JEE Physics Alternating Current MCQs Set 13

Answer: (d) 25 Hz

Explanation:
1. \( \omega_0=\dfrac1{\sqrt{LC}}=\sqrt{25\pi\times100\pi}=50\pi\ \text{rad/s} \).
2. \( f_0=\omega_0/2\pi=25\ \text{Hz} \).

Teacher's Note:
1. With S open, in-phase current means resonance.
2. f = omega/2 pi.

Answer: (c) 200 V and 10 A

Explanation:
1. \( V_s=1000\times\frac15=200\ \text{V} \).
2. Input power \( =9000/0.9=10000 \) W, so \( I_p=10000/1000=10\ \text{A} \).

Teacher's Note:
1. Voltage scales with the turns ratio.
2. Efficiency is output power over input power.

Answer: (a) 2 and 3 only

Explanation:
1. \( X_C=\dfrac1{2\pi\times50\times5\times10^{-4}}=6.4\ \Omega \) and \( X_L=2\pi\times50\times0.01=3.1\ \Omega \), so branch 2 has the smaller impedance and more current.
2. Raising f lowers XC (B1 brightens) and raises XL (B2 dims).

Teacher's Note:
1. A capacitor offers finite opposition to ac.
2. Compare the two branch impedances with the same bulb resistance.

Answer: (c) 3 and 4 only

Explanation:
1. In phase at resonance: \( \omega_0=1/\sqrt{LC}=10^6 \) rad/s (not \( 10^4 \)), independent of R.
2. At very high frequency XL dominates, so it behaves like an inductor.
3. At omega near 0, \( X_C\to\infty \) and the current tends to zero.

Teacher's Note:
1. Resonance depends only on L and C.
2. At high frequency the inductor dominates; at low frequency the capacitor.

Answer: (b) 5000 W

Explanation:
1. At resonance \( Z=R=2\ \Omega \) (R found from the S-closed condition).
2. \( P=\dfrac{V_{rms}^2}{R}=\dfrac{100^2}{2}=5000\ \text{W} \).

Teacher's Note:
1. At resonance the whole source voltage appears across R.
2. P = V squared over R.

Answer: (c) 0.065 H

Explanation:
1. \( R=80/10=8\ \Omega \); \( Z=220/10=22\ \Omega \).
2. \( X_L=\sqrt{22^2-8^2}=20.5\ \Omega \); \( L=\dfrac{20.5}{2\pi\times50}=0.065\ \text{H} \).

Teacher's Note:
1. The lamp must carry 10 A rms with the same resistance.
2. Impedance is the root of R squared plus XL squared.

Answer: (b) 3

Explanation:
1. Over one period \( V=V_0(2t/T-1) \); \( \langle V^2\rangle=V_0^2\int_0^1(2u-1)^2du=V_0^2/3 \).
2. \( V_{rms}=V_0/\sqrt3 \) for any whole number of periods, so x = 3.

Teacher's Note:
1. Average the square of the function over a period, then take the root.
2. A linear ramp symmetric about zero has mean square V0 squared over 3.

Answer: (b) 50 s

Explanation:
1. \( X_L=2\pi(2000)\frac{0.125}{\pi}=500\ \Omega \), \( X_C=\dfrac{1}{2\pi(2000)(500\times10^{-9}/\pi)}=500\ \Omega \): resonance, so \( I=20/20=1\ \text{A} \).
2. Power \( =I^2R=20\ \text{W} \); heat needed \( 100\times10=1000\ \text{J} \), so \( t=1000/20=50\ \text{s} \).

Teacher's Note:
1. Check first whether the circuit is at resonance.
2. Time equals heat required divided by power.

Answer: (c) decrease exponentially with time constant 0.01 s

Explanation:
1. The box is R = 50 ohm in series with C = 200 microfarad.
2. \( \tau=RC=50\times200\times10^{-6}=0.01\ \text{s} \).
3. Charging an initially uncharged capacitor through R gives a current that starts at E/R and decays exponentially.

Teacher's Note:
1. There is no inductor, so no oscillation.
2. Charging current is largest at the start.

Answer: (c) 75

Explanation:
1. \( \omega_0=1/\sqrt{LC}=10^4 \) rad/s, so \( \omega=5000 \): \( X_L=50\ \Omega \), \( X_C=200\ \Omega \).
2. \( Z=\sqrt{150^2+150^2}=150\sqrt2 \); \( I_{rms}=150/(150\sqrt2)=0.707\ \text{A} \); \( P=I^2R=75\ \text{W} \).

Teacher's Note:
1. First get the resonant angular frequency, then halve it.
2. Power is Irms squared times R.

Answer: (b) 1 and 3 only

Explanation:
1. The difference of two sines 120 degrees apart has amplitude \( 2V_0\sin60^\circ=\sqrt3V_0 \).
2. Its rms value is \( V_0\sqrt{3/2} \) for any pair, since all pairs are 120 degrees apart.

Teacher's Note:
1. Use sinA - sinB = 2 cos((A+B)/2) sin((A-B)/2).
2. The three phases are symmetric.

Answer: (d) 0.08 H

Explanation:
1. With dc only the resistance acts: \( R=12/4=3\ \Omega \).
2. With ac, \( Z=12/2.4=5\ \Omega \), so \( X_L=\sqrt{5^2-3^2}=4\ \Omega \).
3. \( L=X_L/\omega=4/50=0.08\ \text{H} \).

Teacher's Note:
1. Direct current sees only resistance, because an inductor has no opposition to a steady current.
2. Then the extra opposition to ac gives the reactance.

Answer: (b) 0.8

Explanation:
1. \( \cos\phi=R/Z=12/15=0.8 \).

Teacher's Note:
1. Power factor is R over Z.
2. It must be at most 1.

Answer: (a) \( 8\cos(100\pi t-37^\circ) \) volt

Explanation:
1. \( X_L=100\pi\cdot\frac1{100\pi}=1\ \Omega \), \( X_C=\dfrac{500\pi}{100\pi}=5\ \Omega \); net reactance \( -4\ \Omega \) (capacitive); \( |Z|=5\ \Omega \).
2. Current amplitude 2 A, leading the source by \( \tan^{-1}(4/3)=53^\circ \): \( i=2\cos(100\pi t+53^\circ) \).
3. Across L and C the amplitude is 2 x 4 = 8 V and the voltage lags the current by 90 degrees: \( 8\cos(100\pi t+53^\circ-90^\circ)=8\cos(100\pi t-37^\circ) \).

Teacher's Note:
1. Find the current first, then multiply by the reactance of the part of the circuit asked about.
2. A net capacitive reactance makes the voltage across it lag the current by 90 degrees.

Answer: (d) \( \dfrac{\omega_0L}{R} \)

Explanation:
1. \( Q=\dfrac{\omega_0}{\Delta\omega}=\dfrac{\omega_0L}{R}=\dfrac1{\omega_0CR} \).

Teacher's Note:
1. Q measures the sharpness of resonance.
2. Larger L or smaller R gives sharper resonance.

Answer: (b) \( 100\ \mu\text{J} \)

Explanation:
1. Power at resonance is 100 W (previous result).
2. Energy \( =100\times10^{-6}=100\ \mu\text{J} \).

Teacher's Note:
1. Heat equals average power times time.
2. Time is given in microseconds.

Answer: (b) 2 and 4 only

Explanation:
1. Current \( I=P/V \) is 100 times smaller at 22000 V.
2. Line loss \( I^2R \) is therefore 10^4 times smaller.

Teacher's Note:
1. High voltage means low current for the same power.
2. Line loss depends on the square of the current.

Answer: (a) 100 W

Explanation:
1. At resonance \( Z=R=100\ \Omega \).
2. \( P=\dfrac{V^2}{R}=\dfrac{100^2}{100}=100\ \text{W} \).

Teacher's Note:
1. At resonance the circuit is purely resistive.
2. P = Vrms squared over R.

Answer: (c) 1, 2 and 3 only

Explanation:
1. \( I_0=\dfrac{10}{\omega L} \): at 100, \( X_L=0.5 \) so 20 A; at 500, \( X_L=2.5 \) so 4 A; at 1000, \( X_L=5 \) so 2 A.

Teacher's Note:
1. Peak current is peak emf divided by reactance.
2. Doubling frequency halves the current in an inductor.

Answer: (c) The frequency must be increased and the current amplitude becomes 10 A

Explanation:
1. A lagging current means the circuit is inductive, so \( \omega>\omega_0 \): the frequency must be increased.
2. At a 45 degree phase angle \( Z=\sqrt2R \), so the amplitude falls by \( \sqrt2 \): \( 10\sqrt2/\sqrt2=10\ \text{A} \).

Teacher's Note:
1. Above resonance the inductive reactance dominates and current lags.
2. At 45 degrees the net reactance equals R.

Answer: (a) \( R=50\ \Omega,\ C=200\ \mu\text{F} \)

Explanation:
1. \( |Z|=\dfrac{100}{\sqrt2}=50\sqrt2\ \Omega \) and \( \tan45^\circ=X_C/R \) give \( R=X_C=50\ \Omega \).
2. \( C=\dfrac{1}{\omega X_C}=\dfrac1{100\times50}=200\ \mu\text{F} \).

Teacher's Note:
1. With 45 degrees, R and XC are equal and Z = R sqrt2.
2. C = 1/(omega XC).

Answer: (d) C3 is removed from the circuit

Explanation:
1. Resonant frequency is \( 1/\sqrt{LC} \), independent of R.
2. Removing C3 reduces the total C and so raises \( \omega_0 \) for certain, while the other combinations change L and C in opposite senses and need not raise the product's reciprocal.

Teacher's Note:
1. Resonant frequency does not depend on resistance.
2. Reducing L or C without raising the other must raise the resonant frequency.

Answer: (b) 8

Explanation:
1. At resonance \( I=60/120=0.5 \) A and \( X_L=40/0.5=80\ \Omega \), so \( L=2\times10^{-4} \) H and \( 1/C=\omega_0^2L=3.2\times10^{7} \).
2. A 45 degree lag needs \( \omega L-\dfrac1{\omega C}=R=120 \): \( \omega^2-6\times10^5\omega-1.6\times10^{11}=0 \).
3. Positive root \( \omega=\dfrac{6\times10^5+10^6}{2}=8\times10^5 \) rad/s.

Teacher's Note:
1. A 45 degree phase angle means net reactance equals resistance.
2. Use the resonance data to get L and C.