JEE Physics Alternating Current MCQs Set 09

Answer: (d) \( \left(\frac{C(V_1^2-V_2^2)}{L}\right)^{1/2} \)

Explanation:
1. Energy conservation: \( \frac12CV_1^2=\frac12CV_2^2+\frac12LI^2 \).
2. \( I=\sqrt{\frac{C(V_1^2-V_2^2)}{L}} \)

Teacher's Note:
1. The electrical energy lost by the capacitor appears as magnetic energy in the inductor.

Answer: (d) in parallel with C, with magnitude \( \frac{1-\omega^2LC}{\omega^2L} \)

Explanation:
1. The current leads, so \( X_C\gt X_L \), i.e. \( \omega^2LC\lt 1 \). To cancel the excess capacitive reactance we must reduce \( X_C \), which needs a larger capacitance, so C' is in parallel.
2. Unity power factor: \( \omega^2L(C+C')=1\Rightarrow C'=\frac{1-\omega^2LC}{\omega^2L} \)

Teacher's Note:
1. Parallel capacitors add; resonance needs omega squared times L times total C to equal 1.

Answer: (a) 0

Explanation:
1. The phase difference is 90 degrees, so \( P_{avg}=\frac12E_0I_0\cos\frac\pi2=0 \).
2. Hence x = 0.

Teacher's Note:
1. Average power is zero when voltage and current are 90 degrees apart.

Answer: (b) 250

Explanation:
1. \( I=\frac{100}{1000}=0.1\ \text{A} \); \( X_L=X_C=\frac1{\omega C}=\frac1{200\times2\times10^{-6}}=2500\ \Omega \)
2. \( V_L=IX_L=250\ \text{V} \)

Teacher's Note:
1. At resonance the current is V/R and the inductor voltage is I times its reactance.

Answer: (b) 2

Explanation:
1. \( \frac{I_s}{I_p}=\frac{N_p}{N_s}\Rightarrow I_s=4\times\frac{140}{280}=2\ \text{A} \)

Teacher's Note:
1. For an ideal transformer the currents are inversely proportional to the turns.

Answer: (c) 200

Explanation:
1. The inductor and capacitor voltages are equal and opposite in phase, so they cancel.
2. \( V_R=\sqrt{200^2-(100-100)^2}=200\ \text{V} \)

Teacher's Note:
1. Equal L and C voltages mean resonance, so the whole supply voltage appears across R.

Answer: (b) 60

Explanation:
1. Total resistance \( =20+40=60\ \Omega \); net reactance \( =100-20=80\ \Omega \).
2. \( Z=\sqrt{60^2+80^2}=100\ \Omega,\ P=0.6 \)

Teacher's Note:
1. Power factor equals total resistance divided by impedance.

Answer: (b) 2

Explanation:
1. \( \frac{q^2}{2C}=\frac12\cdot\frac{Q^2}{2C}\Rightarrow q=\frac{Q}{\sqrt2} \), so x = 2.

Teacher's Note:
1. Equal sharing means the electric energy is half the total.

Answer: (b) 8

Explanation:
1. \( P=\frac RZ=\frac{12}{15}=0.8 \), so 10P = 8.

Teacher's Note:
1. Power factor is R over Z.

Answer: (b) 4

Explanation:
1. \( P_{avg}=\frac{I_0^2R}{2}=\frac{I_0^2R}{\sqrt4} \), so x = 4.

Teacher's Note:
1. Mean power is half of the peak power for a sinusoidal current.

Answer: (a) 0

Explanation:
1. The voltages across L and C are 180 degrees out of phase and equal, so \( V_{LC}=|50-50|=0 \).

Teacher's Note:
1. Inductor and capacitor voltages oppose each other in a series circuit.

Answer: (b) 50

Explanation:
1. \( \omega=50\pi\Rightarrow f=25\ \text{Hz} \).
2. The current is zero twice per cycle: \( 2\times25=50 \) times.

Teacher's Note:
1. A sinusoidal current passes through zero twice in each cycle.

Answer: (b) 242

Explanation:
1. \( \tan30^\circ=\frac{X_L}{R}=\frac{X_C}{R}\Rightarrow X_L=X_C \): resonance, so Z = R.
2. \( P=\frac{V^2}{R}=\frac{220^2}{200}=242\ \text{W} \)

Teacher's Note:
1. Equal phase angles mean resonance, where power is V squared over R.

Answer: (b) 0.065

Explanation:
1. \( R=\frac{80}{10}=8\ \Omega,\ Z=\frac{220}{10}=22\ \Omega \).
2. \( X_L=\sqrt{22^2-8^2}=\sqrt{420}\approx20.5\ \Omega\Rightarrow L=\frac{20.5}{100\pi}\approx0.065\ \text{H} \)

Teacher's Note:
1. Find the needed impedance, then the inductive reactance from Z squared minus R squared.