JEE Physics Alternating Current MCQs Set 06

Answer: (b) \( \frac{i_0}{\sqrt3} \)

Explanation:
1. \( i_{rms}^2=\frac1\tau\int_0^\tau\frac{i_0^2t^2}{\tau^2}dt=\frac{i_0^2}{\tau^3}\cdot\frac{\tau^3}{3}=\frac{i_0^2}{3} \).
2. \( i_{rms}=\frac{i_0}{\sqrt3} \).

Teacher's Note:
1. rms = square root of the mean of the square.
2. Integrate t² over the period.

Answer: (a) 1 : 1

Explanation:
1. \( z_1=\sqrt{4^2+3^2}=5\ \Omega \).
2. \( z_2=\sqrt{4^2+(3-6)^2}=5\ \Omega \).
3. Ratio \( 1:1 \).

Teacher's Note:
1. Inductive and capacitive reactances subtract.
2. (3 − 6)² = 9, the same as 3².

Answer: (a) \( 2R,\ R\sqrt2 \)

Explanation:
1. For the L-R circuit \( \tan45^\circ=\frac{X_L}{R}=1\Rightarrow X_L=R \).
2. With the capacitor, the new phase has the same magnitude: \( \frac{X_L-X_C}{R}=-1\Rightarrow X_C=2R \).
3. Impedance \( Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{R^2+R^2}=R\sqrt2 \).

Teacher's Note:
1. tan φ = (X_L − X_C)/R.
2. A capacitive circuit has the current leading the voltage.

Answer: (d) \( \frac{E_0^2}{4R} \)

Explanation:
1. Impedance \( Z=\sqrt{R^2+R^2}=\sqrt2R \).
2. rms current \( I_{rms}=\frac{E_0/\sqrt2}{\sqrt2R}=\frac{E_0}{2R} \).
3. Power \( P=I_{rms}^2R=\frac{E_0^2}{4R} \).

Teacher's Note:
1. Only the resistor dissipates power.
2. Use rms values for average power.

Answer: (c) 60 volt, 100 Hz

Explanation:
1. Use \( 2\sin x\cos x=\sin2x \): \( V=60\sin200\pi t \).
2. Maximum voltage 60 V; \( \omega=200\pi\Rightarrow f=\frac{\omega}{2\pi}=100\ \text{Hz} \).

Teacher's Note:
1. Rewrite the product as a single sine.
2. f = ω/2π.

Answer: (b) 50 W, 0.5

Explanation:
1. Power factor \( \cos\phi=\cos\frac\pi3=0.5 \).
2. \( P=V_{rms}I_{rms}\cos\phi=100\times1\times0.5=50\ \text{W} \).

Teacher's Note:
1. Average power = V_rms I_rms cos φ.
2. cos 60° = 0.5.

Answer: (a) \( 2\ \mu\text{F} \)

Explanation:
1. Capacitive reactance \( X_C=\frac{V_0}{I_0}=\frac{100}{0.2}=500\ \Omega \).
2. \( X_C=\frac1{\omega C}\Rightarrow C=\frac{1}{1000\times500}=2\times10^{-6}\ \text{F}=2\ \mu\text{F} \).

Teacher's Note:
1. For a pure capacitor, V0 = I0 X_C.
2. ω = 1000 rad/s.

Answer: (d) \( 25\sqrt3;\ 25 \)

Explanation:
1. \( R=Z\cos\phi=50\cos30^\circ=25\sqrt3\ \Omega \).
2. \( X=Z\sin\phi=50\sin30^\circ=25\ \Omega \).

Teacher's Note:
1. In the impedance triangle, R = Z cos φ and X = Z sin φ.
2. π/6 is 30°.

Answer: (c) \( \frac{2V_1^2+V_2^2}{2R} \)

Explanation:
1. Heat per second is the average of \( \frac{V^2}{R} \).
2. \( \langle V^2\rangle=V_1^2+2V_1V_2\langle\cos\omega t\rangle+V_2^2\langle\cos^2\omega t\rangle=V_1^2+\frac{V_2^2}{2} \).
3. \( P=\frac{2V_1^2+V_2^2}{2R} \).

Teacher's Note:
1. The average of cos ωt is zero and of cos² ωt is 1/2.
2. Power is the mean of V²/R.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Resonance needs \( \omega L=\frac1{\omega C} \), which is satisfied by a single frequency \( \omega_0=\frac1{\sqrt{LC}} \).
2. So both statements are true and II explains I.

Teacher's Note:
1. X_L increases with frequency while X_C decreases, so they cross only once.
2. At resonance the impedance equals R.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For d.c., \( f=0 \), so \( X_C=\frac1{2\pi fC}\to\infty \).
2. An infinite reactance blocks the steady current, so both statements are true and II explains I.

Teacher's Note:
1. A capacitor passes a.c. but blocks d.c.
2. X_C decreases as the frequency rises.