This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Advanced Problems on Magnetically Coupled Circuits – 1”.

1. For the circuit given below, the effective inductance of the circuit across the terminal AB is ___________

a) 9 H

b) 21 H

c) 11 H

d) 6 H

View Answer

Explanation: L

_{EFF}across AB = L

_{1}+ L

_{2}+ L

_{3}– 2M

_{12}– 2M

_{13}+ 2M

_{23}

= (4 + 5 + 6 – 2 × 1 -2 × 3 + 2 × 2)

= (15 – 2 – 6 + 4)

= (15 – 8 + 4)

= (15 – 4)

= 11.

2. Two coils are having self-inductance of 5 mH and 10 mH and a mutual inductance of 0.5 mH in a differential connection. The equivalent inductance of the combination is ___________

a) 14 mH

b) 5.85 mH

c) 6 mH

d) 6.15 mH

View Answer

Explanation: When 2 coils are connected in series, then effective inductance,

L

_{EFF}= L

_{1}+ L

_{2}± 2M

For this case, L

_{EFF}= L

_{1}+ L

_{2}– 2M

= 5 + 10 – 2 × 0.5

= 15 – 1

= 14 mH.

3. When two coupled coils of equal self-inductance are connected in series in one way the net inductance is 20 mH and when they are connected in the other way, the net inductance is 12 mH. The maximum value of net inductance when they are connected in parallel is __________

a) 2 mH

b) 5 mH

c) 4 mH

d) 6 mH

View Answer

Explanation: L

_{EFF}= L

_{1}+ L

_{2}± 2M (when the two coils are connected in series)

Now, L

_{1}= L

_{2}= L

∴ L

_{EFF}= 2L ± 2M

Or, 2L + 2M = 20

Or, 2L – 2M = 12

∴ L = 8, M = 2

Now, to get maximum value in parallel connection,

L

_{MAX}= \(\frac{(L+M)(L+M)}{2L+2M}\)

= \(\frac{(8+2)(8+2)}{(16+4)}\)

= \(\frac{100}{20}\) = 5 mH.

4. For 2 coupled inductors L_{A} and L_{B}, their mutual inductance M_{LALB} satisfies ____________

a) M_{LALB} = \(\sqrt{L_A^2 + L_B^2}\)

b) M_{LALB} > \(\frac{(L_A + L_B)}{2}\)

c) M_{LALB} > \(\sqrt{L_A L_B}\)

d) M_{LALB} ≤ \(\sqrt{L_A L_B}\)

View Answer

Explanation: M = K \(\sqrt{L_A L_B}\)

Or, K = \(\frac{M}{\sqrt{L_A L_B}}\)

∴ K≤1

∴ \(\frac{M}{\sqrt{L_A L_B}}\) ≤ 1

Or, M ≤ \(\frac{M}{\sqrt{L_A L_B}}\).

5. The impedance seen by the source in the circuit is given by __________

a) (0.54 + j0.313) Ω

b) (4 – j2) Ω

c) (4.54 – j1.69) Ω

d) (4 + j2) Ω

View Answer

Explanation: Z

_{1}= 10∠30° × (\(\frac{1}{4}\))

^{2}

Z

_{1}= (0.54 + j0.31) Ω

Total impedance= (4 – j2) + (0.56 + j0.31)

= (4.54 – j1.69) Ω.

6. For the circuit given below, the inductance measured across the terminals 1 and 2 was 15 H with open terminals 3 and 4. It was 30 H when terminals 3 and 4 were short-circuited. Both the inductors are having inductances 2H. The coefficient of coupling is ______________

a) 1

b) 0.707

c) 0.5

d) Inderminate due to insufficient data

View Answer

Explanation: When 2 coils are connected in series, then effective inductance,

L

_{EFF}= L

_{1}+ L

_{2}± 2M

For this case, L

_{EFF}= L

_{1}+ L

_{2}– 2M

However, L

_{1}+ L

_{2}+ 2M or L

_{1}+ L

_{2}– 2M cannot be determined.

Hence, data is insufficient to calculate the value of coupling coefficient (k).

7. In the circuit given below, the resonant frequency is ________________

a) \(\frac{1}{2π\sqrt{3}}\) Hz

b) \(\frac{1}{4π\sqrt{3}}\) Hz

c) \(\frac{1}{4π\sqrt{2}}\) Hz

d) \(\frac{1}{π\sqrt{2}}\) Hz

View Answer

Explanation: L

_{EFF}= L

_{1}+ L

_{2}+ 2M

= 2 + 2 + 2 × 1

Or, L

_{EFF}= 6 H

At resonance, ωL = \(\frac{1}{ωC}\)

Or, ω = \(\frac{1}{\sqrt{L_{EFF} C}} = \frac{1}{\sqrt{12}}\)

Or, I = \(\frac{1}{4π\sqrt{3}}\) Hz.

8. The maximum value of mutual inductance of 2 inductively coupled coils with self inductance L_{A} = 49 mH and L_{B} = 81 mH is ______________

a) 130 mH

b) 63 mH

c) 32 mH

d) 3969 mH

View Answer

Explanation: M ≤ K\(\sqrt{L_A L_B}\)

Maximum value of (M)

Or, M

_{max}= K\(\sqrt{L_A L_B}\)

Or, M

_{max}= \(\sqrt{L_A L_B}\)

= \(\sqrt{40 ×81}\) = 63 mH.

9. In the circuit shown below, I_{1} = 4 sin2t A and I_{2} = 0. The value of V_{1} is _______________

a) -16 cos2t V

b) 16 cos2t V

c) 4 cos2t V

d) -4 cos2t V

View Answer

Explanation: V

_{1}= \(\frac{2dI_1}{dt} + \frac{1dI_2}{dt}\)

∴ V

_{1}= \(\frac{2dI_1}{dt}\) = 16 cos2t V.

10. In the circuit shown below, I_{1} = 4 sin2t A and I_{2} = 0. The value of V_{2} is _______________

a) 2 cos2t V

b) -2 cos2t V

c) 8 cos 2t V

d) -8 cos2t V

View Answer

Explanation: V

_{2}= \(\frac{dI_2}{dt} + \frac{dI_1}{dt}\)

Or, V

_{2}= \(\frac{dI_1}{dt}\)

∴ V

_{2}= 8 cos2t V.

11. A parallel resonant circuit has a midband admittance of 20×10^{-3} S, quality factor of 60 and a resonance frequency of 200 k rad/s. The value of R is _____________

a) 50 Ω

b) 56.57 Ω

c) 80 Ω

d) 28.28 Ω

View Answer

Explanation: At midband frequency, Z = R, Y = \(\frac{1}{R}\)

Or, R = \(\frac{1}{20 × 10^{-3}}\) = 50 Ω.

12. A circuit resonates at 1 MHz It also has a Q of 100. Bandwidth between half power points is _____________

a) 10 kHz

b) 100 kHz

c) 10 Hz

d) 100 Hz

View Answer

Explanation: We know that, Q = \(\frac{f}{∆f}\)

Or, ∆f = \(\frac{f}{Q}\)

= \(\frac{10^6}{100}\) = 10 kHz.

13. A series RLC circuit has R = 10 Ω, |X_{L}| = 20 Ω and |X_{C}| = 20 Ω is connected across an AC supply of 200 V_{rms}. The RMS voltage across the capacitor is ____________

a) 200∠-90° V

b) 200∠90° V

c) 400∠90° V

d) 400∠-90° V

View Answer

Explanation: Q = \(\frac{|X_L|}{R} = \frac{|X_C|}{R}\)

So, Q = \(\frac{20}{10}\) = 2

Rms voltage across capacitor, V

_{Crms}= QV∠-90°

Or, V

_{Crms}= 2 X 200∠-90°

Or, V

_{Crms}= 400∠-90° V.

14. For the circuit given below, the value of input frequency which is required to cause a gain equal to 1.5 is _____________

a) 20 rad/s

b) 20 Hz

c) 10 rad/s

d) No such value exists

View Answer

Explanation: H (ω) = \(\frac{V_O}{V_i}\)

= \(\frac{1}{1+jωRC}\)

∴ Gain = \(\frac{1}{\sqrt{1 + ω^2 RC}}\)

For any value of ω, R and C gain is ≤ 1.

15. Two resistances 100 ± 5Ω and 150 ± 15Ω are connected in series. If the error is specified as standard deviations, the resultant error will be ________________

a) ±10 Ω

b) ±10.6 Ω

c) ±15.8 Ω

d) ±20 Ω

View Answer

Explanation: Given, R

_{1}= 100 ± 5 Ω

R

_{2}= 150 ± 15 Ω

Now, R = R

_{1}+ R

_{2}

The probable errors in this case, R = ± (\(R_1^2 + R_2^2)^{0.5}\) = ± 15.8 Ω.

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