JEE Physics Alternating Current MCQs Set 14

Answer: (a) \( 10^{6} \)

Explanation:
1. \( f=1/T=1/10^{-6}=10^6 \) Hz.

Teacher's Note:
1. Frequency is the reciprocal of the period.
2. 1 μs = \( 10^{-6} \) s.

Answer: (b) \( t=\dfrac{1}{300} \) s

Explanation:
1. Maximum when the phase equals \( 90^\circ \): \( 100\pi t+30^\circ=90^\circ \).
2. So \( 100\pi t=\pi/3 \) and \( t=\dfrac1{300} \) s.

Teacher's Note:
1. Set the phase equal to π/2.
2. Convert 60° to radians.

Answer: (a) only a capacitor

Explanation:
1. In a pure capacitor the current leads the voltage by 90°.
2. A resistor gives zero phase difference and a pure inductor makes the current lag by 90°.

Teacher's Note:
1. CIVIL: in a capacitor I leads V; in an inductor V leads I.
2. A resistor in series would reduce the angle below 90°.

Answer: (c) the 40 W bulb

Explanation:
1. Resistance of a bulb \( R=V^2/P \) is largest for the smallest power rating: the 40 W bulb has the largest resistance.
2. In series the same current flows, so power \( i^2R \) is greatest in the 40 W bulb.

Teacher's Note:
1. In series, the largest resistance dissipates the most power.
2. Power rating is for the rated voltage.

Answer: (c) the average value of the current over a complete cycle is zero

Explanation:
1. A DC meter reads the average current, which over a full cycle of an ac is zero.
2. A hot-wire or moving-iron meter responds to the heating effect and can read rms values.

Teacher's Note:
1. The deflection follows the mean value.
2. Use rms-reading instruments for ac.

Answer: (d) 100 mA

Explanation:
1. \( X_L=X_C \): resonance, so \( Z=R=1\,\Omega \).
2. Current \( =\dfrac{0.1}{1}=0.1 \) A = 100 mA.

Teacher's Note:
1. At resonance the reactances cancel.
2. Impedance equals R.

Answer: (d) the net reactance is zero

Explanation:
1. At resonance \( X_L=X_C \), so the net reactance is zero.
2. Impedance equals the resistance, which is not zero.

Teacher's Note:
1. Resonance cancels reactances only.
2. Current is maximum at resonance.

Answer: (d) remain unchanged

Explanation:
1. \( f=\dfrac{1}{2\pi\sqrt{LC}} \); new \( L'C'=1.25L\times0.8C=LC \).
2. So the resonant frequency is unchanged.

Teacher's Note:
1. Check the product LC.
2. 1.25 × 0.8 = 1.

Answer: (b) \( v=50\sin(100\pi t) \) volt

Explanation:
1. \( X_C=\dfrac1{\omega C}=\dfrac{1}{100\pi\times10^{-4}}\approx31.8 \) Ω so \( V_0=I_0X_C=1.57\times31.8\approx50 \) V.
2. Current is maximum at \( t=0 \) and leads the voltage by 90°, so the voltage is zero at \( t=0 \) and rising: \( v=50\sin100\pi t \).

Teacher's Note:
1. Voltage lags the current by 90° in a capacitor.
2. Peak current at t = 0 means zero voltage at t = 0.

Answer: (a) \( 6\ \text V,\ \tan^{-1}(4/3) \)

Explanation:
1. \( V_R=\sqrt{10^2-8^2}=6 \) V.
2. \( \tan\phi=\dfrac{V_C}{V_R}=\dfrac86=\dfrac43 \), the current leading the voltage.

Teacher's Note:
1. V_R and V_C are 90° apart.
2. Use the voltage triangle.

Answer: (b) 311 V

Explanation:
1. \( E_0=\sqrt2\times220=311 \) V.

Teacher's Note:
1. Peak = √2 × rms.
2. 220 V is the rms value.

Answer: (b) \( 4000\,\Omega \)

Explanation:
1. \( X_L=2\pi fL \) is proportional to both \( f \) and \( L \).
2. Doubling both gives 4 times: 4000 Ω.

Teacher's Note:
1. Reactance depends on the product fL.

Answer: (d) \( i=-10\cos(100t) \) A

Explanation:
1. \( X_L=\omega L=100\times0.1=10 \) Ω; \( I_0=100/10=10 \) A.
2. Current lags the voltage by 90°: \( i=10\sin(100t-\pi/2)=-10\cos100t \).

Teacher's Note:
1. In an inductor the current lags by 90°.
2. Peak current is E₀/ωL.

Answer: (b) \( 300\,\Omega \)

Explanation:
1. \( X_L\propto f \): \( 100\times\dfrac{150}{50}=300 \) Ω.

Teacher's Note:
1. Inductive reactance is proportional to frequency.

Answer: (b) \( 50\,\Omega \)

Explanation:
1. \( X_L=2\pi\times50\times\dfrac{20}\pi=2000 \) Ω.
2. \( X_C=\dfrac{1}{2\pi\times50\times(5/\pi)\times10^{-6}}=2000 \) Ω.
3. Resonance: \( Z=R=50 \) Ω.

Teacher's Note:
1. Compute both reactances and compare.
2. Equal reactances mean Z = R.

Answer: (c) 73.2%

Explanation:
1. Power factor \( \cos\phi=R/Z \), so \( Z\propto1/\cos\phi \).
2. Ratio \( \dfrac{Z_2}{Z_1}=\dfrac{0.866}{0.5}=1.732 \): increase 73.2%.

Teacher's Note:
1. With R fixed, Z = R/cos φ.
2. Percentage increase = (ratio − 1) × 100.

Answer: (d) zero

Explanation:
1. For a pure inductance the phase difference is 90°, so \( P=VI\cos90^\circ=0 \).
2. The choke dissipates no power.

Teacher's Note:
1. That is why chokes are used instead of resistors to control ac current.

Answer: (b) 4.8 V

Explanation:
1. Impedance \( Z=5 \) Ω; \( I_0=2 \) A; \( \tan\phi=4/3 \), \( \phi=53^\circ \) (current lags).
2. Voltage across L leads the current by 90°: \( V_L=8\sin(\omega t-53^\circ+90^\circ)=8\sin(\omega t+37^\circ) \).
3. At \( \omega t=180^\circ \): \( V_L=8\sin217^\circ=-4.8 \) V, magnitude 4.8 V.

Teacher's Note:
1. Find the phase angle first.
2. Voltage across L leads the current by 90°.

Answer: (d) 2.5 W

Explanation:
1. \( P=\dfrac{V_0I_0}{2}\cos\phi=\dfrac{100\times0.1}{2}\cos60^\circ=2.5 \) W.

Teacher's Note:
1. Convert mA to A.
2. Average power uses cos φ.

Answer: (c) 6.37 A

Explanation:
1. \( I_{avg}=\dfrac{2I_0}{\pi}=\dfrac{20}{\pi}=6.37 \) A.

Teacher's Note:
1. Average over a half cycle = 0.637 × peak.
2. 7.07 A is the rms value.

Answer: (a) \( 3\sqrt3 \) A

Explanation:
1. \( I_{rms}^2=3^2+\dfrac{6^2}{2}=9+18=27 \).
2. So \( I_{rms}=3\sqrt3 \) A.

Teacher's Note:
1. Square, average, then take the root.
2. The cross term averages to zero.

Answer: (b) \( q=Q_0\sin\left(\dfrac{t}{\sqrt{LC}}+\dfrac\pi2\right) \)

Explanation:
1. The LC circuit oscillates at \( \omega=1/\sqrt{LC} \) with amplitude \( Q_0 \).
2. At \( t=0 \) the upper plate carries \( +Q_0 \) (maximum), so \( q=Q_0\cos\omega t=Q_0\sin(\omega t+\pi/2) \).

Teacher's Note:
1. The charge starts at its maximum value.
2. The sine with phase π/2 is a cosine.