# Network Theory Questions and Answers – Dot Convention in Magnetically Coupled Circuits

«
»

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

Explanation: Q = $$\frac{f_0}{BW}$$
And f0 = 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 ZIN 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

Explanation: ZIN = (-6j) || (ZA)
ZA = j10 + $$\frac{12^2}{(j30+j6-j2+4)}$$
= 0.49 + j5.82
ZIN = $$\frac{(-j6)(0.49+j5.82)}{(-j6+0.49+j5.82)}$$
= 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-125tu(t) mA
b) 20e-1250tu(t) mA
c) 0.2e-1250tu(t) mA
d) 20e-1000tu(t) mA

Explanation: CEQ = $$\frac{0.8 × 0.2}{0.8+0.2}$$ = 0.16
VC (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-1250tu(t) mA.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

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

a) Zero
b) 5 W
c) 10 W
d) 100 W

Explanation: Let the current supplied by a 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

Explanation: I (t) = $$\frac{1.5}{3}$$ = 0.5
LEQ = 15 mH
REQ = 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 |VO| = 5 V, VS = 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

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
I1 = $$\frac{16}{3} × \frac{1}{2} = \frac{8}{3}$$ 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 V1 (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)

Explanation: For u (t) = 1, t>0
V1 (t) = (1-e-3t)
Or, V1 (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, S1 remained closed and S2 open., S1 is initially opened and S2 is initially closed. If the voltage V2 across the capacitor C2 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

Explanation: When S1 is closed and S2 is open,
VC1 (0) = VC1 (0+) = 3V
When S1 is opened and S2 is closed, VC2 (0+) = VC2 (0+) = 3V.

9. In the circuit given below, the switch S1 is initially closed and S2 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

Explanation: By KCL, we get,
$$\frac{V_L}{10} – 10 + \frac{V_L-10}{10}$$ = 0
Hence, 2 VL = 110
∴ VL = 55 V
Or, IC = $$\frac{55-10}{10}$$ = 4.5 A.

10. In the circuit given below, the switch S1 is initially closed and S2 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

Explanation: By KCL, we get,
$$\frac{V_L}{10} – 10 + \frac{V_L-10}{10}$$ = 0
Hence, 2 VL = 110
∴ VL = 55 V.

11. An ideal capacitor is charged to a voltage VO 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) VO
b) VO cos(ωt)
c) VO sin(ωt)
d) VO e-ωtcos(ωt)

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 VO and periodic and is not decaying since both L and C is ideal.
∴ Voltage across the capacitor at time t>0 is VO cos(ωt).

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

a) 2$$\sqrt{6}$$ A
b) 6$$\sqrt{2}$$ A
c) $$\sqrt{\frac{4}{3}}$$ A
d) 1.5 A

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

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 – qo) 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

Explanation: A t=0+,
I (0+) = $$\frac{10}{4||4+3} = \frac{10}{5}$$ = 2A
∴ I2(0+) = $$\frac{2}{2}$$ = 1A
VL(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(-$$\frac{B}{RC}$$)

Explanation: VC = $$\frac{1}{C} ∫Idt$$
= $$\frac{1}{C} ∫_0^B \frac{A}{R} e^{-\frac{t}{RC}}$$ dt
VC = A[1 – exp(-$$\frac{B}{RC}$$)]
Hence, maximum voltage = V [1 – exp (-$$\frac{B}{RC}$$)].

Sanfoundry Global Education & Learning Series – Network Theory.

To practice all areas of Network Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}