JEE Physics Alternating Current MCQs Set 08

Answer: (b) \( 5\sqrt{3/2} \) A

Explanation:
1. \( I_0=5\sqrt2\ \text{A},\ \omega=100\pi \), and \( I=I_0\sin\omega t \) measured from the zero.
2. \( \omega t=100\pi\times\frac1{300}=\frac{\pi}{3}\Rightarrow I=5\sqrt2\times\frac{\sqrt3}{2}=5\sqrt{\tfrac32}\ \text{A} \)

Teacher's Note:
1. Measure time from a zero crossing and use I = I0 sin(wt) with I0 = sqrt(2) times the rms value.

Answer: (c) R = 15 ohm, L = 3.5 H, C = 30 microfarad

Explanation:
1. Sharpness is given by \( Q=\frac1R\sqrt{\frac LC} \).
2. Values: (a) 10.4, (b) 9.4, (c) 22.8, (d) 7.3. The largest is (c).

Teacher's Note:
1. A high quality factor means a narrow resonance peak, so a small R and a large L/C are preferred.

Answer: (a) \( 1/\sqrt2 \) A

Explanation:
1. \( I_{rms}=\frac PV=\frac{12}{24}=0.5\ \text{A} \)
2. \( I_0=\sqrt2\times0.5=\frac1{\sqrt2}\ \text{A} \)

Teacher's Note:
1. Rated voltages and powers are rms values; the peak is sqrt(2) times the rms.

Answer: (c) A resistor and a capacitor, or only a capacitor

Explanation:
1. \( Z_{RC}=\sqrt{R^2+1/\omega^2C^2} \) and \( X_C=1/\omega C \) both decrease with frequency, so the current rises.
2. A resistor alone gives a constant current and an inductor adds a reactance that grows with frequency.

Teacher's Note:
1. Capacitive reactance falls with frequency, so circuits where the capacitor dominates carry more current at higher frequency.

Answer: (a) Reduce the energy loss due to eddy currents

Explanation:
1. Thin insulated laminations break up the paths of the eddy currents in the core.
2. This lowers the I squared R heating loss.

Teacher's Note:
1. Lamination cuts down the eddy current path and so the heating loss in the core.

Answer: (d) \( NAB\omega \)

Explanation:
1. \( \phi=NAB\cos\omega t\Rightarrow \varepsilon=-\frac{d\phi}{dt}=NAB\omega\sin\omega t \)
2. The maximum emf is \( NAB\omega \).

Teacher's Note:
1. The peak emf of a rotating coil is N A B times the angular speed.

Answer: (a) (1), (2) and (3) only

Explanation:
1. In a series LCR circuit \( |\phi|\le\pi/2 \), so \( \cos\phi\ge0 \) and P is never negative.
2. P = 0 for a purely reactive circuit (phi = 90 degrees). Hence (1), (2), (3) are correct and (4) is false.

Teacher's Note:
1. Average power in a series LCR circuit can be zero but never negative.

Answer: (c) L/2

Explanation:
1. \( \omega_0=\frac1{\sqrt{LC}} \) stays the same if the product LC is unchanged.
2. With C doubled, L must be halved.

Teacher's Note:
1. Resonant frequency depends only on the product of L and C.

Answer: (d) A is false, but R is true.

Explanation:
1. At resonance \( X_L=X_C \), so \( Z=R \) is minimum and the current is maximum, so A is false.
2. R is true: the phase difference is zero at resonance.

Teacher's Note:
1. At series resonance the impedance is least, hence the current is largest.

Answer: (a) (1) and (4)

Explanation:
1. The average of a sinusoidal current or voltage over a cycle is zero, so (1) is true and (2) is false (220 V is the rms value).
2. A 90 degree phase would mean no power is delivered, so (3) is false; a phase difference below 90 degrees is possible, so (4) is true.

Teacher's Note:
1. Mains quantities average to zero; 220 V is the rms value.

Answer: (c) \( P\left(\frac RZ\right)^2 \)

Explanation:
1. \( P=\frac{V^2}{R} \); with the inductor \( P'=I^2R=\frac{V^2R}{Z^2} \).
2. \( P'=P\,\frac{R^2}{Z^2} \)

Teacher's Note:
1. Power in the series circuit is I squared R with I = V/Z.

Answer: (c) 1

Explanation:
1. \( \tan\frac\pi3=\frac{X_C}{R}=\frac{X_L}{R}\Rightarrow X_L=X_C \).
2. The full circuit is at resonance, so Z = R and the power factor is 1.

Teacher's Note:
1. Equal phase angles from the RC and RL circuits mean the reactances are equal.

Answer: (d) \( f/2\sqrt2 \)

Explanation:
1. \( f=\frac1{2\pi\sqrt{LC}} \); the product LC becomes \( 8LC \).
2. \( f'=\frac{f}{\sqrt8}=\frac{f}{2\sqrt2} \)

Teacher's Note:
1. Frequency varies inversely as the square root of LC.

Answer: (d) \( \frac{R}{(R^2+\omega^2L^2)^{1/2}} \)

Explanation:
1. \( Z=\sqrt{R^2+\omega^2L^2} \), so \( \cos\phi=\frac RZ \).

Teacher's Note:
1. Power factor is R divided by the impedance.