JEE Physics Alternating Current MCQs Set 05

Answer: (d) 0.80

Explanation:
1. Net reactance \( X = X_L - X_C = 31 - 25 = 6\ \Omega \).
2. Impedance \( Z = \sqrt{R^2 + X^2} = \sqrt{64 + 36} = 10\ \Omega \).
3. Power factor \( \cos\phi = R/Z = 8/10 = 0.80 \).

Teacher's Note:
1. Power factor is always resistance divided by impedance in a series circuit.
2. The source voltage is not needed for this part.

Answer: (d) \( 400\sqrt{2} \) V

Explanation:
1. The rms source voltage is \( 500 \) V. Since \( E_{rms}^2 = V_R^2 + (V_L - V_C)^2 \), \( (V_L - V_C)^2 = 500^2 - 400^2 = 300^2 \).
2. So \( |V_L - V_C| = 300 \) V, and with \( V_L = 700 \) V the capacitor rms voltage is 400 V (the other root, 1000 V, would give a peak value that is not among the options).
3. Peak voltage across the capacitor \( = 400\sqrt{2} \) V.

Teacher's Note:
1. The voltages across L and C are in opposite phase, so only their difference combines with the resistor voltage.
2. Always convert between rms and peak values with a factor of \( \sqrt{2} \).

Answer: (c) \( 1.1\times10^{3} \) rad s\( ^{-1} \)

Explanation:
1. The charge oscillates with \( \omega = \frac{1}{\sqrt{LC}} \).
2. \( LC = 27\times10^{-3}\times30\times10^{-6} = 8.1\times10^{-7} \), so \( \sqrt{LC} = 9\times10^{-4} \) s.
3. \( \omega = 1/(9\times10^{-4}) \approx 1.1\times10^{3} \) rad s\( ^{-1} \).

Teacher's Note:
1. This is the natural frequency of an LC circuit.
2. Take care with the prefixes milli and micro.

Answer: (c) R

Explanation:
1. For an R-L series circuit \( \tan\phi = \frac{X_L}{R} \).
2. With \( \phi = \frac{\pi}{4} \), \( \tan\phi = 1 \), so \( X_L = R \).

Teacher's Note:
1. A phase angle of 45 degrees means resistance and reactance are equal.
2. The same idea works for an R-C circuit.

Answer: (d) 314 V

Explanation:
1. Voltage across an inductor is \( V_L = L\frac{dI}{dt} \).
2. \( \frac{dI}{dt} = 20\times100\pi\cos\left(100\pi t + \frac{\pi}{3}\right) \).
3. At t = 0, \( V_L = 0.1\times2000\pi\times\cos60^\circ = 0.1\times2000\pi\times0.5 = 100\pi \approx 314 \) V.

Teacher's Note:
1. Differentiate the current first, then substitute the instant.
2. The inductor voltage leads the current by 90 degrees, which is why a cosine appears.

Answer: (a) 2.5 ms

Explanation:
1. \( I = I_0\sin\omega t \) and \( 0.707 = \sin\omega t \), so \( \omega t = \frac{\pi}{4} \).
2. Time period \( T = 1/50 = 20 \) ms, and \( \omega t = \frac{\pi}{4} \) corresponds to \( t = T/8 \).
3. \( t = 20/8 = 2.5 \) ms.

Teacher's Note:
1. 70.7 percent of the peak is the rms value, reached at 45 degrees.
2. One-eighth of a cycle is 45 degrees.

Answer: (a) 2.5 mJ

Explanation:
1. The inductor branch (1 \( \Omega \) in series with L) is directly across the 2 V battery, so \( 2 = i\times1 + e_L \) where \( e_L \) is the induced emf.
2. The emf is maximum (2 V) at the instant of closing, when i = 0. At 50% of maximum, \( e_L = 1 \) V, so \( i\times1 = 1 \) V and i = 1 A.
3. Energy \( U = \frac{1}{2}Li^2 = \frac{1}{2}\times5\times10^{-3}\times1^2 = 2.5 \) mJ.

Teacher's Note:
1. The 2 ohm resistor is directly across the battery and does not affect the inductor branch.
2. Induced emf is largest when the current is changing fastest, that is, at the start.

Answer: (c) about 310 V

Explanation:
1. The quoted 220 V is the rms value.
2. \( V_0 = \sqrt{2}\,V_{rms} = 1.414\times220 \approx 311 \) V.

Teacher's Note:
1. Household voltage ratings are rms values.
2. Peak value is about 1.41 times rms.

Answer: (c) 15 V

Explanation:
1. Move from A to B along the current: \( V_B = V_A - IR + \varepsilon - L\frac{dI}{dt} \), where the battery is crossed from its negative to its positive terminal.
2. \( IR = 5\times1 = 5 \) V, \( \varepsilon = 15 \) V, and \( L\frac{dI}{dt} = 5\times10^{-3}\times(-10^{3}) = -5 \) V.
3. \( V_B - V_A = -5 + 15 + 5 = 15 \) V.

Teacher's Note:
1. A falling current makes the inductor push to keep the current going, so it acts like a source here.
2. Fix the direction of travel first and then add each element with its correct sign.

Answer: (b) 503 kHz

Explanation:
1. \( f_0 = \frac{1}{2\pi\sqrt{LC}} \) with \( L = 200\times10^{-6} \) H and \( C = 5\times10^{-10} \) F.
2. \( LC = 10^{-13} \), \( \sqrt{LC} = 3.162\times10^{-7} \) s.
3. \( f_0 = \frac{1}{2\pi\times3.162\times10^{-7}} \approx 5.03\times10^{5} \) Hz = 503 kHz.

Teacher's Note:
1. Resistance does not affect the resonant frequency.
2. Convert micro-henry and micro-farad to SI units before substituting.

Answer: (a) effective resistance due to capacitor

Explanation:
1. Capacitive reactance \( X_C = \frac{1}{\omega C} \) is measured in ohms.
2. It is the opposition a capacitor offers to alternating current, so it is the effective resistance due to the capacitor.

Teacher's Note:
1. Reactance and resistance share the same unit, the ohm.
2. Unlike resistance, reactance does not dissipate energy.

Answer: (c) zero

Explanation:
1. In a pure capacitor the current leads the voltage by 90 degrees.
2. Average power \( P = VI\cos90^\circ = 0 \).
3. Energy is stored in one quarter cycle and returned in the next, so there is no net dissipation.

Teacher's Note:
1. An ideal capacitor or inductor draws wattless current.
2. The expressions with \( CV^2 \) give stored energy, not power.

Answer: (a) 8 V

Explanation:
1. \( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{9 + 16} = 5\ \Omega \).
2. Current \( I = 10/5 = 2 \) A.
3. Voltage across L and C together is \( I(X_L - X_C) = 2\times4 = 8 \) V.

Teacher's Note:
1. The combined L and C act as a single reactance equal to the difference of the two.
2. Check with \( \sqrt{6^2 + 8^2} = 10 \) V.

Answer: (a) \( \tan^{-1}(\tan\phi_1 - \tan\phi_2) \)

Explanation:
1. With R and L, \( \tan\phi_1 = \frac{X_L}{R} \). With R and C, taking the angle as a magnitude, \( \tan\phi_2 = \frac{X_C}{R} \).
2. For all three in series, \( \tan\phi = \frac{X_L - X_C}{R} = \tan\phi_1 - \tan\phi_2 \).
3. So \( \phi = \tan^{-1}(\tan\phi_1 - \tan\phi_2) \).

Teacher's Note:
1. Tangent of the phase angle is net reactance divided by resistance.
2. Reactances of L and C subtract because they act in opposite directions.

Answer: (b) 2 A, 0 V

Explanation:
1. In the L-C branch \( X_L = X_C = 2\ \Omega \), so its net impedance is zero (series resonance).
2. The voltmeter is across this branch, so it reads 0 V, and the whole 110 V falls across the resistor.
3. Current \( I = \frac{110}{55} = 2 \) A.

Teacher's Note:
1. At resonance the L-C combination acts like a short circuit for the rest of the loop.
2. The ammeter reads the rms current of the series loop.

Answer: (c) \( \frac{10}{\sqrt{2}} \) V

Explanation:
1. Since \( V_L = V_C \), they cancel and the source voltage equals \( V_R = 10 \) V. Also \( X_L = X_C = R \) because the voltages are equal for the same current.
2. After removing C, \( Z = \sqrt{R^2 + X_L^2} = \sqrt{2}R \), so \( I = \frac{10}{\sqrt{2}R} \).
3. \( V_L = IX_L = \frac{10}{\sqrt{2}R}\times R = \frac{10}{\sqrt{2}} \) V.

Teacher's Note:
1. Equal voltages across R, L and C in the same current mean equal resistance and reactances.
2. The source voltage is the phasor sum, not the arithmetic sum.

Answer: (d) \( \frac{V_0^2}{10R} \)

Explanation:
1. Net reactance \( X = 3R - R = 2R \), so \( Z = \sqrt{R^2 + 4R^2} = \sqrt{5}R \).
2. \( I_{rms} = \frac{V_0}{\sqrt{2}\sqrt{5}R} \).
3. \( P = I_{rms}^2 R = \frac{V_0^2}{10R^2}R = \frac{V_0^2}{10R} \).

Teacher's Note:
1. Power uses the rms current, so remember the factor of 2 from the peak value.
2. Only the resistor dissipates energy.

Answer: (b) 1.7 A

Explanation:
1. \( X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi\times60\times30\times10^{-6}} \approx 88.4\ \Omega \).
2. \( I_{rms} = \frac{V_{rms}}{X_C} = \frac{150}{88.4} \approx 1.7 \) A.

Teacher's Note:
1. Capacitive reactance falls as frequency or capacitance rises.
2. Always check that the answer has a sensible size: a few amperes for a household supply.

Answer: (d) Current, EMF and phase difference

Explanation:
1. Average power \( P = E_{rms}I_{rms}\cos\phi \).
2. So it depends on the emf, the current and the phase difference between them.

Teacher's Note:
1. The factor \( \cos\phi \) is the power factor.
2. Current and emf alone do not tell the whole story.

Answer: (b) f = 250\( \pi \) Hz

Explanation:
1. Equal readings mean \( X_L = X_C \), which is resonance: \( \omega^2 = \frac{1}{LC} = \frac{\pi^2}{16\times10^{-6}} \), so \( \omega = 250\pi \) rad/s.
2. Then \( f = \frac{\omega}{2\pi} = 125 \) Hz, so statement (a) is true and statement (b), which writes 250\( \pi \) as a frequency in hertz, is wrong (250\( \pi \) is the angular frequency in rad/s).
3. At resonance \( I = \frac{200}{100} = 2 \) A and \( X_L = \omega L = 250\pi\times\frac{2}{\pi} = 500\ \Omega \), so \( V_1 = V_2 = 2\times500 = 1000 \) V, which makes (c) and (d) true.

Teacher's Note:
1. Do not mix up angular frequency and frequency.
2. At resonance voltages across L and C can be much larger than the source voltage.

Answer: (b) \( \sqrt{2}\sin\left(100t - \frac{\pi}{4}\right) \)

Explanation:
1. \( \omega = 100 \) rad/s, so \( X_L = \omega L = 100\ \Omega \) and \( Z = \sqrt{100^2 + 100^2} = 100\sqrt{2}\ \Omega \).
2. Peak current \( I_0 = \frac{200}{100\sqrt{2}} = \sqrt{2} \) A.
3. \( \tan\phi = \frac{X_L}{R} = 1 \), so the current lags the voltage by \( \frac{\pi}{4} \): \( I = \sqrt{2}\sin\left(100t - \frac{\pi}{4}\right) \) A.

Teacher's Note:
1. The current in an inductive circuit lags the voltage.
2. The angular frequency inside the sine does not change.

Answer: (d) \( \frac{\varepsilon^2 R}{R^2 + \left(L\omega - \frac{1}{C\omega}\right)^2} \)

Explanation:
1. \( P = I_{rms}^2 R \) with \( I_{rms} = \frac{\varepsilon}{Z} \).
2. So \( P = \frac{\varepsilon^2 R}{Z^2} \).
3. With \( Z^2 = R^2 + \left(L\omega - \frac{1}{C\omega}\right)^2 \) this is option (d).

Teacher's Note:
1. Check units: \( \varepsilon^2/R \) is power, and the ratio \( R^2/Z^2 \) is dimensionless.
2. Power is maximum at resonance, where it becomes \( \varepsilon^2/R \).

Answer: (d) Power factor is zero

Explanation:
1. At resonance \( X_L = X_C \), so \( Z = R \) and \( \phi = 0 \).
2. The power factor \( \cos\phi = 1 \), average power equals apparent power, and the wattless component \( I\sin\phi \) is zero.
3. The peak energies \( \frac{1}{2}LI_0^2 \) and \( \frac{1}{2}CV_{C0}^2 \) are equal. So the false statement is that the power factor is zero.

Teacher's Note:
1. Resonance means the circuit behaves as purely resistive.
2. Power factor is 1 at resonance and 0 for a pure L or C.

Answer: (d) 250 V

Explanation:
1. Current \( I = \frac{V_R}{R} = \frac{100}{1000} = 0.1 \) A.
2. \( X_C = \frac{1}{\omega C} = \frac{1}{200\times2\times10^{-6}} = 2500\ \Omega \), and at resonance \( X_L = X_C \).
3. \( V_L = IX_L = 0.1\times2500 = 250 \) V.

Teacher's Note:
1. At resonance \( V_L \) and \( V_C \) are equal and opposite.
2. Both can exceed the voltage across the resistor.

Answer: (b) 8.4 A

Explanation:
1. \( Z = \sqrt{30^2 + 40^2} = 50\ \Omega \).
2. \( I_0 = \frac{V_0}{Z} = \frac{420}{50} = 8.4 \) A.

Teacher's Note:
1. Use the peak voltage to get the peak current.
2. The 30-40-50 triangle is a handy shortcut.

Answer: (a) 4 times

Explanation:
1. Resonance needs \( \omega^2 LC = 1 \), so \( LC \) must stay constant for the same frequency.
2. If C becomes \( \frac{C}{4} \), then L must become 4L.

Teacher's Note:
1. L and C are inversely related at a fixed resonant frequency.
2. The resonant frequency formula is \( f = \frac{1}{2\pi\sqrt{LC}} \).

Answer: (d) The net magnetic field produced is zero

Explanation:
1. The deflecting torque on the coil is proportional to the instantaneous current, which changes direction every half cycle.
2. Over one full cycle the average current, and hence the net effect (average torque), is zero.
3. So the pointer only vibrates about zero and gives no steady reading; a hot-wire or moving-iron meter is used instead.

Teacher's Note:
1. A dc meter responds to the average of the current, and the average of a sine wave is zero.
2. AC meters are built to respond to the heating effect or the square of the current.