This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Advanced Problems Involving Complex Circuit Diagram – 2”.

1. In the circuit given below, the KVL for first loop is ___________

a) V (t) = R_{1}i_{1} + L_{1} \(\frac{di_1}{dt}\) + M \(\frac{di_2}{dt}\)

b) V (t) = R_{1}i_{1} – L_{1} \(\frac{di_1}{dt}\) – M \(\frac{di_2}{dt}\)

c) V (t) = R_{1}i_{1} + L_{1} \(\frac{di_1}{dt}\) – M \(\frac{di_2}{dt}\)

d) V (t) = R_{1}i_{1} – L_{1} \(\frac{di_1}{dt}\) + M \(\frac{di_2}{dt}\)

View Answer

Explanation: We know that, in general, the KVL is of the form V (t) = R

_{1}i

_{1}+ L

_{1}\(\frac{di_1}{dt}\) + M \(\frac{di_2}{dt}\)

But here, M term is negative because i

_{1}, is entering the dotted terminal and i

_{2}, is leaving the dotted terminal.

So, V (t) = R

_{1}i

_{1}+ L

_{1}\(\frac{di_1}{dt}\) – M \(\frac{di_2}{dt}\).

2. In a parallel RL circuit, 12 A current enters into the resistor R and 16 A current enters into the Inductor L. The total current I the sinusoidal source is ___________

a) 25 A

b) 4 A

c) 20 A

d) Cannot be determined

View Answer

Explanation: Currents in resistance and inductance are out of phase by 90°.

Hence, I = \(I_1^2 + I_2^2\)

Or, I = [12

^{2}+ 16

^{2}]

^{0.5}

Or, I = \(\sqrt{144+256} = \sqrt{400}\)

= 20 A.

3. Consider a series RLC circuit having resistance = 1Ω, capacitance = 1 F, considering that the capacitor gets charged to 10 V. At t = 0 the switch is closed so that i = e^{-2t}. When i = 0.37 A, the voltage across capacitor is _____________

a) 1 V

b) 6.7 V

c) 0.37 V

d) 0.185 V

View Answer

Explanation: We know that, during discharge of capacitor,

V

_{C}= V

_{R}

Now, V

_{R}= 0.67 X 10 = 6.7 V

So, V

_{C}= 6.7 V.

4. A waveform is of the form of a trapezium, which increases linearly with the linear slope till θ = \(\frac{π}{3}\), constant till θ = \(\frac{π}{2}\) and again linearly decreases to 0 till θ = π. The average value of this waveform is ______________

a) 2 V

b) 0 V

c) 4 V

d) 3 V

View Answer

Explanation: The average value of the waveform = \(\frac{2 X Area \,of \,1st \,triangle\, + \,Area \,of\, 2nd \,triangle}{π}\)

= \(\frac{2 X \frac{π}{3} X \frac{1}{2} X 6 + 6(\frac{π}{2} – \frac{π}{3})}{π}\)

= \(\frac{2π + π}{π}\) = 3 Volt.

5. For a series RLC circuit excited by a unit step voltage, V_{c} is __________

a) 1 – e^{-t/RC}

b) e^{-t/RC}

c) e^{t/RC}

d) 1

View Answer

Explanation: At t = 0, V

_{c}= 0 and at t = ∞, V

_{c}= 1.

This condition can be satisfied only by (1 – e

^{-t/RC}).

6. In the circuit given below, a dc circuit fed by a current source. With respect to terminals AB, Thevenin’s voltage and Thevenin’s resistance are ____________

a) V_{TH} = 5 V, R_{TH} = 0.75 Ω

b) V_{TH} = 0.5 V, R_{TH} = 0.75 Ω

c) V_{TH} = 2.5 V, R_{TH} = 1 Ω

d) V_{TH} = 5 V, R_{TH} = 1 Ω

View Answer

Explanation: V

_{TH}= \(\frac{1 X 20}{40}\) X 1 = 0.5 V

Also, R

_{TH}= \(\frac{1 X 30}{40}\) = 0.75 Ω.

7. In the circuit given below, the value of R is ____________

a) 12 Ω

b) 6 Ω

c) 3 Ω

d) 1.5 Ω

View Answer

Explanation: The resistance of parallel combination is given by,

R

_{eq}= \(\frac{40}{3} \) – 10 = 3.33 Ω

Or, \(\frac{1}{3.33} = \frac{1}{12} + \frac{1}{15} + \frac{1}{15}\)

Or, R = 6 Ω.

8. A circuit consists of an excitation voltage V_{S}, a resistor network and a resistor R. For different values of R, the values of V and I are as given, R = ∞, V = 5 volt; R = 0, I = 2.5 A; when R = 3 Ω, the value of V is __________

a) 1 V

b) 2 V

c) 3 V

d) 5 V

View Answer

Explanation: When R = ∞, V = 5v,

Then, V

_{oc}= 5V and the circuit is open

When R = 0, I = 2.5A

Then, I

_{sc}= 2.5 and the circuit is short circuited.

So, R

_{eq}= \(\frac{V_{OC}}{I_{SC}}\)

= \(\frac{5}{2.5}\) = 2 Ω

Hence the voltage across 3 Ω is 3 volt.

9. Three inductors each 30 mH are connected in delta. The value of inductance or each arm of equivalent star is _____________

a) 10 mH

b) 15 mH

c) 30 mH

d) 90 mH

View Answer

Explanation: We know that if an inductor L is connected in delta, then the equivalent star of each arm = \(\frac{L X L}{L+L+L}\)

Given that, L = 30 mH

= \(\frac{30 X 30}{30+30+30}\)

= \(\frac{900}{90}\) = 10 mH.

10. In a series RLC circuit having resistance R = 2 Ω, and excited by voltage V = 1 V, the average power is 250 mW. The phase angle between voltage and current is ___________

a) 75°

b) 60°

c) 15°

d) 45°

View Answer

Explanation: VI cos θ = 0.25 or I cos θ = 0.25

Or, Z cosθ = 2

Or, \(\frac{V}{I}\) cosθ = 2

Or, cos θ = \(\frac{1}{\sqrt{2}}\)

So, from the above equations, cos θ = 0.707 and θ = 45°.

11. In the circuit given below, the equivalent capacitance is _________________

a) 1.6 F

b) 3.1 F

c) 0.5 F

d) 4.6 F

View Answer

Explanation: C

_{CB}= \(\left(\frac{C_2 C_3}{C_2+ C_3}\right)\) + C

_{5}= 7.5 F

Now, C

_{AB}= \(\left(\frac{C_1 C_{CB}}{C_1 + C_{CB}}\right)\) + C

_{6}= 8 F

C

_{XY}= \(\frac{C_{AB} × C_4}{C_{AB} + C_4}\) = 3.1 F.

12. In the circuit given below, the equivalent capacitance is ______________

a) 5.43 μF

b) 4.23 μF

c) 3.65 μF

d) 5.50 μF

View Answer

Explanation: The 2 μF capacitor is in parallel with 1 μF capacitor and this combination is in series with 0.5 μF.

Hence, C

_{1}= \(\frac{0.5(2+1)}{0.5+2+1}\)

= \(\frac{1.5}{3.5}\) = 0.43

Now, C

_{1}is in parallel with the 5 μF capacitor.

∴ C

_{EQ}= 0.43 + 5 = 5.43 μF.

13. In the circuit given below, the voltage across AB is _______________

a) 250 V

b) 150 V

c) 325 V

d) 100 V

View Answer

Explanation: Loop current I

_{1}= \(\frac{50}{20}\) = 2.5 A

I

_{2}= \(\frac{100}{20}\) = 5 A

V

_{AB}= (50) (2.5) + 100 + (5) (20)

= 125 + 100 + 100

= 325 V.

14. The number of non-planar graph of independent loop equations is ______________

a) 8

b) 12

c) 3

d) 5

View Answer

Explanation: The total number of independent loop equations are given by L = B – N + 1 where,

L = number of loop equations

B = number of branches = 10

N = number of nodes = 8

∴ L = 10 – 8 + 1 = 3.

15. In the circuit given below, M = 20. The resonant frequency is _______________

a) 4.1 Hz

b) 41 Hz

c) 0.41 Hz

d) 0.041 Hz

View Answer

Explanation: I

_{EQ}= L

_{1}+ L

_{2}+ 2M

L

_{EQ}= 10 + 20 + 2 × \(\frac{1}{20}\) = 30.1 H

∴ F

_{O}= \(\frac{1}{2π\sqrt{LC}}\)

= \(\frac{1}{2π\sqrt{30.1 × 0.5}}\)

= 0.041 Hz.

**Sanfoundry Global Education & Learning Series – Network Theory.**

To practice all areas of Network Theory, __here is complete set of 1000+ Multiple Choice Questions and Answers__.