1 = =0.4Ω
2+ 1+ 2
2X1
R2= =0.4Ω
2+ 1+ 2
2X2
R3 = =0.8Ω
2+ 1+ 2
Simplifying the network,
RTH = 2.85 Ω
Step III: Value of R L
For maximum power transfer
R L = RTH = 2.85 Ω
Step IV: Calculation of Pmax
V Th 2 (15.85)2
PMAX= = =22.04W
4 × R Th 4 X 2.85
3 Find the value of resistance RL for maximum power transfer and calculate the maximum power
Step I: Calculation of VTh
Removing the variable resistor RL from the network,
Applying KVL to Mesh 1,
120 – 10I1 – 5(I1 – I2) = 0
15I1 – 5I2 = 120 (1)
Writing current equation for Mesh 2,
I2 = – 6 (2)
Solving Eqs (1) and (2),
I1 = 6 A
Writing VTh equation,
120 – 10I1 – VTh = 0
VTh = 120 – 10 (6)
= 60 V
Step II: Calculation of RTh
Replacing the voltage source by a short circuit and the current source by an open
circuit,
RTh = 10 | | 5 = 3.33 W
Step III: Value of RL
For maximum power transfer
RL = RTh = 3.33 W
Step IV: Calculation of Pmax
V Th 2 (60)2
PMAX = =270.27W
4 × R Th 4 X 3.33
4 Find the value of resistance RL for maximum power transfer and calculate maximum
power.
Solution
Step I : Calculation of VTh
Removing the variable resistor RL from the network,
Meshes 1 and 2 will form a supermesh.
Writing the current equation for the supermesh,
I2 – I1 = 1 (1)
Writing the voltage equation for the supermesh,
5 – 1I1 – 10 (I2 – I3) = 0
I1 + 10 I2 – 10 I3 = 5 (2)
Applying KVL to Mesh 3,
– 10 (I3 – I2) – 2 I3 – 3 I3 = 0
– 10 I2 + 15 I3 = 0 (3)
Solving Eqs (1), (2) and (3),
I1 = 0.38 A
I2 = 1.38 A
I3 = 0.92 A
Writing VTh equation,
V Th = 3 I3 = 2.76 V
Step II: Calculation of RTh
Replacing voltage source by a short circuit and current source by an open circuit,
RTh = 6.48 W
Step III: Value of RL
For maximum power transfer
RL = RTh = 6.48 W
Step IV: Calculation of Pmax
V Th 2 (2.67)2
PMAX= = =0.29W
4 × R Th 4 X 6.48
5 Find the value of resistance the RL for maximum power transfer and calculate the maximum
power.
Solution
Step I: Calculation of VTh
Removing the variable resistor RL from the network,
By star-delta transformation,
100
I= =2.08A
5+5+20+9+ 9
Writing VTh equation,
100 – 5I – VTh – 9I = 0
VTh = 100 – 14I
= 100 – 14(2.08)
= 70.88 V
Step II: Calculation of RTh
Replacing the voltage source by a short circuit
RTh = 23.92 Ω
Step III: Value of RL
For maximum power transfer
RL = RTh = 23.92 W
Step IV: Calculation of Pmax
V Th 2 (70.88)2
PMAX= = =52.51W
4 × R Th 4 X 23.92
