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 ___________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q1
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 \(\frac{di_2}{dt}\)
View Answer

Answer: c
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 \(\frac{di_2}{dt}\).
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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

Answer: c
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
View Answer

Answer: b
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
View Answer

Answer: d
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
View Answer

Answer: a
Explanation: At t = 0, Vc = 0 and at t = ∞, Vc = 1.
This condition can be satisfied only by (1 – e-t/RC).
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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 ____________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q6
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
View Answer

Answer: b
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 ____________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q7
a) 12 Ω
b) 6 Ω
c) 3 Ω
d) 1.5 Ω
View Answer

Answer: b
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
View Answer

Answer: c
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
View Answer

Answer: a
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.
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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

Answer: d
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 _________________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q11
a) 1.6 F
b) 3.1 F
c) 0.5 F
d) 4.6 F
View Answer

Answer: b
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 ______________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q12
a) 5.43 μF
b) 4.23 μF
c) 3.65 μF
d) 5.50 μF
View Answer

Answer: a
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 _______________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q13
a) 250 V
b) 150 V
c) 325 V
d) 100 V
View Answer

Answer: c
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.
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14. The number of non-planar graph of independent loop equations is ______________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q14
a) 8
b) 12
c) 3
d) 5
View Answer

Answer: c
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 _______________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q15
a) 4.1 Hz
b) 41 Hz
c) 0.41 Hz
d) 0.041 Hz
View Answer

Answer: d
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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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn