This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Dot Convention in Magnetically Coupled Circuits”.

1. A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100. If each of R, L and C is doubled from its original value, the new Q of the circuit is _____________

a) 25

b) 50

c) 100

d) 200

View Answer

Explanation: Q = \(\frac{f_0}{BW}\)

And f

_{0}= 1/2π (LC)

^{0.5}

BW = R/L

Or, Q = \(\frac{1}{R} (\frac{L}{C})^{0.5}\)

When R, L and C are doubled, Q’ = 50.

2. In the circuit given below, the input impedance Z_{IN} of the circuit is _________

a) 0.52 – j4.30 Ω

b) 0.52 + j15.70 Ω

c) 64.73 + j17.77 Ω

d) 0.3 – j33.66 Ω

View Answer

Explanation: Z

_{IN}= (-6j) || (Z

_{A})

Z

_{A}= j10 + \(\frac{12^2}{(j30+j6-j2+4)}\)

= 0.49 + j5.82

Z

_{IN}= \(\frac{(-j6)(0.49+j5.82)}{(-j6+0.49+j5.82)}\)

= 64.73 + j17.77 Ω.

3. The switch in the circuit shown was on position X for a long time. The switch is then moved to position Y at time t=0. The current I(t) for t>0 is ________

a) 0.2e^{-125t}u(t) mA

b) 20e^{-1250t}u(t) mA

c) 0.2e^{-1250t}u(t) mA

d) 20e^{-1000t}u(t) mA

View Answer

Explanation: C

_{EQ}= \(\frac{0.8 × 0.2}{0.8+0.2}\) = 0.16

V

_{C}(t=0

^{–}) = 100 V

At t≥0,

The discharging current I (t) = \(\frac{V_O}{R} e^{-\frac{t}{RC}}\)

= \(\frac{100}{5000} e^{- \frac{t}{5×10^3×0.16×10^{-6}}}\)

= 0.2e

^{-1250t}u(t) mA.

4. In the circuit shown, the voltage source supplies power which is _____________

a) Zero

b) 5 W

c) 10 W

d) 100 W

View Answer

Explanation: Let the current supplied by voltage source.

Applying KVL in outer loop,

10 – (I+3) × (I+1) – (I+2) × 2 = 0

10 – 2(I+3) – 2(I-2) = 0

Or, I = 0

∴ Power VI = 0.

5. In the circuit shown below the current I(t) for t≥0^{+} (assuming zero initial conditions) is ___________

a) 0.5-0.125e^{-1000t} A

b) 1.5-0.125e^{-1000t} A

c) 0.5-0.5e^{-1000t} A

d) 0.375e^{-1000t} A

View Answer

Explanation: I (t) = \(\frac{1.5}{3}\) = 0.5

L

_{EQ}= 15 mH

R

_{EQ}= 5+10 = 15Ω

L = \(\frac{L_{EQ}}{R_{EQ}}\)

= \(\frac{15 × 10^3}{15} = \frac{1}{1000}\)

I (t) A – (A – B) e

^{-t}= 0.5 – (0.5-B) e

^{-1000t}

= 0.5(0.5 – 0.375) e

^{-1000t}

= 0.5 – 0.125 e

^{-1000t}

I (t) = 0.5-0.125e

^{-1000t}.

6. Initial voltage on capacitor VO as marked |V_{O}| = 5 V, V_{S} = 8 u (t), where u (t) is the unit step. The voltage marked V at t=0^{+} is _____________

a) 1 V

b) -1 V

c) \(\frac{13}{3}\) V

d) –\(\frac{13}{3}\) V

View Answer

Explanation: Applying voltage divider method, we get,

I = \(\frac{V}{R_{EQ}} \)

= \(\frac{8}{1+1||1} \)

= \(\frac{8}{1+\frac{1}{2}} = \frac{16}{3}\) A

I

_{1}= \(\frac{16}{3} × \frac{1}{2} = \frac{8}{3}\) A

And \(I’_2 = \frac{V}{R_{EQ}} = \frac{5}{1+1||1}\)

= \(\frac{5}{1+\frac{1}{2}} = \frac{10}{3}\) A

Now, \(I’_1 = I’_2 × \frac{1}{1+1} \)

= \(\frac{10}{3} × \frac{1}{2} = \frac{5}{3}\) A

Hence, the net current in 1Ω resistance = I1 + \(I’_1\)

= \(\frac{8}{3} + \frac{5}{3} = \frac{13}{3}\) A

∴ Voltage drop across 1Ω = \(\frac{13}{3} × 1 = \frac{13}{3}\) V.

7. For a unit step signal u (t), the response is V_{1} (t) = (1-e^{-3t}) for t>0. If a signal 3u (t) + δ(t) is applied, the response will be (considering zero initial conditions)?

a) (3-6e^{-3t})u(t)

b) (3-3e^{-3t})u(t)

c) 3u(t)

d) (3+3e^{-3t})u(t)

View Answer

Explanation: For u (t) = 1, t>0

V

_{1}(t) = (1-e

^{-3t})

Or, V

_{1}(s) = \(\left(\frac{1}{s} + \frac{1}{s+3}\right) = \frac{3}{s(s+3)}\)

And T(s) = \(\frac{V_1 (S)}{u(S)} = \frac{3}{s+3}\)

Now, for R(s) = (\(\frac{3}{s}\) + 1)

Response, H(s) = R(s) T(s) = \((\frac{3+s}{s}) (\frac{3}{s+3}) = \frac{3}{s}\)

Or, h (t) = 3 u (t).

8. In the circuit given below, for time t<0, S_{1} remained closed and S_{2} open., S_{1} is initially opened and S_{2} is initially closed. If the voltage V_{2} across the capacitor C_{2} at t=0 is zero, the voltage across the capacitor combination at t=0^{+} will be ____________

a) 1 V

b) 2 V

c) 1.5 V

d) 3 V

View Answer

Explanation: When S

_{1}is closed and S

_{2}is open,

V

_{C1}(0

^{–}) = V

_{C1}(0

^{+}) = 3V

When S

_{1}is opened and S

_{2}is closed, V

_{C2}(0

^{+}) = V

_{C2}(0

^{+}) = 3V.

9. In the circuit given below, the switch S_{1} is initially closed and S_{2} is opened. The inductor L carries a current of 10A and the capacitor is charged to 10 V with polarities as shown. The current through C during t=0^{+} is _____________

a) 55 A

b) 5.5 A

c) 45 A

d) 4.5 A

View Answer

Explanation: By KCL, we get,

\(\frac{V_L}{10} – 10 + \frac{V_L-10}{10}\) = 0

Hence, 2 V

_{L}= 110

∴ V

_{L}= 55 V

Or, I

_{C}= \(\frac{55-10}{10}\) = 4.5 A.

10. In the circuit given below, the switch S_{1} is initially closed and S_{2} is opened. The inductor L carries a current of 10A and the capacitor is charged to 10 V with polarities as shown. The current through L during t=0^{+} is _____________

a) 55 A

b) 5.5 A

c) 45 A

d) 4.5 A

View Answer

Explanation: By KCL, we get,

\(\frac{V_L}{10} – 10 + \frac{V_L-10}{10}\) = 0

Hence, 2 V

_{L}= 110

∴ V

_{L}= 55 V.

11. An ideal capacitor is charged to a voltage V_{O} and connected at t=0 across an ideal inductor L. If ω = \(\frac{1}{\sqrt{LC}}\), the voltage across the capacitor at time t>0 is ____________

a) V_{O}

b) V_{O} cos(ωt)

c) V_{O} sin(ωt)

d) V_{O} e^{-ωt}cos(ωt)

View Answer

Explanation: Voltage across capacitor will discharge through inductor up to voltage across the capacitor becomes zero. Now, inductor will start charging capacitor.

Voltage across capacitor will be decreasing from V

_{O}and periodic and is not decaying since both L and C is ideal.

∴ Voltage across the capacitor at time t>0 is V

_{O}cos(ωt).

12. In the figure given below, the RMS value of the periodic waveform is _____________

a) 2\(\sqrt{6}\) A

b) 6\(\sqrt{2}\) A

c) \(\sqrt{\frac{4}{3}}\) A

d) 1.5 A

View Answer

Explanation: The rms value for any waveform is = \(\sqrt{\frac{1}{T} \int_0^T f^2 (t)dt)}\)

= \([\frac{1}{T}(\int_0^{\frac{T}{2}}(mt)^2 dt + \int_{\frac{T}{2}}^T 6^2 dt]^{1/2}\)

= \([\frac{1}{T}(\frac{144}{T^2} × \frac{T^3}{3×8} + 36 × \frac{T}{2})]^{\frac{1}{2}}\)

= \([6+18]^{\frac{1}{2}} = \sqrt{24} = 2\sqrt{6}\) A.

13. In the circuit given below, the capacitor initially has a charge of 10 C. The current in the circuit at t=1 sec after the switch S is closed will be ___________

a) 14.7 A

b) 18.5 A

c) 40 A

d) 50 A

View Answer

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

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

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

Or, 100C – q = (100C – q

_{o}) e

^{-t/RC}

I = \(\frac{dq}{dt} = \frac{(100C – q_0)}{RC} e^{-1/1}\)

= 40e

^{-1}= 14.7 A.

14. In the circuit given below, the switch is closed at time t=0. The voltage across the inductance just at t=0^{+} is ____________

a) 2 V

b) 4 V

c) -6 V

d) 8 V

View Answer

Explanation: A t=0

^{+},

I (0

^{+}) = \(\frac{10}{4||4+3} = \frac{10}{5}\) = 2A

∴ I

_{2(0+)}= \(\frac{2}{2}\) = 1A

V

_{L(0+)}= 1 × 4 = 4 V.

15. A rectangular voltage wave of magnitude A and duration B is applied to a series combination of resistance R and capacitance C. The voltage developed across the capacitor is ____________

a) A[1 – exp(-\(\frac{B}{RC}\))]

b) \(\frac{AB}{RC}\)

c) A

d) A exp(-\(\frac{B}{RC}\))

View Answer

Explanation: V

_{C}= \(\frac{1}{C} ∫Idt\)

= \(\frac{1}{C} ∫_0^B \frac{A}{R} e^{-\frac{t}{RC}}\) dt

V

_{C}= A[1 – exp(-\(\frac{B}{RC}\))]

Hence, maximum voltage = V [1 – exp (-\(\frac{B}{RC}\))].

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