# Network Theory Questions and Answers – Advanced Problems Involving Complex Circuit Diagram – 2

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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) = R1i1 + L1 $$\frac{di_1}{dt}$$ + M $$\frac{di_2}{dt}$$
b) V (t) = R1i1 – L1 $$\frac{di_1}{dt}$$ – M $$\frac{di_2}{dt}$$
c) V (t) = R1i1 + L1 $$\frac{di_1}{dt}$$ – M $$\frac{di_2}{dt}$$
d) V (t) = R1i1 – L1 $$\frac{di_1}{dt}$$ + M $$\frac{di_2}{dt}$$

Explanation: We know that, in general, the KVL is of the form V (t) = R1i1 + L1 $$\frac{di_1}{dt}$$ + M $$\frac{di_2}{dt}$$
But here, M term is negative because i1, is entering the dotted terminal and i2, is leaving the dotted terminal.
So, V (t) = R1i1 + L1 $$\frac{di_1}{dt}$$ – 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

Explanation: Currents in resistance and inductance are out of phase by 90°.
Hence, I = $$I_1^2 + I_2^2$$
Or, I = [122 + 162]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

Explanation: We know that, during discharge of capacitor,
VC = VR
Now, VR = 0.67 X 10 = 6.7 V
So, VC = 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

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, Vc is __________
a) 1 – e-t/RC
b) e-t/RC
c) et/RC
d) 1

Explanation: At t = 0, Vc = 0 and at t = ∞, Vc = 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) VTH = 5 V, RTH = 0.75 Ω
b) VTH = 0.5 V, RTH = 0.75 Ω
c) VTH = 2.5 V, RTH = 1 Ω
d) VTH = 5 V, RTH = 1 Ω

Explanation: VTH = $$\frac{1 X 20}{40}$$ X 1 = 0.5 V
Also, RTH = $$\frac{1 X 30}{40}$$ = 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

Explanation: The resistance of parallel combination is given by,
Req = $$\frac{40}{3}$$ – 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 VS, 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

Explanation: When R = ∞, V = 5v,
Then, Voc = 5V and the circuit is open
When R = 0, I = 2.5A
Then, Isc = 2.5 and the circuit is short circuited.
So, Req = $$\frac{V_{OC}}{I_{SC}}$$
= $$\frac{5}{2.5}$$ = 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

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°

Explanation: VI cos θ = 0.25 or I cos θ = 0.25
Or, Z cosθ = 2
Or, $$\frac{V}{I}$$ cos⁡θ = 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

Explanation: CCB = $$\left(\frac{C_2 C_3}{C_2+ C_3}\right)$$ + C5 = 7.5 F
Now, CAB = $$\left(\frac{C_1 C_{CB}}{C_1 + C_{CB}}\right)$$ + C6 = 8 F
CXY = $$\frac{C_{AB} × C_4}{C_{AB} + C_4}$$ = 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

Explanation: The 2 μF capacitor is in parallel with 1 μF capacitor and this combination is in series with 0.5 μF.
Hence, C1 = $$\frac{0.5(2+1)}{0.5+2+1}$$
= $$\frac{1.5}{3.5}$$ = 0.43
Now, C1 is in parallel with the 5 μF capacitor.
∴ CEQ = 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

Explanation: Loop current I1 = $$\frac{50}{20}$$ = 2.5 A
I2 = $$\frac{100}{20}$$ = 5 A
VAB = (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

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

Explanation: IEQ = L1 + L2 + 2M
LEQ = 10 + 20 + 2 × $$\frac{1}{20}$$ = 30.1 H
∴ FO = $$\frac{1}{2π\sqrt{LC}}$$
= $$\frac{1}{2π\sqrt{30.1 × 0.5}}$$
= 0.041 Hz.

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