This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Advanced Problems Involving Complex Circuit Diagram – 1”.

1. The current wave shape is in the form of a square terminating at t = 4sec. The voltage across the element increases linearly till t = 4 sec and then becomes constant. The element is ____________

a) Resistance

b) Inductance

c) Capacitance

d) Semi-conductor

View Answer

Explanation: We know that, when a current pulse is applied to a capacitor, the voltage will have a waveform which rises linearly and then becomes constant towards the end of pulse. Hence, the element is a capacitor.

2. An infinite ladder is constructed with 1 Ω and 2 Ω resistor shown below. The current I flowing through the circuit is ___________

a) 8.18 A

b) 0 A

c) 9 A

d) 10 A

View Answer

Explanation: Equivalent Resistance, R

_{EQ}= R = 1 + (2 || R)

Or, R

_{EQ}= 2

So, I = \(\frac{18}{2}\) A = 9 A.

3. In the circuit given below, the phase angle of the current I with respect to the voltage V_{1} is __________

a) 0°

b) +45°

c) -45°

d) -90°

View Answer

Explanation: Net voltage applied to the circuit is 200∠0° V

I

_{1}= \(\frac{200∠0°}{10.0}\)

= 20∠0° = 20

I

_{2}= \(\frac{200∠0°}{10∠90°}\)

= 20∠-90° = -j20

I = I

_{1}+ I

_{2}= 20(1-j) = 20\(\sqrt{2}\)∠45°

Voltage V

_{1}= 100(1+j)

= 100\(\sqrt{2}\)∠45°

∴ Required phase angle = -45° – 45° = -90°.

4. Consider a circuit having 3 identical Ammeters A_{1}, A_{2}, A_{3} parallel to one another. The 1^{st} Ammeter is in series with a resistance, the 2^{nd} Ammeter is in series with a capacitor and the circuit is excited by a voltage V. If A_{1} and A_{3} read 5 and 13 A respectively, reading of A_{2} will be?

a) 8 A

b) 13 A

c) 18 A

d) 12 A

View Answer

Explanation: We can infer from the circuit,

A

_{2}= \(\sqrt{13^2 – 5^2}\)

Or, A

_{2}= \(\sqrt{169 – 25}\)

Or, A

_{2}= \(\sqrt{144}\)

Or, A

_{2}= 12 A.

5. In the circuit given below, the value of V_{1} is __________

a) 32.2 V

b) -25.23 V

c) 29.25 V

d) -29.25 V

View Answer

Explanation: \(\frac{V_A}{30} + \frac{V_A-40}{12} + \frac{V_A-V_B}{8}\) = 0

Or, 29 V

_{A}– 15 V

_{B}= 400

Also, \(\frac{V_B-V_A}{8} + \frac{V_B-120}{8}\) + 6 = 0

Or, V

_{A}= 65.23 V, V

_{B}= 99.44 V

V

_{1}= 40-65.23 = -25.23 V.

6. For the three coupled coils shown in figure, KVL equation is ____________

a) V = (L_{1} + L_{2} + L_{3}) \(\frac{di}{dt}\)

b) V = (L_{1} – L_{2} – L_{3} – M_{13}) \(\frac{di}{dt}\)

c) V = (L_{1} + L_{2} + L_{3} + 2M_{12} – 2M_{23} – 2M_{13}) \(\frac{di}{dt}\)

d) V = (L_{1} + L_{2} + L_{3} – 2M_{12} + 2M_{23} + 2M_{13}) \(\frac{di}{dt}\)

View Answer

Explanation: M

_{12}is positive while M

_{23}and M

_{13}are negative because of dots shown in figure.

So, the KVL equation is given by,

V = (L

_{1}+ L

_{2}+ L

_{3}+ 2M

_{12}– 2M

_{23}– 2M

_{13}) \(\frac{di}{dt}\).

7. A circuit excited by voltage V has a resistance R which is in series with an inductor and capacitor, which are connected in parallel. The voltage across the resistor at the resonant frequency is ___________

a) 0

b) \(\frac{V}{2}\)

c) \(\frac{V}{3}\)

d) V

View Answer

Explanation: Dynamic resistance of the tank circuit, Z

_{DY}= \(\frac{L}{RLC}\)

But given that R

_{L}= 0

So, Z

_{DY}= \(\frac{L}{0XC}\) = ∞

Therefore current through circuit, I = \(\frac{V}{∞}\) = 0

∴ V

_{D}= 0.

8. In the circuit given below, the value of resistance R is _________

a) 10 Ω

b) 18 Ω

c) 24 Ω

d) 12 Ω

View Answer

Explanation: Using KVL in loop 1, we get, 100 – 14I – 30 = 0

Or, I = \(\frac{70}{14}\) = 5 A

Then, V

_{P}– V

_{Q}= 14I – (15-I).1

= 70 – 10 = 60 V

∴ R = \(\frac{60}{10-I} = \frac{60}{5}\) = 12 Ω.

9. The current flowing through the resistance R in the circuit in the figure has the form 2 cos 4t, where R is ____________

a) (0.18 + j0.72)

b) (0.46 + j1.90)

c) – (0.18 + j1.90)

d) (0.23 – 0.35 j)

View Answer

Explanation: Inductor is not given, hence ignoring the inductance. Let I

_{1}and I

_{2}are currents in the loop then,

I

_{1}= \(\frac{2 cos4t}{3}\)

= 0.66 cos 4t

Again, I

_{2}= \(\frac{-j X 4 X 0.75 I_1}{3.92-2.56j}\)

= (0.23 – 0.35j) cos 4t

10. In the circuit given below, the voltage V_{AB} is _________

a) 6 V

b) 25 V

c) 10 V

d) 40 V

View Answer

Explanation: For finding the Thevenin Equivalent circuit across A-B we remove the 5 Ω resistor.

Then, I = \(\frac{10+50}{15}\) = 4 A

V

_{OC}= 50 – (10×4) =10 V

And R

_{EQ}= \(\frac{10×5}{10+5} = \frac{10}{3}\) Ω

Current I

_{1}= \(\frac{10}{10/3+5} = \frac{6}{5}\)

Hence, V

_{AB}= \(\frac{6}{5 × 5}\) = 6 V.

11. In the circuit given below, the magnitudes of V_{L} and V_{C} are twice that of V_{K}. Calculate the inductance of the coil, given that f = 50.50 Hz.

a) 6.41 mH

b) 5.30 mH

c) 3.18 mH

d) 2.31 mH

View Answer

Explanation: V

_{L}= V

_{C}= 2 VR

∴ Q = \(\frac{V_L}{V_R}\) = 2

But we know, Q = \(\frac{ωL}{R} = \frac{1}{ωCR}\)

∴ 2 = \(\frac{2πf × L}{5}\)

Or, L = 3.18 mH.

12. In the circuit given below, the current source is 1 A, voltage source is 5 V, R_{1} = R_{2} = R_{3} = 1 Ω, L_{1} = L_{2} = L_{3} = 1 H, C_{1} = C_{2} = 1 F. The current through R_{3} is _________

a) 1 A

b) 5 A

c) 6 A

d) 8 A

View Answer

Explanation: At steady state, the circuit becomes,

∴ The current through R

_{3}= \(\frac{5}{1}\) = 5 A.

13. In the circuit given below, the capacitor is initially having a charge of 10 C. 1 second after the switch is closed, the current in the circuit is ________

a) 14.7 A

b) 18.5 A

c) 40.0 A

d) 50.0 A

View Answer

Explanation: Using KVL, 100 = \(R\frac{dq}{dt} + \frac{q}{C}\)

100 C = \(RC\frac{dq}{dt}\) + q

Or, \(\int_{q_o}^q \frac{dq}{100C-q} = \frac{1}{RC} ∫_0^t \,dt\)

100C – q = (100C – q

_{o})e

^{-t/RC}

I = \(\frac{dq}{dt} = \frac{(100C – q_o)}{RC} e^{-1/1}\)

∴ e

^{-t/RC}= 40e

^{-1}= 14.7 A.

14. 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

View Answer

Explanation: We, draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’.

V

_{xx’}= V

_{N}= \(\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}\) = 5V

∴ R

_{N}= 8 || (16 + 8)

= \(\frac{8×24}{8+24}\) = 6 Ω

∴ I

_{N}= \(\frac{V_N}{R_N} = \frac{5}{6}\) = 0.833 A.

15. In the circuit given below, the current source is 1 A, voltage source is 5 V, R_{1} = R_{2} = R_{3} = 1 Ω, L_{1} = L_{2} = L_{3} = 1 H, C_{1} = C_{2} = 1 F. The current through the voltage source V is _________

a) 1 A

b) 3 A

c) 2 A

d) 4 A

View Answer

Explanation: At steady state, the circuit becomes,

∴ The current through the voltage source V = 5 – 1 = 4 A.

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