JEE Physics Alternating Current MCQs Set 18

Answer: (b) 242 W

Explanation:
1. Without C: \( \tan30^\circ=\dfrac{X_L}{R} \). Without L: \( \tan30^\circ=\dfrac{X_C}{R} \). So \( X_L=X_C \): the circuit is tuned (X_L=X_C) and \( Z=R \).
2. Power \( P=\dfrac{V^2}{R}=\dfrac{220^2}{200}=242 \) W.

Teacher's Note:
1. Equal reactances mean the circuit is tuned, so the impedance is purely resistive.
2. At the tuned condition the power factor is 1.

Answer: (d) \( \dfrac{V}{R_2} \) at t = 0 and \( \dfrac{V(R_1+R_2)}{R_1R_2} \) at t = infinity

Explanation:
1. At t = 0 the inductor opposes the current, so the branch with L carries none; the battery current is \( V/R_2 \).
2. After a long time the inductor has no voltage across it, so the two resistors are in parallel: \( I=\dfrac V{R_1}+\dfrac V{R_2}=\dfrac{V(R_1+R_2)}{R_1R_2} \).

Teacher's Note:
1. An inductor acts as an open circuit at the instant of switching and as a plain wire in steady state.
2. Parallel branches add their currents.

Answer: (a) -1

Explanation:
1. After the switch the loop has no battery, so the two voltage drops add to zero at all times: \( V_R+V_L=0 \).
2. Hence \( \dfrac{V_R}{V_L}=-1 \) at every instant, including \( t=L/R \).

Teacher's Note:
1. In a source-free series loop, the sum of voltage drops around the loop is zero.
2. The ratio does not depend on time.

Answer: (c) 0.67 mA

Explanation:
1. Steady current \( I_0=\dfrac{15}{150}=0.1 \) A and \( \tau=\dfrac LR=\dfrac{0.03}{150}=0.2 \) ms.
2. At \( t=1 \) ms \( =5\tau \): \( I=I_0e^{-5}=\dfrac{0.1}{150}=6.7\times10^{-4} \) A = 0.67 mA.

Teacher's Note:
1. Time constant of an L-R circuit is \( L/R \).
2. Convert kilohm to ohm.

Answer: (b) 0.065 H

Explanation:
1. Lamp resistance \( R=\dfrac{80}{10}=8 \) ohm. At 10 A and 220 V the impedance must be \( Z=\dfrac{220}{10}=22 \) ohm.
2. Reactance \( X_L=\sqrt{22^2-8^2}=\sqrt{420}=20.5 \) ohm.
3. Inductance \( L=\dfrac{X_L}{2\pi f}=\dfrac{20.5}{314}=0.065 \) H.

Teacher's Note:
1. Use the impedance triangle: \( Z^2=R^2+X_L^2 \).
2. The lamp resistance is found from the DC rating.

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

Explanation:
1. Quality factor is the ratio of the reactance at the tuned frequency to the resistance: \( Q=\dfrac{\omega_0L}{R}=\dfrac{1}{\omega_0CR} \).

Teacher's Note:
1. A high Q means a sharp response curve.
2. At this frequency \( \omega_0L=\dfrac1{\omega_0C} \).

Answer: (b) \( \dfrac{1000}{\sqrt2} \) W, 10 A

Explanation:
1. Average power \( P=\dfrac{E_0I_0}{2}\cos\phi=\dfrac{100\times20}{2}\cos45^\circ=\dfrac{1000}{\sqrt2} \) W.
2. Wattless current \( I_{rms}\sin\phi=\dfrac{20}{\sqrt2}\times\dfrac1{\sqrt2}=10 \) A.

Teacher's Note:
1. Power factor is the cosine of the phase angle between emf and current.
2. The wattless component is the one that multiplies \( \sin\phi \).

Answer: (a) After a time 2\( \tau \), the charge on the capacitor is \( CV(1-e^{-2}) \)

Explanation:
1. Charging: \( q=CV(1-e^{-t/\tau}) \). At \( t=2\tau \): \( q=CV(1-e^{-2}) \).
2. At \( t=\tau \) the charge is \( CV(1-e^{-1})\approx0.63CV \), not CV/2.
3. Work done by the source \( =CV^2 \) while heat dissipated \( =\tfrac12CV^2 \), so the work is twice the heat.

Teacher's Note:
1. Substitute the time in units of the time constant.
2. Source work in charging is twice the energy stored.

Answer: (c) At some instants during the oscillation the instantaneous current in the circuit can be \( V\sqrt{C/L} \)

Explanation:
1. At t = 0 all energy is electrical (charged capacitor), so the first statement is wrong; the current reverses periodically, and energy is exchanged continuously.
2. Total energy \( \tfrac12CV^2=\tfrac12LI_{max}^2 \) gives \( I_{max}=V\sqrt{C/L} \).

Teacher's Note:
1. In an LC circuit energy moves back and forth between electric and magnetic forms.
2. Maximum current occurs when the capacitor is momentarily uncharged.

Answer: (b) \( Q=-LC\dfrac{d^2Q}{dt^2} \)

Explanation:
1. Loop law: \( L\dfrac{di}{dt}+\dfrac QC=0 \) with \( i=\dfrac{dQ}{dt} \).
2. So \( \dfrac{d^2Q}{dt^2}=-\dfrac{Q}{LC} \) i.e. \( Q=-LC\dfrac{d^2Q}{dt^2} \).

Teacher's Note:
1. Kirchhoff's loop rule gives the differential equation of simple harmonic motion.
2. Angular frequency is \( 1/\sqrt{LC} \).

Answer: (a) the bulb glows brighter

Explanation:
1. Capacitive reactance \( X_C=\dfrac1{\omega C} \) decreases as \( \omega \) increases.
2. Impedance \( Z=\sqrt{R^2+X_C^2} \) decreases, the current increases and the bulb glows brighter.

Teacher's Note:
1. Capacitors pass high frequencies more easily.
2. Brightness depends on the rms current.

Answer: (c) (2) and (3) only

Explanation:
1. With the dielectric the capacitance is four times larger, so \( X_C=\dfrac{1}{\omega C} \) is smaller and the impedance is smaller. The current is larger in case B: \( I_R^A<I_R^B \).
2. The voltage across the capacitor is \( V_C=\dfrac{VX_C}{\sqrt{R^2+X_C^2}} \), which increases with \( X_C \). Case A has the larger \( X_C \), so \( V_C^A>V_C^B \).

Teacher's Note:
1. A larger capacitance means a smaller reactance.
2. The share of the source voltage across C grows with its reactance.

Answer: (d) 4

Explanation:
1. \( Z^2=R^2+X_C^2=1.25R^2 \), so \( X_C=\dfrac R2 \).
2. Then \( \dfrac{1}{\omega C}=\dfrac R2 \), \( RC=\dfrac2\omega=\dfrac{2}{500}=4\times10^{-3} \) s = 4 ms.

Teacher's Note:
1. Find the reactance from the impedance.
2. Time constant of an RC circuit is RC.

Answer: (a) 30

Explanation:
1. Current \( I=\dfrac{P}{V}=\dfrac{600\times10^3}{4000}=150 \) A. Cable resistance \( R=0.4\times20=8 \) ohm.
2. Loss \( I^2R=150^2\times8=180 \) kW, which is \( \dfrac{180}{600}=30\% \).

Teacher's Note:
1. Power loss in a cable is \( I^2R \).
2. Find the current from power over voltage.

Answer: (c) 200 : 1

Explanation:
1. Step-up output: \( 4000\times10=40000 \) V.
2. Step-down transformer: primary voltage 40000 V and secondary 200 V, so \( \dfrac{N_p}{N_s}=\dfrac{40000}{200}=200 \).

Teacher's Note:
1. Turn ratio equals voltage ratio for an ideal transformer.
2. The cable drop is small (about 120 V of 40 kV).

Answer: (d) (3) and (4) only

Explanation:
1. Tuned angular frequency \( \omega_0=\dfrac1{\sqrt{LC}}=\dfrac{1}{\sqrt{10^{-6}\times10^{-6}}}=10^6 \) rad/s, so (1) is false; and the tuned frequency does not depend on R, so (3) is true.
2. For very large \( \omega \) the inductive reactance dominates, so the circuit behaves like an inductor, not a capacitor: (2) false.
3. At \( \omega\to0 \) the capacitor reactance tends to infinity and the current is nearly zero: (4) true.

Teacher's Note:
1. Current and voltage are in phase in the tuned condition, where \( X_L=X_C \).
2. A capacitor blocks DC.

Answer: (b) (1) and (4) only

Explanation:
1. The phasors are 120 degrees apart, so the difference has amplitude \( 2V_0\sin60^\circ=\sqrt3V_0 \).
2. RMS value \( =\dfrac{\sqrt3V_0}{\sqrt2}=V_0\sqrt{\dfrac32} \) for every pair of terminals. So (1) and (4) are true; (2) and (3) are false.

Teacher's Note:
1. Three-phase voltages are symmetric.
2. Difference of two equal phasors 120 degrees apart has magnitude \( \sqrt3 \) times each.

Answer: (a) 2

Explanation:
1. Acceleration \( a=\dfrac{qE_0}{m}\sin\omega t \).
2. Velocity from rest: \( v=\dfrac{qE_0}{m\omega}(1-\cos\omega t) \).
3. Maximum when \( \cos\omega t=-1 \): \( v_{max}=\dfrac{2qE_0}{m\omega}=\dfrac{2\times1}{10^{-3}\times10^3}=2 \) m/s.

Teacher's Note:
1. Integrate the acceleration with the initial condition v = 0.
2. The maximum of \( 1-\cos\omega t \) is 2.