250+ TOP MCQs on Dot Convention in Magnetically Coupled Circuits and Answers

Network Theory Multiple Choice Questions 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

Answer: B
Clarification: 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 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

Answer: C
Clarification: 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^-125tu(t) mA
B. 20e^-1250tu(t) mA
C. 0.2e^-1250tu(t) mA
D. 20e^-1000tu(t) mA

Answer: C
Clarification: 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^-1250tu(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

Answer: A
Clarification: 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

Answer: A
Clarification: 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

Answer: C
Clarification: 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)

Answer: C
Clarification: 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

Answer: D
Clarification: When S₁ is closed and S₂ is open,
V_C1 (0^–) = V_C1 (0⁺) = 3V
When S₁ is opened and S₂ is closed, V_C2 (0⁺) = V_C2 (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

Answer: D
Clarification: 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 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

Answer: A
Clarification: 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^-ωtcos(ωt)

Answer: B
Clarification: 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, 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

Answer: A
Clarification: 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} = 2sqrt{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

Answer: A
Clarification: 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

Answer: B
Clarification: 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}))

Answer: A
Clarification: 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}))].