250+ TOP MCQs on Thevenin Theorem Involving Dependent and Independent Sources and Answers

Network Theory Multiple Choice Questions on “Thevenin Theorem Involving Dependent and Independent Sources”.

1. A circuit is given in the figure below. The Thevenin equivalent as viewed from terminals x and x’ is ___________

A. 8 V and 6 Ω
B. 5 V and 6 Ω
C. 5 V and 32 Ω
D. 8 V and 32 Ω

2. For the circuit given in figure below, the Thevenin equivalent as viewed from terminals y and y’ is _________

A. 8 V and 32 Ω
B. 4 V and 32 Ω
C. 5 V and 6 Ω
D. 7 V and 6 Ω

3. In the following circuit, when R = 0 Ω, the current IR equals to 10 A. The value of R, for which maximum power is absorbed by it is ___________

A. 4 Ω
B. 3 Ω
C. 2 Ω
D. 1 Ω

4. In the following circuit, when R = 0 Ω, the current I_R equals to 10 A. The maximum power will be?

A. 50 W
B. 100 W
C. 200 W
D. 400 W

Answer: A
Clarification: The Thevenin equivalent of the circuit is as shown below.
I = 10 A, R_th = 2 Ω
∴ P_max = ((frac{10}{2}))² × 2
= 5×5×2 = 50 W.

5. For the circuit given below, the Thevenin resistance across the terminals A and B is _____________

A. 5 Ω
B. 7 kΩ
C. 1.5 kΩ
D. 1.1 kΩ

Answer: B
Clarification: Let V_AB = 1 V
5 V_AB = 5
Or, 1 = 1 × I₁ or, I₁ = 1
Also, 1 = -5 + 1(I – I₁)
∴ I = 7
Hence, R = 0.2 kΩ.

6. For the circuit given below, the Thevenin voltage across the terminals A and B is ____________

A. 1.25 V
B. 0.25 V
C. 1 V
D. 0.5 V

Answer: D
Clarification: Current through 1 Ω = (frac{5}{2}) – I₁
Using source transformation to 5 V sources, V_OC = 1 × I₁
V_OC = -5 V_OC + ((frac{5}{2}) – I₁) × 1
Eliminating I1, we get, V_OC = 0.5 V.

7. In the following circuit, the value of open circuit voltage and the Thevenin resistance between terminals a and b are ___________

A. V_OC = 100 V, R_TH = 1800 Ω
B. V_OC = 0 V, R_TH = 270 Ω
C. V_OC = 100 V, R_TH = 90 Ω
D. V_OC = 0 V, R_TH = 90 Ω

Answer: D
Clarification: By writing loop equations for the circuit, we get,
V_S = V_X, I_S = I_X
V_S = 600(I₁ – I₂) + 300(I₁ – I₂) + 900 I₁
= (600+300+900) I₁ – 600I₂ – 300I₃
= 1800I₁ – 600I₂ – 300I₃
I₁ = I_S, I₂ = 0.3 V_S
I₃ = 3I_S + 0.2V_S
V_S = 1800IS – 600(0.01V_S) – 300(3I_S + 0.01V_S)
= 1800I_S – 6V_S – 900I_S – 3V_S
10V_S = 900I_S
For Thevenin equivalent, V_S = R_TH I_S + V_OC
So, Thevenin voltage V_OC = 0
Resistance R_TH = 90Ω.

8. In the circuit given below, it is given that V_AB = 4 V for R_L = 10 kΩ and V_AB = 1 V for R_L = 2kΩ. The values of the Thevenin resistance and voltage for the network N are ____________

A. 16 kΩ and 30 V
B. 30 kΩ and 16 V
C. 3 kΩ and 6 V
D. 50 kΩ and 30 V

Answer: B
Clarification: When R_L = 10 kΩ and V_AB = 4 V
Current in the circuit I = (frac{V_{AB}}{R_L} = frac{4}{10}) = 0.4 mA
Thevenin voltage is given by V_TH = I (R_TH + R_L)
= 0.4(R_TH + 10)
= 0.4R_TH + 4
Similarly, for R_L = 2 kΩ and V_AB = 1 V
So, I = (frac{1}{2}) = 0.5 mA
V_TH = 0.5(R_TH + 2)
= 0.5 R_TH + 1
∴ 0.1R_TH = 3
Or, R_TH = 30 kΩ
And V_TH = 12 + 4 = 16 V.

9. For the circuit shown in figure below, the value of the Thevenin resistance is _________

A. 100 Ω
B. 136.4 Ω
C. 200 Ω
D. 272.8 Ω

Answer: A
Clarification: I_X = 1 A, V_X = V_test
V_test = 100(1-2I_X) + 300(1-2I_X – 0.01V_S) + 800
Or, V_test = 1200 – 800I_X – 3V_test
Or, 4V_test = 1200 – 800 = 400
Or, V_test = 100V
∴ R_TH = (frac{V_{test}}{1}) = 100 Ω.

10. For the circuit shown in the figure below, the Thevenin voltage and resistance looking into X-Y are __________

A. (frac{4}{3}) V and 2 Ω
B. 4V and (frac{2}{3}) Ω
C. (frac{4}{3}) V and (frac{2}{3}) Ω
D. 4 V and 2 Ω

Answer: D
Clarification: (R_{TH} = frac{V_{OC}}{I_{SC}})
V_TH = V_OC
Applying KCL at node A, (frac{2I-V_{TH}}{1} + 2 = I + frac{V_{TH}}{2})
Or, I = (frac{V_{TH}}{1})
Putting, 2V_TH – V_TH + 2 = V_TH + (frac{V_{TH}}{2})
Or, V_TH = 4 V.
∴ R_TH = 4/2 = 2Ω.

11. In the figure given below, the value of the source voltage is ___________

A. 12 V
B. 24 V
C. 30 V
D. 44 V

Answer: C
Clarification: By applying KCL, (frac{V_P-E}{6} + frac{V_P}{6}) – 1 = 0
Or, 2 V_P – E = 6
Where, ((frac{-V_P+E}{6})) = 2
∴ E – V_P = 12
Or, V_P = 18 V
∴ E = 30V.

12. In the figure given below, the value of Resistance R by Thevenin Theorem is ___________

A. 10
B. 20
C. 30
D. 40

Answer: B
Clarification: (frac{V_P-100}{10} + frac{V_P}{10}) + 2 = 0
Or, 2V_P – 100 + 20 = 0
∴ V_P = 80/2 = 40V
∴ R = 20Ω.

13. In the figure given below, the Thevenin’s equivalent pair, as seen at the terminals P-Q, is given by __________

A. 2 V and 5 Ω
B. 2 V and 7.5 Ω
C. 4 V and 5 Ω
D. 4 V and 7.5 Ω

14. The Thevenin equivalent impedance Z between the nodes P and Q in the following circuit is __________

A. 1
B. 1 + s + (frac{1}{s})
C. 2 + s + (frac{1}{s})
D. 3 + s + (frac{1}{s})

Answer: A
Clarification: To calculate the Thevenin resistance, all the current sources get open-circuited and voltage source short-circuited.
∴ R_TH = ((frac{1}{s}) + 1) || (1+s)
= (frac{left(frac{1}{s} + 1right)×(1+s)}{left(frac{1}{s} + 1right)+(1+s)})
= (frac{frac{1}{s}+1+1+s}{frac{1}{s}+1+1+s}) = 1
So, R_TH = 1.

15. While computing the Thevenin equivalent resistance and the Thevenin equivalent voltage, which of the following steps are undertaken?
A. Both the dependent and independent voltage sources are short-circuited and both the dependent and independent current sources are open-circuited
B. Both the dependent and independent voltage sources are open-circuited and both the dependent and independent current sources are short-circuited
C. The dependent voltage source is short-circuited keeping the independent voltage source untouched and the dependent current source is open-circuited keeping the independent current source untouched
D. The dependent voltage source is open-circuited keeping the independent voltage source untouched and the dependent current source is short-circuited keeping the independent current source untouched

Answer: C
Clarification: While computing the Thevenin equivalent voltage consisting of both dependent and independent sources, we first find the equivalent voltage called the Thevenin voltage by opening the two terminals. Then while computing the Thevenin equivalent resistance, we short-circuit the dependent voltage sources keeping the independent voltage sources untouched and open-circuiting the dependent current sources keeping the independent current sources untouched.