250+ TOP MCQs on Advanced Problems on Network Theorems – 2 and Answers

Advanced Network Theory Questions on “Advanced Problems on Network Theorems – 2”.

1. A network contains linear resistors and ideal voltage source s. If values of all the resistors are doubled, then voltage across each resistor is __________
A. Halved
B. Doubled
C. Increases by 2 times
D. Remains same

Answer: D
Clarification: V / R ratio is a constant R. If R is doubled then, electric current will become half. So voltage across each resistor is same.

2. A voltage waveform V(t) = 12t² is applied across a 1 H inductor for t ≥ 0, with initial electric current through it being zero. The electric current through the inductor for t ≥ 0 is given by __________
A. 12 t
B. 24 t
C. 12 t³
D. 4 t³

Answer: D
Clarification: We know that, I = (frac{1}{L} int_0^t V ,dt)
= 1(int_0^t 12 t^2 ,dt)
= 4 t³.

3. The linear circuit element among the following is ___________
A. Capacitor
B. Inductor
C. Resistor
D. Capacitor & Inductor

Answer: C
Clarification: A linear circuit element does not change its value with voltage or current. The resistance is only one among the others does not change its value with voltage or current.

4. In the circuit shown, V_C is zero at t=0 s. For t>0, the capacitor current I_C(t), where t is in second, is ___________

A. 0.50 e^-25t mA
B. 0.25 e^-25t mA
C. 0.50 e^-12.5t mA
D. 0.25 e^-6.25t mA

Answer: A
Clarification: The capacitor voltage V_C (t) = V_C (∞) – [V_C (∞)-V_C (0)]e^-t/RC
R = 20 || 20 = (frac{20×20}{20+20} = frac{400}{40}) = 10 kΩ
V_C (∞) = 10 × (frac{20}{20+20}) = 5 V
Given, V_C (0) = 0
∴ V_C (t) = 5 – (5-0)e^{-t/10×4×10^(-6)×10^3}
= 5(1 – e^-25t)
I_C (t) = C(frac{dV_C (t)}{dt} = 4×10^{-6} frac{d}{dt}5(1-e^{-25t}))
= 4 × 10^-6 × 5 × 25e^-25t
∴ I_L (t) = 0.50e^-2.5t mA.

5. In the circuit given below, the phasor voltage V_AB (in volt) is _________

A. Zero
B. 5∠30°
C. 12.5∠30°
D. 17∠30°

Answer: D
Clarification: Equivalent impedance = (5 + j3) || (5 – √3)
= (frac{(5+j3)×(5 – sqrt{3})}{(5+j3) + (5 – sqrt{3})} )
= (frac{25+9}{10}) = 3.4 Ω
V_AB = Current × Impedance
= 5∠30° × 3.4
= 17∠30°.

6. For the circuit given below, the driving point impedance is given by, Z(s) = (frac{0.2s}{s^2+0.1s+2} ). The component values are _________

A. L = 5 H, R = 0.5 Ω, C = 0.1 F
B. L = 0.1 H, R = 0.5 Ω, C = 5 F
C. L = 5 H, R = 2 Ω, C = 0.1 F
D. L = 0.1 H, R = 2 Ω, C = 5 F

Answer: D
Clarification: Dividing point impedance = R || sL || (frac{1}{sC} )
= (Big{frac{(R)(frac{1}{sC})}{R + frac{1}{sC}}Big}) || sL
= (frac{frac{R}{1+sRC}}{frac{R}{1+sRC}+sL} )
= (frac{sRL}{s^2 RLC+sL+R} )
Given that, Z(s) = (frac{0.2s}{s^2+0.1s+2} )
∴ On comparing, we get L = 0.1 H, R = 2 Ω, C = 5 F.

7. What is the power loss in the 10 Ω resistor in the network shown in the figure below?

A. 15.13 W
B. 11.23 W
C. 16 W
D. 14 W

Answer: A
Clarification: In mesh aef, 8(I₁ – I₃) + 3(I₁ – I₂) = 15
Or, 11 I₁ – 3 I₂ – 8I₃ = 15
In mesh efd, 5(I₂ – I₃) + 2I₂ + 3(I₂ – I₁) = 0
Or, -3 I₁ + 10I₂ – 5I₃ = 0
In mesh abcde, 10I₃ + 5(I₃ – I₂) + 8(I₃ – I₁) = 0
Or, -8I₁ – 5I₂ +23I₃ = 0
Thus loop equations are,
11 I₁ – 3 I₂ – 8I₃ = 15
-3 I₁ + 10I₂ – 5I₃ = 0
-8I₁ – 5I₂ + 23I₃ = 0
Solving by Cramer’s rule, I₃ = current through the 10Ω resistor = 1.23 A
∴ Current through 10 Ω resistor = 1.23 A
Power loss (P) = (I_3^2 r) = (1.23)²×10 = 15.13 W.

8. The switch S is the circuit shown in the figure is ideal. If the switch is repeatedly closed for 1 ms and opened for 1 ms, the average value of i(t) is ____________

A. 0.25 mA
B. 0.35 mA
C. 0.5 mA
D. 1 mA

Answer: A
Clarification: Since i = (frac{5}{10 × 10^{-3}}) = 0.5 × 10³ = 0.5 mA
As the switch is repeatedly close, then i(t) will be a square wave.
So average value of electric current is ((frac{0.5}{2})) = 0.25 mA.

9. In the circuit given below the value of resistance, R_eq is ___________

A. 10 Ω
B. 11.86 Ω
C. 11.18 Ω
D. 25 Ω

10. A particular electric current is made up of two component: a 10 A and a sine wave of peak value 14.14 A. The average value of electric current is __________
A. 0
B. 24.14 A
C. 10 A
D. 14.14 A

Answer: C
Clarification: Average dc electric current = 10 A
Average ac electric current = 0 A as it is alternating in nature.
Average electric current = 10 + 0 = 10 A.

11. Given that, R₁ = 36 Ω and R₂ = 75 Ω, each having tolerance of ±5% are connected in series. The value of resultant resistance is ___________
A. 111 ± 0 Ω
B. 111 ± 2.77 Ω
C. 111 ± 5.55 Ω
D. 111 ± 7.23 Ω

Answer: C
Clarification: R₁ = 36 ± 5% = 36 ± 1.8 Ω
R₂ = 75 ± 5% = 75 ± 3.75 Ω
∴ R₁ + R₂ = 111 ± 5.55 Ω.

12. In the circuit of figure below a charge of 600 C is delivered to 100 V source in a 1 minute. The value of V₁ is ___________

A. 30 V
B. 60 V
C. 120 V
D. 240 V

Answer: D
Clarification: In order for 600 C charges to be delivered to 100 V source, the electric current must be anti-clockwise.
(i = frac{dQ}{dt} = frac{600}{60}) = 10A
Applying KVL we get
V₁ + 60 – 100 = 10 × 20 ⇒ V₁ = 240 V.

13. The energy required to charge a 10 μF capacitor to 100 V is ____________
A. 0.01 J
B. 0.05 J
C. 5 X 10^-9 J
D. 10 X 10^-9 J

Answer: B
Clarification: E = (frac{1}{2} CV^2)
= 5 X 10^-6 X 100²
= 0.05 J.

14. Among the following, the active element of electrical circuit is ____________
A. Voltage source
B. Current source
C. Resistance
D. Voltage and current source both

Answer: D
Clarification: We know that active elements are the ones that are used to drive the circuit. They also cause the electric current to flow through the circuit or the voltage drop across the element. Here only the voltage and current source are the ones satisfying the above conditions.

15. In the circuit given below, the source current is _________

A. 2 A
B. 3 A
C. 0.54 A
D. 1.5 A

Answer: C
Clarification: In the given figure, R_X = R_Y = R_Z = (frac{5×5}{5+5+5}) = 1.67 Ω
Here, R = [(R_X + 2) || (R_Y + 3)] + R_X
= (3.67 || 4.67) + R_Z
= (frac{3.67 × 4.67}{3.67 + 4.67}) + 1.67
= (frac{17.1389}{8.34}) + 1.67
= 3.725 Ω
∴ I = (frac{V}{R} = frac{2}{3.725}) = 0.54 A.