JEE Physics Alternating Current MCQs Set 23

Answer: (b) 45 A

Explanation:
1. Input power \( =2300\times5=11500 \) W; output power \( =0.9\times11500=10350 \) W.
2. \( I_s=\dfrac{10350}{230}=45 \) A.

Teacher's Note:
1. Compute the reactances or the time constant first, then use the impedance or exponential law.
2. Check the phase relation (current lags in an inductor, leads in a capacitor) against the answer.

Answer: (d) 30

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

Teacher's Note:
1. Compute the reactances or the time constant first, then use the impedance or exponential law.
2. Check the phase relation (current lags in an inductor, leads in a capacitor) against the answer.

Answer: (b) \( V\sqrt{\dfrac CL} \)

Explanation:
1. Energy conservation: \( \tfrac12CV^2=\tfrac12LI_{max}^2 \).
2. \( I_{max}=V\sqrt{\dfrac CL} \).

Teacher's Note:
1. Compute the reactances or the time constant first, then use the impedance or exponential law.
2. Check the phase relation (current lags in an inductor, leads in a capacitor) against the answer.

Answer: (d) \( 10^4\ \text{A/s} \)

Explanation:
1. \( L\dfrac{dI}{dt}=\dfrac QC \).
2. \( \left|\dfrac{dI}{dt}\right|=\dfrac{Q}{LC}=\dfrac{100\times10^{-6}}{2\times10^{-3}\times5\times10^{-6}}=10^4 \) A/s.

Teacher's Note:
1. Compute the reactances or the time constant first, then use the impedance or exponential law.
2. Check the phase relation (current lags in an inductor, leads in a capacitor) against the answer.

Answer: (b) \( 1.732\times10^{-4}\ \text C \)

Explanation:
1. Total energy \( \dfrac{Q_0^2}{2C} \). With \( I=\dfrac{I_{max}}2 \) the magnetic part is one quarter of the total, leaving three quarters in the capacitor.
2. \( \dfrac{Q^2}{2C}=\dfrac34\dfrac{Q_0^2}{2C}\Rightarrow Q=\dfrac{\sqrt3}{2}\times200\ \mu C=1.732\times10^{-4} \) C.

Teacher's Note:
1. Compute the reactances or the time constant first, then use the impedance or exponential law.
2. Check the phase relation (current lags in an inductor, leads in a capacitor) against the answer.

Answer: (c) \( 5.17\times10^2\ \text J \)

Explanation:
1. \( X_L=2\pi(50)(0.02)=6.28\ \Omega \), \( X_C=\dfrac1{2\pi(50)(120\times10^{-6})}=26.5\ \Omega \), \( Z^2=60^2+(20.2)^2=4010\ \Omega^2 \).
2. \( P=\dfrac{V^2R}{Z^2}=\dfrac{576\times60}{4010}=8.62 \) W; energy in 60 s \( =517 \) J.

Teacher's Note:
1. Compute the reactances or the time constant first, then use the impedance or exponential law.
2. Check the phase relation (current lags in an inductor, leads in a capacitor) against the answer.

Answer: (a) RC circuit with \( R=1\ \text k\Omega \) and \( C=10\ \mu \)F

Explanation:
1. A phase angle of \( \pi/4 \) needs \( \tan\phi=1 \), i.e. equal reactance and resistance.
2. RC: \( X_C=\dfrac1{\omega C}=\dfrac1{100\times10^{-5}}=1000\ \Omega=R \), giving \( 45^\circ \). For RL the reactance \( \omega L \) is only 0.1 or 1 \( \Omega \), far below 1 k\( \Omega \).

Teacher's Note:
1. Compute the reactances or the time constant first, then use the impedance or exponential law.
2. Check the phase relation (current lags in an inductor, leads in a capacitor) against the answer.

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

Explanation:
1. Quality factor \( Q=\dfrac{\omega_0}{2\Delta\omega}=\dfrac{\omega_0}{R/L} \).
2. \( Q=\dfrac{\omega_0L}{R}=\dfrac{1}{\omega_0CR} \).

Teacher's Note:
1. Compute the reactances or the time constant first, then use the impedance or exponential law.
2. Check the phase relation (current lags in an inductor, leads in a capacitor) against the answer.

Answer: (a) the bulb glows brighter

Explanation:
1. \( Z=\sqrt{R^2+\dfrac1{\omega^2C^2}} \) decreases as \( \omega \) increases.
2. The current \( V_0/Z \) rises, so the bulb glows brighter.

Teacher's Note:
1. Compute the reactances or the time constant first, then use the impedance or exponential law.
2. Check the phase relation (current lags in an inductor, leads in a capacitor) against the answer.

Answer: (c) (ii) and (iii)

Explanation:
1. The difference of two sinusoids with phase gap \( 2\pi/3 \): \( V_{XY}=2V_0\sin\dfrac\pi3\times\)(a sinusoid), amplitude \( \sqrt3V_0 \).
2. rms value \( =\dfrac{\sqrt3V_0}{\sqrt2}=V_0\sqrt{\dfrac32} \), the same for every pair, since each pair has the same \( 2\pi/3 \) phase gap.

Teacher's Note:
1. Compute the reactances or the time constant first, then use the impedance or exponential law.
2. Check the phase relation (current lags in an inductor, leads in a capacitor) against the answer.

Answer: (a) 4

Explanation:
1. \( Z^2=R^2+X_C^2=1.25R^2\Rightarrow X_C=\dfrac R2 \).
2. \( \dfrac1{\omega C}=\dfrac R2\Rightarrow RC=\dfrac2\omega=\dfrac2{500}=4\times10^{-3} \) s = 4 ms.

Teacher's Note:
1. Compute the reactances or the time constant first, then use the impedance or exponential law.
2. Check the phase relation (current lags in an inductor, leads in a capacitor) against the answer.

Answer: (b) (i) and (ii)

Explanation:
1. At \( \omega\to0 \) the capacitor reactance is huge, so the current is nearly zero (i). The resonant frequency is \( \omega_0=1/\sqrt{LC} \), independent of R (ii).
2. \( \omega_0=\dfrac1{\sqrt{10^{-6}\times10^{-6}}}=10^6 \) rad/s, not \( 10^4 \) (iii false). At \( \omega\gg\omega_0 \) the inductive reactance dominates, so the circuit behaves like an inductor (iv false).

Teacher's Note:
1. Compute the reactances or the time constant first, then use the impedance or exponential law.
2. Check the phase relation (current lags in an inductor, leads in a capacitor) against the answer.