JEE Physics Alternating Current MCQs Set 19

Answer: (d) \( \frac{V_0}{2} \)

Explanation:
1. \( V_{rms}^2 = \frac{1}{1}\int_0^1V_0^2t^3\,dt = \frac{V_0^2}{4} \).
2. \( V_{rms} = \frac{V_0}{2} \).

Teacher's Note:
1. Square, average over the interval, then take the root.
2. Integrate t³ to get t⁴/4.

Answer: (b) 5 A sin(ωt + 53°)

Explanation:
1. \( I_3 = 3\sin\omega t + 4\cos\omega t = 5\sin(\omega t+\phi) \) with \( \tan\phi = \frac43 \).
2. So φ = 53°.

Teacher's Note:
1. Write a sin + b cos as R sin(ωt + φ) with R = √(a² + b²).
2. tan φ = b/a.

Answer: (a) \( \sqrt{\frac{17}{2}} \) A

Explanation:
1. \( I_{rms}^2 = 2^2 + \frac{3^2}{2} = 4+4.5 = \frac{17}{2} \).

Teacher's Note:
1. For a DC plus AC current, the mean squares add.
2. The mean square of sin is ½.

Answer: (d) \( 5.5\times10^{-2} \) H

Explanation:
1. Lamp resistance \( R = 10\ \Omega \). Needed impedance \( Z = \frac{200}{10} = 20\ \Omega \).
2. \( X_L = \sqrt{20^2-10^2} = 17.3\ \Omega \), \( L = \frac{17.3}{2\pi\times50} = 5.5\times10^{-2} \) H.

Teacher's Note:
1. The lamp must still carry 10 A.
2. Z² = R² + X_L².

Answer: (d) 0.35 A

Explanation:
1. X is a resistor, \( R = \frac{220}{0.5} = 440\ \Omega \). Y is a capacitor, \( X_C = 440\ \Omega \).
2. Series: \( Z = 440\sqrt2 \approx 622\ \Omega \), \( I = \frac{220}{622} = 0.35 \) A.

Teacher's Note:
1. Current leading by 90° means a pure capacitor.
2. Series impedance is the vector sum of R and X_C.

Answer: (c) 30 A

Explanation:
1. \( I_{rms}^2 = 10^2+\frac{40^2}{2} = 100+800 = 900 \).
2. So I_rms = 30 A.

Teacher's Note:
1. Mean squares add for DC plus AC.
2. AC part contributes I₀²/2.

Answer: (c) At a very high frequency of the source, the brightness of B₁ is more than that of B₂

Explanation:
1. Capacitor branch: \( X_C = \frac{1}{\omega C} \) falls as ω rises. Inductor branch: \( X_L = \omega L \) rises with ω.
2. At very high frequency B₁ has the larger current and B₂ the smaller, so B₁ is brighter.

Teacher's Note:
1. Capacitors pass high frequencies; inductors pass low frequencies.
2. Brightness follows the current in each branch.

Answer: (d) 75 times

Explanation:
1. \( \omega = 75\pi \Rightarrow f = \frac{\omega}{2\pi} = 37.5 \) Hz.
2. Current becomes zero twice per cycle: 2 × 37.5 = 75 times.

Teacher's Note:
1. A sinusoidal current crosses zero twice per cycle.
2. Frequency is ω/2π.

Answer: (b) zero

Explanation:
1. V_L and V_C are equal in magnitude and opposite in phase.
2. So the voltage across the L–C combination is 50 − 50 = 0.

Teacher's Note:
1. Resonance has the voltages across L and C cancelling.
2. At resonance the supply voltage equals V_R.

Answer: (c) 0 V and 8 A

Explanation:
1. X_L = X_C so the circuit is at resonance: \( Z = R = 30\ \Omega \), \( I = \frac{240}{30} = 8 \) A.
2. The voltage across L and C together is zero.

Teacher's Note:
1. At resonance impedance equals R.
2. Voltmeter across L–C reads 0.

Answer: (d) 3 A

Explanation:
1. \( 50^2 = V_R^2+(100-60)^2 \Rightarrow V_R^2 = 2500-1600 = 900 \), \( V_R = 30 \) V.
2. \( I = \frac{30}{10} = 3 \) A.

Teacher's Note:
1. Use the phasor relation V² = V_R² + (V_C − V_L)².
2. Current = V_R / R.

Answer: (d) either capacitor, resistor and inductor together (with capacitive reactance larger), or only a capacitor and a resistor

Explanation:
1. The current leads the voltage by 45°, so the net behaviour is capacitive with a phase angle of 45°.
2. A series R and C (with X_C = R) gives this, and so does an LCR combination in which X_C exceeds X_L.

Teacher's Note:
1. Leading current means capacitive; lagging current means inductive.
2. A pure capacitor would give a lead of 90°, not 45°.

Answer: (a) zero and zero

Explanation:
1. At f = 0 the capacitor reactance \( \frac{1}{\omega C} \to \infty \), so the current is zero (a capacitor blocks DC).
2. At f → ∞ the inductive reactance \( \omega L \to \infty \), so the current is again zero.

Teacher's Note:
1. Reactances: X_C = 1/ωC, X_L = ωL.
2. Current peaks only at resonance.

Answer: (c) 2W

Explanation:
1. Sine wave: \( W = \frac{V_0^2}{2R} \). Square wave: \( V_{rms} = V_0 \), so \( P = \frac{V_0^2}{R} = 2W \).

Teacher's Note:
1. The rms value of a square wave equals its amplitude.
2. The rms of a sine is amplitude/√2.

Answer: (d) \( \frac{10}{\sqrt2} \) V

Explanation:
1. V_L = V_C, so the source voltage equals V_R = 10 V and X_L = R.
2. After shorting C: \( Z = \sqrt{R^2+X_L^2} = \sqrt2R \), \( I = \frac{10}{\sqrt2R} \), so \( V_L = IX_L = \frac{10}{\sqrt2} \) V.

Teacher's Note:
1. First find the source voltage from the resonance condition.
2. R and X_L are equal, so the voltages divide equally in quadrature.