JEE Physics Alternating Current MCQs Set 17

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Multiple Choice Questions

Question 1: In an ac circuit an alternating voltage \( e = 200\sqrt2\sin100t \) volts is connected to a capacitor of capacity 10 µF. The r.m.s. value of the current in the circuit is
(a) 10 mA
(b) 100 mA
(c) 200 mA
(d) 20 mA
Show Answer & Explanation

Answer: (c) 200 mA

Explanation:
1. \( X_C = 1/(100\times10\times10^{-6}) = 1000 \) Ω.
2. \( I_{rms} = 200/1000 = 0.2 \) A = 200 mA.

Teacher's Note:
1. Use the rms voltage, 200 V.

Question 2: In a given series LCR circuit R = 4 Ω, X_L = 5 Ω and X_C = 9 Ω, the current
(a) leads the voltage by 45°
(b) leads the voltage by 53°
(c) lags the voltage by 45°
(d) lags the voltage by 53°
Show Answer & Explanation

Answer: (a) leads the voltage by 45°

Explanation:
1. \( X_L - X_C = -4 \) Ω, so the circuit is capacitive.
2. \( \tan\phi = 4/4 = 1 \), so the current leads by 45°.

Teacher's Note:
1. Net capacitive reactance means current leads.

Question 3: A series R-L-C circuit (R = 10 Ω, X_L = 20 Ω, X_C = 20 Ω) is supplied by \( V = 10\sin\omega t \) volt. Then power dissipation in circuit is
(a) zero
(b) 10 watt
(c) 5 watt
(d) 2.5 watt
Show Answer & Explanation

Answer: (c) 5 watt

Explanation:
1. \( X_L = X_C \), so \( Z = R = 10 \) Ω.
2. \( P = V_{rms}^2/R = (10/\sqrt2)^2/10 = 5 \) W.

Teacher's Note:
1. Use the rms voltage.

Question 4: In a series LCR circuit voltage across resistor, inductor and capacitor are 1 V, 3 V and 2 V respectively. At the instant when the source voltage is \( V = V_0\cos\omega t \), the current in the circuit will be
(a) \( I_0\cos(\omega t+\pi/4) \)
(b) \( I_0\cos(\omega t-\pi/4) \)
(c) \( I_0\cos(\omega t+\pi/3) \)
(d) \( I_0\cos(\omega t-\pi/3) \)
Show Answer & Explanation

Answer: (b) \( I_0\cos(\omega t-\pi/4) \)

Explanation:
1. \( \tan\phi = (V_L - V_C)/V_R = (3-2)/1 = 1 \), so \( \phi = 45^\circ \).
2. The circuit is inductive, so the current lags: \( I = I_0\cos(\omega t - \pi/4) \).

Teacher's Note:
1. V_L larger than V_C means the current lags.

Question 5: In LCR circuit, the voltage across the terminals of a resistance, inductance and capacitance are 40 V, 30 V and 60 V, then the voltage across the main source will be
(a) 130 volt
(b) 100 volt
(c) 70 volt
(d) 50 volt
Show Answer & Explanation

Answer: (d) 50 volt

Explanation:
1. \( V = \sqrt{V_R^2 + (V_C - V_L)^2} = \sqrt{40^2 + 30^2} = 50 \) V.

Teacher's Note:
1. Subtract L and C voltages before squaring.

Question 6: In an A.C. circuit inductance, capacitance and resistance are connected. If the effective voltage across inductance is V_L, across capacitance is V_C and across resistance is V_R, then the total effective value of voltage is
(a) V_R + V_L + V_C
(b) V_R + V_L − V_C
(c) \( \sqrt{V_R^2 + (V_L - V_C)^2} \)
(d) \( \sqrt{V_R^2 - (V_L - V_C)^2} \)
Show Answer & Explanation

Answer: (c) \( \sqrt{V_R^2 + (V_L - V_C)^2} \)

Explanation:
1. The voltages are phasors: V_R along the current, V_L 90° ahead, V_C 90° behind.
2. \( V = \sqrt{V_R^2 + (V_L - V_C)^2} \).

Teacher's Note:
1. Voltages in an AC circuit do not add arithmetically.

Question 7: For a series LCR circuit the power loss at resonance is
(a) \( \dfrac{V^2}{\omega L - \frac1{\omega C}} \)
(b) \( I^2L\omega \)
(c) \( I^2R \)
(d) \( \dfrac{V^2}{C\omega} \)
Show Answer & Explanation

Answer: (c) \( I^2R \)

Explanation:
1. At resonance \( Z = R \) and the power factor is 1.
2. \( P = I^2R \).

Teacher's Note:
1. Power is dissipated only in the resistor.

Question 8: For the circuit with R = 1100 Ω and L = 3.5 H connected in series to a 50 Hz source, the power factor is
(a) 0
(b) 1/2
(c) \( 1/\sqrt2 \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( 1/\sqrt2 \)

Explanation:
1. \( X_L = 2\pi(50)(3.5)\approx1100 \) Ω.
2. \( \cos\phi = R/\sqrt{R^2+X_L^2} = 1/\sqrt2 \).

Teacher's Note:
1. Here X_L equals R.

Question 9: An inductance L, a capacitance C and resistance R may be connected to an AC source of angular frequency ω, in three different combinations of RC, RL and RLC in series. Assume that \( \omega L = 1/\omega C \). The power drawn by the three combinations are P₁, P₂, P₃ respectively. Then
(a) P₁ > P₂ > P₃
(b) P₁ = P₂ < P₃
(c) P₁ = P₂ > P₃
(d) P₁ = P₂ = P₃
Show Answer & Explanation

Answer: (b) P₁ = P₂ < P₃

Explanation:
1. Let \( X_L = X_C = X \). For RC and RL, \( Z = \sqrt{R^2 + X^2} \), so \( P_1 = P_2 \).
2. For RLC, \( Z = R \), which is smaller, so the power is larger: \( P_3 > P_1 = P_2 \).

Teacher's Note:
1. Resonance gives the lowest impedance.

Question 10: A resistor R, an inductor L and a capacitor C with \( R = X_L = X_C = 10\ \Omega \) are each connected directly across a 300 V a.c. supply (in parallel). Voltmeters V₁, V₂ and V₃ are connected across R, L and C respectively. The correct set of readings is
(a) V₁ = 100 V, V₂ = 100 V, V₃ = 100 V
(b) V₁ = 150 V, V₂ = 0 V, V₃ = 150 V
(c) V₁ = 300 V, V₂ = 100 V, V₃ = 100 V
(d) V₁ = 300 V, V₂ = 300 V, V₃ = 300 V
Show Answer & Explanation

Answer: (d) V₁ = 300 V, V₂ = 300 V, V₃ = 300 V

Explanation:
1. In a parallel connection each element has the full source voltage across it.
2. All three voltmeters read 300 V.

Teacher's Note:
1. Parallel elements share the same voltage.

Question 11: In a series L-C-R circuit connected to a 220 V, 50 Hz source with R = 200 Ω, the readings of voltmeters V₁ and V₂ (across L and C) are 400 V each. The readings of the voltmeter V₃ (across R) and the ammeter A are respectively
(a) 100 V, 1 A
(b) 150 V, 1.1 A
(c) 220 V, 1.1 A
(d) 220 V, 1 A
Show Answer & Explanation

Answer: (c) 220 V, 1.1 A

Explanation:
1. Equal voltages across L and C mean \( X_L = X_C \), so the circuit is at resonance.
2. The whole 220 V falls across R, and \( I = 220/200 = 1.1 \) A.

Teacher's Note:
1. L and C voltages cancel at resonance.

Question 12: Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?
(a) R = 15 Ω, L = 4 H, C = 30 µF
(b) R = 25 Ω, L = 1 H, C = 45 µF
(c) R = 20 Ω, L = 0.5 H, C = 35 µF
(d) R = 25 Ω, L = 2 H, C = 45 µF
Show Answer & Explanation

Answer: (a) R = 15 Ω, L = 4 H, C = 30 µF

Explanation:
1. Good tuning needs a large quality factor \( Q = \dfrac1R\sqrt{L/C} \).
2. The values are about 24.3, 6.0, 6.0 and 8.4 respectively.
3. The first combination has the largest Q.

Teacher's Note:
1. A sharper resonance gives better tuning.

Question 13: The potential differences across the resistance, capacitance and inductance are 40 V, 20 V and 50 V respectively in an L-C-R circuit. The power factor of this circuit is
(a) 0.8
(b) 1.0
(c) 0.4
(d) 0.5
Show Answer & Explanation

Answer: (a) 0.8

Explanation:
1. \( V = \sqrt{40^2 + (50-20)^2} = 50 \) V.
2. \( \cos\phi = V_R/V = 40/50 = 0.8 \).

Teacher's Note:
1. Power factor is the resistor voltage over the source voltage.

Question 14: A 160 V, 60 W lamp is run from a 200 V AC mains using a capacitor in series with the lamp instead of a resistor. Then the voltage across the capacitor is about
(a) 40 V
(b) 120 V
(c) 300 V
(d) 311 V
Show Answer & Explanation

Answer: (b) 120 V

Explanation:
1. The lamp and capacitor voltages are 90° apart.
2. \( V_C = \sqrt{200^2 - 160^2} = 120 \) V.

Teacher's Note:
1. Use the vector sum of voltages.

Question 15: In an a.c. circuit V and I are given by V = 200 sin(100t) volts, \( I = 50\sin(100t + \pi/3) \) mA. The power dissipated in the circuit is
(a) \( 10^4 \) watt
(b) 10 watt
(c) 5.0 watt
(d) 2.5 watt
Show Answer & Explanation

Answer: (d) 2.5 watt

Explanation:
1. \( P = \tfrac12V_0I_0\cos\phi \).
2. \( = \tfrac12(200)(0.05)\cos60^\circ = 2.5 \) W.

Teacher's Note:
1. Convert mA to A.

Question 16: If V = 100 sin 100t volt and \( I = 100\sin(100t + \pi/6) \) A, then the watt-less power in watt is
(a) \( 10^4 \)
(b) \( 10^3 \)
(c) \( 10^2 \)
(d) \( 2.5\times10^3 \)
Show Answer & Explanation

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

Explanation:
1. Watt-less power \( = V_{rms}I_{rms}\sin\phi = \tfrac12V_0I_0\sin\phi \).
2. \( = \tfrac12(100)(100)\sin30^\circ = 2.5\times10^3 \) W.

Teacher's Note:
1. The wattless component involves sin φ.

Question 17: Power dissipated in pure inductance will be
(a) \( LI^2/2 \)
(b) \( 2LI^2 \)
(c) \( LI^2/4 \)
(d) zero
Show Answer & Explanation

Answer: (d) zero

Explanation:
1. The current lags the voltage by 90°, so \( \cos\phi = 0 \).
2. The average power is zero.

Teacher's Note:
1. Energy is stored and returned, not dissipated.

Question 18: What is the value of inductance L for which the current is a maximum in a series LCR circuit with C = 10 µF and ω = 200 s⁻¹?
(a) 25 mH
(b) 200 mH
(c) 2.5 H
(d) cannot be calculated unless R is known
Show Answer & Explanation

Answer: (c) 2.5 H

Explanation:
1. Maximum current occurs at resonance, \( \omega^2 = 1/LC \).
2. \( L = 1/(\omega^2C) = 1/(4\times10^4\times10^{-5}) = 2.5 \) H.

Teacher's Note:
1. R does not affect the resonance condition.

Question 19: What is the frequency of power in an AC circuit connected to a capacitor? Given f = 30 Hz, V = 200 V
(a) 50 Hz
(b) 100 Hz
(c) zero
(d) 60 Hz
Show Answer & Explanation

Answer: (d) 60 Hz

Explanation:
1. \( P = VI\propto\sin\omega t\cos\omega t = \tfrac12\sin2\omega t \).
2. The power oscillates at \( 2f = 60 \) Hz.

Teacher's Note:
1. Instantaneous power has twice the supply frequency.

Question 20: A 100 V, 50 Hz source is connected to a resistor and an inductor L in series. The potential difference across the resistor is 80 V and the power consumed by it is 160 W. The value of L is
(a) 0.3 H
(b) \( \dfrac{0.3}{\pi} \) H
(c) 10 H
(d) \( \dfrac{0.7}{\pi} \) H
Show Answer & Explanation

Answer: (b) \( \dfrac{0.3}{\pi} \) H

Explanation:
1. \( I = P/V_R = 160/80 = 2 \) A.
2. \( V_L = \sqrt{100^2 - 80^2} = 60 \) V, so \( X_L = 30\ \Omega \).
3. \( L = 30/(2\pi\times50) = 0.3/\pi \) H.

Teacher's Note:
1. Power in the resistor gives the current.

Question 21: In an alternating circuit applied voltage and flowing current are \( E = E_0\sin\omega t \) and \( I = I_0\sin(\omega t + \pi/2) \) respectively. Then the power consumed in the circuit will be
(a) zero
(b) \( E_0I_0/2 \)
(c) \( E_0I_0/\sqrt2 \)
(d) \( E_0I_0/4 \)
Show Answer & Explanation

Answer: (a) zero

Explanation:
1. The phase difference is 90°, so \( \cos\phi = 0 \).
2. The average power is zero.

Teacher's Note:
1. 90° means a purely reactive circuit.

Question 22: An inductor L and resistor R are joined in series and connected by a source of frequency ω. Power dissipated in the circuit is (V is the rms source voltage)
(a) \( \dfrac{R^2+\omega^2L^2}{V} \)
(b) \( \dfrac{V^2R}{R^2+\omega^2L^2} \)
(c) \( \dfrac{V}{R^2+\omega^2L^2} \)
(d) \( \dfrac{\sqrt{R^2+\omega^2L^2}}{V^2} \)
Show Answer & Explanation

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

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

Teacher's Note:
1. Power is I²R with the rms current.

Question 23: In an a.c. circuit the e.m.f. (e) and the current (i) at any instant are given respectively by \( e = E_0\sin\omega t \), \( i = I_0\sin(\omega t - 30^\circ) \). The average power in the circuit over one cycle of a.c. is
(a) \( \dfrac{\sqrt3E_0I_0}{4} \)
(b) \( E_0I_0 \)
(c) \( \dfrac{\sqrt3E_0I_0}{2} \)
(d) \( \dfrac{E_0I_0}{4} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{\sqrt3E_0I_0}{4} \)

Explanation:
1. \( P = \tfrac12E_0I_0\cos30^\circ \).
2. \( = \tfrac12E_0I_0\times\tfrac{\sqrt3}2 = \dfrac{\sqrt3}4E_0I_0 \).

Teacher's Note:
1. Average power is half the peak product times cos φ.

Question 24: A resistance of 20 Ω draws power 100 W when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 50 Ω, the power drawn will be
(a) \( 100\sqrt{2/5} \) W
(b) 40 W
(c) 100 W
(d) 16 W
Show Answer & Explanation

Answer: (d) 16 W

Explanation:
1. \( V^2/R = 100 \), so \( V^2 = 2000 \).
2. \( I^2 = V^2/Z^2 = 2000/2500 = 0.8 \).
3. \( P = I^2R = 0.8\times20 = 16 \) W.

Teacher's Note:
1. The source voltage stays the same.

Question 25: An inductor 20 mH, a capacitor 50 µF and a resistor 40 Ω are connected in series across a source of emf V = 10 sin 340t. The power loss in A.C. circuit is
(a) 0.46 W
(b) 0.67 W
(c) 0.76 W
(d) 0.89 W
Show Answer & Explanation

Answer: (a) 0.46 W

Explanation:
1. \( X_L = 340\times0.02 = 6.8 \) Ω and \( X_C = 1/(340\times50\times10^{-6}) = 58.8 \) Ω.
2. \( Z^2 = 40^2 + (52)^2 = 4306 \).
3. \( P = \tfrac12V_0^2R/Z^2 = 0.5\times100\times40/4306 = 0.46 \) W.

Teacher's Note:
1. Power uses the rms current.

Question 26: An inductor 20 mH, a capacitor 100 µF and a resistor 50 Ω are connected in series across a source of emf V = 10 sin 314t. The power loss in the circuit is
(a) 0.79 W
(b) 0.43 W
(c) 2.74 W
(d) 1.13 W
Show Answer & Explanation

Answer: (a) 0.79 W

Explanation:
1. \( X_L = 314\times0.02 = 6.28 \) Ω and \( X_C = 1/(314\times10^{-4}) = 31.8 \) Ω.
2. \( Z^2 = 50^2 + 25.6^2 = 3155 \).
3. \( P = 0.5\times100\times50/3155 = 0.79 \) W.

Teacher's Note:
1. Average power is half V₀² R over Z².

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