# Network Theory Questions and Answers – Advanced Problems Involving Complex Circuit Diagram – 1

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

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

Explanation: Equivalent Resistance, REQ = R = 1 + (2 || R)
Or, REQ = 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 V1 is __________

a) 0°
b) +45°
c) -45°
d) -90°

Explanation: Net voltage applied to the circuit is 200∠0° V
I1 = $$\frac{200∠0°}{10.0}$$
= 20∠0° = 20
I2 = $$\frac{200∠0°}{10∠90°}$$
= 20∠-90° = -j20
I = I1 + I2 = 20(1-j) = 20$$\sqrt{2}$$∠45°
Voltage V1 = 100(1+j)
= 100$$\sqrt{2}$$∠45°
∴ Required phase angle = -45° – 45° = -90°.

4. Consider a circuit having 3 identical Ammeters A1, A2, A3 parallel to one another. The 1st Ammeter is in series with a resistance, the 2nd Ammeter is in series with a capacitor and the circuit is excited by a voltage V. If A1 and A3 read 5 and 13 A respectively, reading of A2 will be?
a) 8 A
b) 13 A
c) 18 A
d) 12 A

Explanation: We can infer from the circuit,
A2 = $$\sqrt{13^2 – 5^2}$$
Or, A2 = $$\sqrt{169 – 25}$$
Or, A2 = $$\sqrt{144}$$
Or, A2 = 12 A.

5. In the circuit given below, the value of V1 is __________

a) 32.2 V
b) -25.23 V
c) 29.25 V
d) -29.25 V

Explanation: $$\frac{V_A}{30} + \frac{V_A-40}{12} + \frac{V_A-V_B}{8}$$ = 0
Or, 29 VA – 15 VB = 400
Also, $$\frac{V_B-V_A}{8} + \frac{V_B-120}{8}$$ + 6 = 0
Or, VA = 65.23 V, VB = 99.44 V
V1 = 40-65.23 = -25.23 V.

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

a) V = (L1 + L2 + L3) $$\frac{di}{dt}$$
b) V = (L1 – L2 – L3 – M13) $$\frac{di}{dt}$$
c) V = (L1 + L2 + L3 + 2M12 – 2M23 – 2M13) $$\frac{di}{dt}$$
d) V = (L1 + L2 + L3 – 2M12 + 2M23 + 2M13) $$\frac{di}{dt}$$

Explanation: M12 is positive while M23 and M13 are negative because of dots shown in figure.
So, the KVL equation is given by,
V = (L1 + L2 + L3 + 2M12 – 2M23 – 2M13) $$\frac{di}{dt}$$.

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

Explanation: Dynamic resistance of the tank circuit, ZDY = $$\frac{L}{RLC}$$
But given that RL = 0
So, ZDY = $$\frac{L}{0XC}$$ = ∞
Therefore current through circuit, I = $$\frac{V}{∞}$$ = 0
∴ VD = 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

Explanation: Using KVL in loop 1, we get, 100 – 14I – 30 = 0
Or, I = $$\frac{70}{14}$$ = 5 A
Then, VP – VQ = 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)

Explanation: Inductor is not given, hence ignoring the inductance. Let I1 and I2 are currents in the loop then,
I1 = $$\frac{2 cos⁡4t}{3}$$
= 0.66 cos 4t
Again, I2 = $$\frac{-j X 4 X 0.75 I_1}{3.92-2.56j}$$
= (0.23 – 0.35j) cos 4t

10. In the circuit given below, the voltage VAB is _________

a) 6 V
b) 25 V
c) 10 V
d) 40 V

Explanation: For finding the Thevenin Equivalent circuit across A-B we remove the 5 Ω resistor.
Then, I = $$\frac{10+50}{15}$$ = 4 A
VOC = 50 – (10×4) =10 V
And REQ = $$\frac{10×5}{10+5} = \frac{10}{3}$$ Ω
Current I1 = $$\frac{10}{10/3+5} = \frac{6}{5}$$
Hence, VAB = $$\frac{6}{5 × 5}$$ = 6 V.

11. In the circuit given below, the magnitudes of VL and VC are twice that of VK. 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

Explanation: VL = VC = 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, R1 = R2 = R3 = 1 Ω, L1 = L2 = L3 = 1 H, C1 = C2 = 1 F. The current through R3 is _________

a) 1 A
b) 5 A
c) 6 A
d) 8 A

Explanation: At steady state, the circuit becomes,

∴ The current through R3 = $$\frac{5}{1}$$ = 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

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 – qo)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

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

Vxx’ = VN = $$\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}$$ = 5V
∴ RN = 8 || (16 + 8)
= $$\frac{8×24}{8+24}$$ = 6 Ω
∴ IN = $$\frac{V_N}{R_N} = \frac{5}{6}$$ = 0.833 A.

15. In the circuit given below, the current source is 1 A, voltage source is 5 V, R1 = R2 = R3 = 1 Ω, L1 = L2 = L3 = 1 H, C1 = C2 = 1 F. The current through the voltage source V is _________

a) 1 A
b) 3 A
c) 2 A
d) 4 A

Explanation: At steady state, the circuit becomes,

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

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