250+ TOP MCQs on Norton’s Theorem Involving Dependent and Independent Sources and Answers

Network Theory Questions & Answers for Exams on “Norton’s Theorem Involving Dependent and Independent Sources”.

1. The circuit shown in figure has a load equivalent to _________

A. (frac{4}{3}) Ω
B. (frac{8}{3}) Ω
C. 4 Ω
D. 2 Ω

Answer: B
Clarification: Applying KCL in the given circuit, we get, (frac{V}{4} + frac{V-2I}{2}) = I
Or, (frac{3V-4I}{4}) = I
Or, 3V = 8I
∴ (frac{V}{I} = frac{8}{3}) Ω.

2. In the following circuit, the value of Norton’s resistance between terminals a and b are ___________

A. R_N = 1800 Ω
B. R_N = 270 Ω
C. R_N = 90 Ω
D. R_N = 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 Voltage, V_S = R_N I_S + V_OC
Here V_OC = 0
So, Resistance R_N = 90Ω.

3. For the circuit shown in figure below, the value of Norton’s 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_N = (frac{V_{test}}{1}) = 100 Ω.

4. For the circuit shown in the figure below, the Norton Resistance looking into X-Y is __________

A. 2 Ω
B. (frac{2}{3})
C. (frac{5}{3})
D. 2 Ω

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

5. In the figure given below, the value of Resistance R by Norton’s Theorem is ___________

A. 40
B. 20
C. 50
D. 80

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 (By Norton’s Theorem).

6. In the figure given below, the Norton Resistance, as seen at the terminals P-Q, is given by __________

A. 5 Ω
B. 7.5 Ω
C. 5 Ω
D. 7.5 Ω

7. The Norton 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 Norton resistance, all the current sources get open-circuited and voltage sources get short-circuited.
∴ R_N = ((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_N = 1.

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 value of Norton resistance for the network N is ____________

A. 16 kΩ
B. 30 kΩ
C. 3 kΩ
D. 50 kΩ

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

9. For the circuit given below, the Norton’s 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 = 7 kΩ.

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

A. 6 Ω and 1.333 A
B. 6 Ω and 0.833 A
C. 32 Ω and 0.156 A
D. 32 Ω and 0.25 A

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

A. 32 Ω and 0.25 A
B. 32 Ω and 0.125 A
C. 6 Ω and 0.833 A
D. 6 Ω and 1.167 A

12. In the figure given below, the power loss in 1 Ω resistor using Norton’s Theorem is ________

A. 9.76 W
B. 9.26 W
C. 10.76 W
D. 11.70 W

Answer: B
Clarification: Let us remove the 1 Ω resistor and short x-y.
At Node 1, assuming node potential to be V, (frac{V-10}{5}) + I_SC = 5
But I_SC = (frac{V}{2})
∴ (frac{V-10}{5} + frac{V}{2}) = 5
Or, 0.7 V = 7
That is, V= 10 V
∴ I_SC = (frac{V}{2}) = 5 A
To find R_int, all constant sources are deactivated. R_int = (frac{(5+2)×2}{5+2+2} = frac{14}{9}) = 1.56 Ω
R_int = 1.56 Ω; I_SC = I_N = 5A
Here, I = I_N (frac{R_{int}}{R_{int}+1} = 5 × frac{1.56}{1.56+1}) = 3.04 A
∴ Power loss = (3.04)² × 1 = 9.26 W.

13. The value of R_N from the circuit given below is ________

A. 3 Ω
B. 1.2 Ω
C. 5 Ω
D. 12.12 Ω

Answer: D
Clarification: V_X = 3(frac{V_X}{6}) + 4
Or, V_X = 8 V = V_OC
If terminal is short-circuited, V_X = 0.
I_SC = (frac{4}{3+3}) = 0.66 A
∴ R_N = (frac{V_{OC}}{I_{SC}}) = (frac{8}{0.66}) = 12.12 Ω.

14. The current I, as shown in the figure below, is ________

A. 3 A
B. 2 A
C. 1 A
D. 0

Answer: C
Clarification: The 3 Ω resistance is an extra element because voltage at node B is independent of the 3 Ω resistance.
I₁ = (frac{3}{2+1}) = 1 A (B -> A.
The net current in 2 Ω resistance is I = 1 – I₁
= 2 – 1 = 1 A (A -> B..

15. While computing the Norton equivalent resistance and the Norton equivalent current, 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 open-circuited keeping the independent voltage source untouched and the dependent current source is short-circuited keeping the independent current source untouched
D. 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

Answer: D
Clarification: While computing the Norton equivalent voltage consisting of both dependent and independent sources, we first find the equivalent resistance called the Norton resistance by opening the two terminals. Then while computing the Norton current, 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.