Alternating Current — Top 20 Questions

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Step-down transformer: input 2.3 kV, primary current 5 A, efficiency 90%, output 230 V. Output current (A):

A50

B45

C40

D55

✅ Correct: B

Output power \(= 0.9\times V_1 I_1 = 0.9\times2300\times5 = 10350\) W. Output current \(= P_{out}/V_2 = 10350/230 = \mathbf{45\text{ A}}\).

\(I = I_A\sin\omega t + I_B\cos\omega t\). RMS current:

A\(\sqrt{I_A^2+I_B^2}/2\)

B\(\sqrt{(I_A^2+I_B^2)/2}\)

C\(|I_A+I_B|/\sqrt{2}\)

D\(\sqrt{I_A^2+I_B^2}\)

✅ Correct: B

Peak current \(I_0 = \sqrt{I_A^2+I_B^2}\). RMS \(= I_0/\sqrt{2} = \sqrt{(I_A^2+I_B^2)/2}\). \(\mathbf{\sqrt{(I_A^2+I_B^2)/2}}\).

Coil: \(L = 2\) H, \(R = 4\Omega\), 10 V DC applied. Energy stored in steady state (J):

A3.125

B6.25

C12.5

D1.5

✅ Correct: B

Steady state: inductor acts as wire. \(I = V/R = 10/4 = 2.5\) A. Energy \(= \frac{1}{2}LI^2 = \frac{1}{2}\times2\times6.25 = \mathbf{6.25\text{ J}}\).

\(V = 260\sin(628t)\) applied to inductor \(L = 5\) mH. Inductive reactance (\(\Omega\)):

A6.28

B0.318

C0.5

D3.14

✅ Correct: D

\(\omega = 628\text{ rad/s}\). \(X_L = \omega L = 628\times5\times10^{-3} = \mathbf{3.14\Omega}\).

LCR series: \(L = 0.5\) mH, \(C = 20\) μF, \(R = 20\Omega\). At resonance, amplitude of current is \(\sqrt{x}\) A if \(V_0 = 220\) V. Find \(x\):

A100

B121

C242

D484

✅ Correct: B

At resonance: \(I_0 = V_0/R = 220/20 = 11\text{ A}\). \(I_0 = \sqrt{121}\) → \(x = \mathbf{121}\).

Capacitor 200 pF at 230 V, \(\omega = 300\) rad/s. RMS conduction and displacement current:

A14.3 μA and 143 μA

B13.8 μA and 13.8 μA

C13.8 μA and 138 μA

D1.38 μA and 13.8 μA

✅ Correct: B

\(X_C = 1/\omega C = 1/(300\times200\times10^{-12}) \approx 1.67\times10^7\Omega\). \(I_{rms} = V_{rms}/X_C = 230/1.67\times10^7 = 13.8\mu A\). Displacement current = conduction current = \(\mathbf{13.8\mu A}\).

\(V = 100\sin(100t)\) V, \(I = 100\sin(100t+\pi/3)\) mA. Average power (W):

A2.5 W

B5 W

C10 W

D1.25 W

✅ Correct: A

\(P = \frac{1}{2}V_0I_0\cos\phi = \frac{1}{2}\times100\times0.1\times\cos(\pi/3) = 5\times0.5 = \mathbf{2.5\text{ W}}\).

In purely resistive AC circuit, phase difference between current and voltage is:

A\(\pi/2\)

B\(\pi\)

D\(\pi/4\)

✅ Correct: C

In purely resistive circuit: \(V = IR\), both in phase. Phase difference \(= \mathbf{0}\). In purely inductive: current lags by \(\pi/2\). In purely capacitive: current leads by \(\pi/2\).

Quality factor \(Q\) of series LCR circuit:

A\(\omega_0 L/R\)

B\(R/\omega_0 L\)

C\(\omega_0 RC\)

D\(\omega_0/R\)

✅ Correct: A

\(Q = \omega_0 L/R = 1/(\omega_0 CR) = \sqrt{L/C}/R\). Higher \(Q\) → sharper resonance peak, better frequency selectivity. \(\mathbf{\omega_0 L/R}\).

Q10

Transformer turns ratio \(N_1:N_2 = 1:10\). Primary voltage 230 V. Secondary voltage:

A23 V

B2300 V

C230 V

D460 V

✅ Correct: B

\(V_2/V_1 = N_2/N_1 = 10/1\). \(V_2 = 230\times10 = \mathbf{2300\text{ V}}\). This is a step-up transformer.

Q11

Power factor of pure inductor:

C0.5

D\(1/\sqrt{2}\)

✅ Correct: B

Pure inductor: phase difference \(\phi = 90°\). Power factor \(= \cos90° = \mathbf{0}\). No power is dissipated in pure inductor — energy alternately stored and released.

Q12

Resonant frequency of series LCR circuit (\(L = 4\) mH, \(C = 4\) μF):

A2500 Hz

B5000 Hz

C250 Hz

D1250 Hz

✅ Correct: A

\(f_0 = \dfrac{1}{2\pi\sqrt{LC}} = \dfrac{1}{2\pi\sqrt{4\times10^{-3}\times4\times10^{-6}}} = \dfrac{1}{2\pi\times4\times10^{-4}} = \dfrac{10^4}{8\pi} \approx \mathbf{\frac{10000}{25.1} \approx 2500\text{ Hz}}\).

Q13

In AC circuit, wattless current means:

ACurrent in phase with voltage

BCurrent 90° out of phase with voltage

CCurrent with maximum power

DDC component of current

✅ Correct: B

Wattless current (idle current) is the component of current 90° out of phase with voltage. It contributes zero average power (\(P = VI\cos90° = 0\)) but causes reactive power and VA loading.

Q14

Impedance of series LCR at resonance:

B\(\sqrt{R^2+(X_L-X_C)^2}\)

D\(X_L + X_C\)

✅ Correct: C

At resonance: \(X_L = X_C\). Impedance \(Z = \sqrt{R^2+(X_L-X_C)^2} = \sqrt{R^2+0} = \mathbf{R}\). Minimum impedance, maximum current.

Q15

Peak value of AC voltage if RMS value = 220 V:

A220 V

B311 V

C156 V

D440 V

✅ Correct: B

\(V_0 = V_{rms}\times\sqrt{2} = 220\times1.414 = \mathbf{311\text{ V}}\). Standard mains supply: 220 V RMS ≈ 311 V peak.

Q16

Mutual inductance of two coils is 1 H. Rate of change of current in primary = 2 A/s. EMF in secondary:

A0.5 V

B1 V

C2 V

D4 V

✅ Correct: C

\(\varepsilon_2 = -M\dfrac{dI_1}{dt} = -1\times2 = -2\text{ V}\). Magnitude \(= \mathbf{2\text{ V}}\). Negative sign indicates opposition (Lenz's law).

Q17

Self-inductance of a coil is 2 H. Current changes from 2 A to 4 A in 0.5 s. Induced EMF:

A4 V

B8 V

C2 V

D16 V

✅ Correct: B

\(|\varepsilon| = L|dI/dt| = 2\times(4-2)/0.5 = 2\times4 = \mathbf{8\text{ V}}\).

Q18

For maximum power transfer in AC circuit, condition is:

A\(X_L = X_C\) and \(Z = 0\)

B\(X_L = X_C\) and \(Z = R\)

C\(Z = \infty\)

D\(R = 0\)

✅ Correct: B

Maximum power at resonance: \(X_L = X_C\) so impedance \(Z = R\) (purely resistive). All power goes to \(R\), power factor \(= 1\). \(P_{max} = V_{rms}^2/R\). \(\mathbf{X_L = X_C,\; Z = R}\).

Q19

Energy density in magnetic field \(B\):

A\(B^2/2\mu_0\)

B\(B/2\mu_0\)

C\(B^2\mu_0/2\)

D\(\mu_0/2B^2\)

✅ Correct: A

Energy density \(u_B = \dfrac{B^2}{2\mu_0}\) (J/m³). Compare with electric field energy density: \(u_E = \dfrac{\varepsilon_0 E^2}{2}\). Both are important for EM wave energy. \(\mathbf{B^2/2\mu_0}\).

Q20

A choke coil is used in AC circuits instead of resistor because:

AIt allows more current

BIt dissipates less power

CIt increases frequency

DIt reduces inductance

✅ Correct: B

Choke coil (inductor) has high reactance but low resistance → dissipates very little power (\(P = I^2R \approx 0\)) while reducing current. A resistor would waste power as heat. Used in tube lights and fans.

Roshan

Expert · 5 Years Experience

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