JEE Physics Alternating Current MCQs Set 03

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Multiple Choice Questions

Question 1: The peak value of an alternating emf E = E0 cos ωt is 10 V and its frequency is 50 Hz. At t = 1/600 s the instantaneous emf is
(a) 10 V
(b) \( 5\sqrt3 \) V
(c) 5 V
(d) 1 V
Show Answer & Explanation

Answer: (b) \( 5\sqrt3 \) V

Explanation:
1. \( \omega t=2\pi\times50\times\frac1{600}=\frac\pi6 \).
2. \( E=10\cos30^\circ=5\sqrt3 \) V.

Teacher's Note:
1. Work in radians: ωt = π/6.
2. cos 30° = √3/2.

Question 2: The rms value of the current i = i1 cos ωt + i2 sin ωt is
(a) \( \frac1{\sqrt2}(i_1+i_2) \)
(b) \( \frac1{\sqrt2}(i_1+i_2)^2 \)
(c) \( \frac1{\sqrt2}(i_1^2+i_2^2)^{1/2} \)
(d) \( \frac12(i_1^2+i_2^2)^{1/2} \)
Show Answer & Explanation

Answer: (c) \( \frac1{\sqrt2}(i_1^2+i_2^2)^{1/2} \)

Explanation:
1. Combine into one sinusoid with amplitude \( i_0=\sqrt{i_1^2+i_2^2} \).
2. \( I_{rms}=\frac{i_0}{\sqrt2} \).

Teacher's Note:
1. Two quadrature components add as vectors.
2. Divide the amplitude by √2.

Question 3: A hot-wire ammeter measures
(a) DC current only
(b) AC current only
(c) neither
(d) both DC and AC current
Show Answer & Explanation

Answer: (d) both DC and AC current

Explanation:
1. It works on the heating effect, which depends on I², independent of direction.
2. So it reads rms for AC and the steady value for DC.

Teacher's Note:
1. Heating effect is non-polar.
2. The scale is non-uniform (∝ I²).

Question 4: A resonant AC circuit contains a capacitor of capacitance 10⁻⁶ F and an inductor of 10⁻⁴ H. The frequency of electrical oscillations is
(a) \( 10^5 \) Hz
(b) 10 Hz
(c) \( \frac{10^5}{2\pi} \) Hz
(d) \( \frac{10}{2\pi} \) Hz
Show Answer & Explanation

Answer: (c) \( \frac{10^5}{2\pi} \) Hz

Explanation:
1. \( f=\frac1{2\pi\sqrt{LC}}=\frac1{2\pi\sqrt{10^{-10}}}=\frac{10^5}{2\pi} \) Hz.

Teacher's Note:
1. √(10⁻¹⁰) = 10⁻⁵.
2. Do not forget the 2π.

Question 5: A 110 V, 60 W lamp is run from 220 V AC mains using a capacitor in series with the lamp instead of a resistor. The voltage across the capacitor is about
(a) 110 V
(b) 190 V
(c) 220 V
(d) 311 V
Show Answer & Explanation

Answer: (b) 190 V

Explanation:
1. Lamp resistance \( R=\frac{110^2}{60}=201.7 \) Ω, current \( I=\frac{110}{201.7}=0.545 \) A.
2. \( Z=\frac{220}{0.545}=403.3 \) Ω, \( X_C=\sqrt{Z^2-R^2}=349 \) Ω.
3. \( V_C=IX_C\approx190 \) V.

Teacher's Note:
1. Voltages across R and C add in quadrature.
2. Check: √(110² + 190²) ≈ 220 V.

Question 6: A student connects a long air-cored coil of manganin wire to a 100 V DC supply and records a current of 25 A. When connected across 100 V, 50 Hz AC the current falls to 20 A. The reactance of the coil is
(a) 4 Ω
(b) 3 Ω
(c) 5 Ω
(d) none of these
Show Answer & Explanation

Answer: (b) 3 Ω

Explanation:
1. DC: \( R=\frac{100}{25}=4 \) Ω.
2. AC: \( Z=\frac{100}{20}=5 \) Ω, so \( X=\sqrt{Z^2-R^2}=\sqrt{25-16}=3 \) Ω.

Teacher's Note:
1. DC current is limited by resistance alone.
2. Then Z² = R² + X².

Question 7: A bulb and a capacitor are connected in series to an AC source. If the frequency is increased while the source voltage is kept constant,
(a) the bulb will give more intense light
(b) the bulb will give less intense light
(c) the bulb will give light of the same intensity
(d) the bulb will stop radiating light
Show Answer & Explanation

Answer: (a) the bulb will give more intense light

Explanation:
1. \( X_C=\frac1{\omega C} \) falls as ω rises, so the impedance falls and the current rises.
2. More current through the bulb gives more intense light.

Teacher's Note:
1. A capacitor blocks low frequencies and passes high ones.
2. Bulb brightness follows the rms current.

Question 8: The impedance of a circuit with a resistance R and an inductor L in series in an AC circuit of frequency f is
(a) \( \sqrt{R+4\pi fL^2} \)
(b) \( \sqrt{R+4\pi^2f^2L^2} \)
(c) \( \sqrt{R^2+4\pi^2f^2L^2} \)
(d) \( \sqrt{R^2+2\pi^2f^2L^2} \)
Show Answer & Explanation

Answer: (c) \( \sqrt{R^2+4\pi^2f^2L^2} \)

Explanation:
1. \( X_L=2\pi fL \), \( Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+4\pi^2f^2L^2} \).

Teacher's Note:
1. Add R² and X_L², not R and X_L.
2. X_L = ωL = 2πfL.

Question 9: The inductive reactance of a 1/π henry coil at 50 Hz is
(a) \( \frac{50}{\pi} \) Ω
(b) \( \frac{\pi}{50} \) Ω
(c) 100 Ω
(d) 50 Ω
Show Answer & Explanation

Answer: (c) 100 Ω

Explanation:
1. \( X_L=2\pi fL=2\pi\times50\times\frac1\pi=100 \) Ω.

Teacher's Note:
1. The π cancels.
2. X_L = ωL.

Question 10: A capacitor of 100 µF and a resistance of 100 Ω are connected in series to a 220 V, 50 Hz supply. The current leads the voltage by
(a) \( \tan^{-1}\frac1{2\pi} \)
(b) \( \tan^{-1}\frac1\pi \)
(c) \( \tan^{-1}\frac2\pi \)
(d) \( \tan^{-1}\frac4\pi \)
Show Answer & Explanation

Answer: (b) \( \tan^{-1}\frac1\pi \)

Explanation:
1. \( X_C=\frac1{2\pi\times50\times100\times10^{-6}}=\frac{100}{\pi} \) Ω.
2. \( \tan\phi=\frac{X_C}{R}=\frac1\pi \).

Teacher's Note:
1. Convert µF to F.
2. In a capacitive circuit the current leads.

Question 11: A 50 Hz AC source of 20 V is connected across R and C in series. The voltage across R is 12 V. The voltage across C is
(a) 8 V
(b) 16 V
(c) 10 V
(d) not possible to determine unless values of R and C are given
Show Answer & Explanation

Answer: (b) 16 V

Explanation:
1. \( V^2=V_R^2+V_C^2\Rightarrow V_C=\sqrt{400-144}=16 \) V.

Teacher's Note:
1. R and C voltages are 90° out of phase.
2. Do not add them arithmetically.

Question 12: A resistor and an inductor are in series across a 220 V AC supply. The voltage across the inductor is 176 V. The voltage across the resistor is
(a) 132 V
(b) 396 V
(c) 185 V
(d) \( \sqrt{220\times176} \) V
Show Answer & Explanation

Answer: (a) 132 V

Explanation:
1. \( V^2=V_R^2+V_L^2\Rightarrow V_R=\sqrt{220^2-176^2}=\sqrt{17424}=132 \) V.

Teacher's Note:
1. The voltages add as perpendicular phasors.
2. 220 : 176 : 132 is a 5 : 4 : 3 triangle.

Question 13: If L, C and R represent inductance, capacitance and resistance respectively, the combination having the dimensions of frequency is
(a) LC
(b) \( (LC)^{-1/2} \)
(c) \( \left(\frac LC\right)^{-1/2} \)
(d) \( \frac CL \)
Show Answer & Explanation

Answer: (b) \( (LC)^{-1/2} \)

Explanation:
1. Resonance \( \omega=\frac1{\sqrt{LC}} \) has dimensions of frequency.

Teacher's Note:
1. Check: LC has dimensions of time squared.
2. √(L/C) is an impedance (Ω).

Question 14: At resonance in a series LCR circuit, which statement is true?
(a) current is maximum and the phase difference between E and I is π/2
(b) current is maximum and the phase difference between E and I is zero
(c) voltage is maximum and the phase difference is π/2
(d) current is minimum and the phase difference is zero
Show Answer & Explanation

Answer: (b) current is maximum and the phase difference between E and I is zero

Explanation:
1. At resonance X_L = X_C, so Z = R is the minimum impedance.
2. Hence the current is maximum and in phase with the emf.

Teacher's Note:
1. Impedance is purely resistive.
2. Power factor is 1.

Question 15: For a series R-L-C circuit with R = 10 Ω, X_L = 20 Ω and X_C = 20 Ω supplied by V = 10 sin ωt volt, the power dissipated is
(a) zero
(b) 10 W
(c) 5 W
(d) 2.5 W
Show Answer & Explanation

Answer: (c) 5 W

Explanation:
1. \( X_L=X_C \): resonance, so \( Z=R=10 \) Ω and \( I_{rms}=\frac{10/\sqrt2}{10}=\frac1{\sqrt2} \) A.
2. \( P=I_{rms}^2R=\frac12\times10=5 \) W.

Teacher's Note:
1. Use rms values for average power.
2. Peak voltage is 10 V, not rms.

Question 16: The quality factor of a series resonant circuit is given by
(a) \( \frac{\omega L}{R} \)
(b) \( \frac{\omega}{RC} \)
(c) \( \sqrt{LC} \)
(d) \( \frac LR \)
Show Answer & Explanation

Answer: (a) \( \frac{\omega L}{R} \)

Explanation:
1. \( Q=\frac{\omega_0L}{R}=\frac1{\omega_0CR}=\frac1R\sqrt{\frac LC} \).
2. Option (b) would be dimensionally wrong.

Teacher's Note:
1. Q measures sharpness of resonance.
2. Large Q means small R.

Question 17: A 50 Ω resistor, an inductor of reactance 400 Ω and a capacitor of reactance 400 Ω are connected in series to a 100 V, 50 Hz supply. The voltage across the resistor and the current are respectively
(a) 400 V, 2 A
(b) 800 V, 2 A
(c) 100 V, 2 A
(d) 100 V, 4 A
Show Answer & Explanation

Answer: (c) 100 V, 2 A

Explanation:
1. \( X_L=X_C \): resonance, \( Z=R=50 \) Ω.
2. \( I=\frac{100}{50}=2 \) A and \( V_R=IR=100 \) V.

Teacher's Note:
1. At resonance the whole supply voltage appears across R.
2. V_L and V_C (800 V each) cancel.

Question 18: A sinusoidal AC current flows through a resistor R. If the peak current is Ip, the power dissipated is
(a) \( I_p^2R\cos\theta \)
(b) \( \frac12I_p^2R \)
(c) \( \frac4\pi I_p^2R \)
(d) \( \frac1{\pi^2}I_p^2R \)
Show Answer & Explanation

Answer: (b) \( \frac12I_p^2R \)

Explanation:
1. \( P=I_{rms}^2R=\left(\frac{I_p}{\sqrt2}\right)^2R=\frac12I_p^2R \).

Teacher's Note:
1. Use the rms current.
2. cos θ applies to reactive circuits.

Question 19: In an AC circuit V = 100 sin(100t) volt and I = 100 sin(100t + π/3) mA. The power dissipated in the circuit is
(a) \( 10^4 \) W
(b) 10 W
(c) 2.5 W
(d) 5.0 W
Show Answer & Explanation

Answer: (c) 2.5 W

Explanation:
1. \( P=\frac12V_0I_0\cos\phi=\frac12\times100\times0.1\times\cos60^\circ \).
2. \( =2.5 \) W.

Teacher's Note:
1. Convert mA to A.
2. φ = π/3.

Question 20: If V = 100 sin 100t volt and I = 100 sin(100t + π/6) A, the wattless power is
(a) \( 10^4 \) W
(b) \( 10^3 \) W
(c) \( 10^2 \) W
(d) \( 2.5\times10^3 \) W
Show Answer & Explanation

Answer: (d) \( 2.5\times10^3 \) W

Explanation:
1. Wattless power \( =V_{rms}I_{rms}\sin\phi=\frac{100\times100}{2}\sin30^\circ \).
2. \( =2500=2.5\times10^3 \) W.

Teacher's Note:
1. Real power uses cos φ; wattless (reactive) power uses sin φ.
2. rms = peak/√2.

Question 21: An inductor L and a resistor R are joined in series to a source of rms voltage V and angular frequency ω. The power dissipated is
(a) \( \frac{R^2+\omega^2L^2}{V} \)
(b) \( \frac{V^2R}{R^2+\omega^2L^2} \)
(c) \( \frac{V}{\omega^2L^2+R^2} \)
(d) \( \frac{\sqrt{R^2+\omega^2L^2}}{V^2} \)
Show Answer & Explanation

Answer: (b) \( \frac{V^2R}{R^2+\omega^2L^2} \)

Explanation:
1. \( I=\frac{V}{\sqrt{R^2+\omega^2L^2}} \).
2. \( P=I^2R=\frac{V^2R}{R^2+\omega^2L^2} \).

Teacher's Note:
1. P = I²R with the rms current.
2. Check dimensions: V²/R.

Question 22: In a purely capacitive circuit, the average power dissipated is
(a) \( V_{rms}I_{rms} \)
(b) dependent on capacitance
(c) infinite
(d) zero
Show Answer & Explanation

Answer: (d) zero

Explanation:
1. The current leads the voltage by 90°, so cos φ = 0.
2. Energy is stored and returned each cycle, so the average power is zero.

Teacher's Note:
1. The same holds for a pure inductor.
2. Average power = V I cos φ.

Question 23: If an alternating current of rms value a flows through a resistance R, the power loss in the resistance is
(a) zero
(b) \( a^2R \)
(c) \( \frac{a^2R}2 \)
(d) \( 2a^2R \)
Show Answer & Explanation

Answer: (b) \( a^2R \)

Explanation:
1. \( P=I_{rms}^2R=a^2R \).

Teacher's Note:
1. The rms value already accounts for the average of the square.
2. Do not divide by 2 again.

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