JEE Physics Alternating Current MCQs Set 16

Physics Objective Questions and Answers: Alternating Current

Access targeted multiple-choice questions for Alternating Current designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: What is the r.m.s. value of an alternating current which when passed through a resistor produces heat which is four times of that produced by a direct current of 2 amperes in the same resistor?
(a) 6 A
(b) 4 A
(c) 3.46 A
(d) 0.66 A
Show Answer & Explanation

Answer: (b) 4 A

Explanation:
1. Heat is proportional to \( I_{rms}^2 \).
2. \( I_{rms}^2 = 4\times2^2 = 16 \), so \( I_{rms} = 4 \) A.

Teacher's Note:
1. Heating depends on the square of the current.

Question 2: Correct statement is: (a) A.C. meters can measure D.C. also. (b) If A.C. meter measures D.C. their scale must be linear and uniform. (c) A.C. meters are based on heating effect of current and D.C. meters based on average current method. (d) A.C. meter reads rms value of current.
(a) a, b
(b) b, c
(c) c, d
(d) a, c, d
Show Answer & Explanation

Answer: (d) a, c, d

Explanation:
1. Hot wire (heating effect) AC meters respond to the square of the current, so they can read DC too: (a) is true.
2. Because of the square-law response their scale is not linear, so (b) is false.
3. (c) and (d) are true.

Teacher's Note:
1. Heating effect depends on I².

Question 3: The r.m.s. value of current for a variable current \( i = 4\sin\omega t + 3\cos\omega t \) is
(a) 5 A
(b) \( 5\sqrt2 \) A
(c) \( 5/\sqrt2 \) A
(d) 2.5 A
Show Answer & Explanation

Answer: (c) \( 5/\sqrt2 \) A

Explanation:
1. Amplitude \( = \sqrt{4^2+3^2} = 5 \) A.
2. \( I_{rms} = 5/\sqrt2 \) A.

Teacher's Note:
1. Combine sine and cosine into a single amplitude.

Question 4: The relation between an A.C. voltage source and time in SI units is \( V = 100\sin(50\pi t)\cos(50\pi t) \) volt. The value of peak voltage and frequency will be respectively
(a) 100 volt and 100 Hz
(b) \( 100/\sqrt2 \) volt and 100 Hz
(c) 50 volt and 50 Hz
(d) 50 volt and 100 Hz
Show Answer & Explanation

Answer: (c) 50 volt and 50 Hz

Explanation:
1. \( V = 50\sin(100\pi t) \).
2. Peak voltage = 50 V and \( \omega = 100\pi \), so \( f = 50 \) Hz.

Teacher's Note:
1. Use \( 2\sin\theta\cos\theta = \sin2\theta \).

Question 5: A voltage that is V₀ for the first half of every period T and −V₀/2 for the second half is applied across a resistor. The r.m.s. value of this voltage is
(a) \( V_0/\sqrt3 \)
(b) \( V_0 \)
(c) \( V_0/\sqrt2 \)
(d) \( \dfrac{V_0}2\sqrt{\dfrac52} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{V_0}2\sqrt{\dfrac52} \)

Explanation:
1. \( V_{rms}^2 = \tfrac12\left[V_0^2 + (V_0/2)^2\right] = \tfrac58V_0^2 \).
2. \( V_{rms} = V_0\sqrt{5/8} = \dfrac{V_0}{2}\sqrt{5/2} \).

Teacher's Note:
1. Average V² over a period, then take the root.

Question 6: The graph shows variation of I with f for a series R-L-C network. Keeping L and C constant, if R decreases: (a) Maximum current (I_m) increases (b) Sharpness of the graph increases (c) Quality factor increases (d) Band width increases
(a) a, b, c
(b) b, c, d
(c) c, d, a
(d) all
Show Answer & Explanation

Answer: (a) a, b, c

Explanation:
1. At resonance \( I_m = V/R \) increases when R decreases.
2. \( Q = \omega_0L/R \) increases, so the resonance curve becomes sharper.
3. A sharper peak means a smaller band width, so (d) is false.

Teacher's Note:
1. Lower resistance means sharper resonance.

Question 7: Alternating current is flowing in capacitance C and resistance R in series. The frequency of source is ω/2π. Which of the following statements is correct?
(a) For low frequency the limiting value of impedance is C.
(b) For high frequency the limiting value of impedance is 1/ωC.
(c) For high frequency the limiting value of impedance is R.
(d) For low frequency the limiting value of impedance is 1/C.
Show Answer & Explanation

Answer: (c) For high frequency the limiting value of impedance is R.

Explanation:
1. \( Z = \sqrt{R^2 + 1/\omega^2C^2} \).
2. As ω becomes large, \( 1/\omega C\to0 \), so \( Z\to R \).

Teacher's Note:
1. At high frequency a capacitor behaves like a short circuit.

Question 8: A bulb and an inductor are connected in series to a source of alternating current. If its frequency is increased, while keeping the voltage of the source constant, then
(a) bulb will give more intense light
(b) bulb will give less intense light
(c) bulb will give light of same intensity as before
(d) bulb will stop radiating light
Show Answer & Explanation

Answer: (b) bulb will give less intense light

Explanation:
1. \( X_L = \omega L \) increases, so the impedance increases.
2. The current falls and the bulb becomes dimmer.

Teacher's Note:
1. Inductors choke high frequencies.

Question 9: An a.c. source of voltage V and of frequency 50 Hz is connected to an inductor of 2 H and negligible resistance. A current of r.m.s value I flows in the coil. When the frequency of the voltage is changed to 400 Hz keeping the magnitude of V the same, the current is now
(a) 8 I in phase with V
(b) 4 I and leading by 90° from V
(c) I/4 and lagging by 90° from V
(d) I/8 and lagging by 90° from V
Show Answer & Explanation

Answer: (d) I/8 and lagging by 90° from V

Explanation:
1. \( I = V/\omega L\propto1/f \).
2. The frequency becomes 8 times, so the current is I/8.
3. In a pure inductor the current lags the voltage by 90°.

Teacher's Note:
1. Current in an inductor varies inversely with frequency.

Question 10: A circuit has a source of 200 V, a resistor and an inductor in series. The voltage across the inductor is \( 100\sqrt3 \) V. The voltage across the resistor V_R is
(a) 100 V
(b) \( 100\sqrt3 \) V
(c) 200 V
(d) 50 V
Show Answer & Explanation

Answer: (a) 100 V

Explanation:
1. \( V^2 = V_R^2 + V_L^2 \).
2. \( V_R = \sqrt{200^2 - (100\sqrt3)^2} = \sqrt{40000 - 30000} = 100 \) V.

Teacher's Note:
1. Resistor and inductor voltages are 90° apart.

Question 11: In an electrical circuit R, L, C and an a.c. voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is π/4. If instead, C is removed from the circuit, the phase difference is again π/4. The power factor of the circuit is
(a) 1
(b) \( \sqrt3/2 \)
(c) 1/2
(d) \( 1/\sqrt2 \)
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. Without L: \( \tan(\pi/4) = X_C/R \), so \( X_C = R \).
2. Without C: \( X_L = R \).
3. So \( X_L = X_C \), the circuit is at resonance and the power factor is 1.

Teacher's Note:
1. At resonance the phase angle is zero.

Question 12: A series R-C circuit is connected to an alternating voltage source. Consider two situations: (a) when capacitor is air filled, (b) when capacitor is mica filled. Current through resistor is i and voltage across capacitor is V then
(a) i_a = i_b
(b) V_a < V_b
(c) i_b > i_a
(d) i_a > i_b
Show Answer & Explanation

Answer: (c) i_b > i_a

Explanation:
1. Mica has K > 1, so the capacitance increases.
2. \( X_C = 1/\omega C \) falls, so the impedance falls and the current increases: \( i_b > i_a \).

Teacher's Note:
1. Dielectric lowers the reactance.

Question 13: A capacitor C and a resistor R are connected in series to an AC source. An ammeter A is in the circuit, voltmeter V₁ is across the capacitor and V₂ across the resistor. Consider the following statements: (I) Readings in A and V₂ are always in phase. (II) Reading in V₁ is ahead of reading in V₂. (III) Readings in A and V₁ are always in phase. Which of these statements is/are correct?
(a) I only
(b) II only
(c) I and II only
(d) II and III only
Show Answer & Explanation

Answer: (a) I only

Explanation:
1. The current and the voltage across the resistor are in phase, so (I) is true.
2. The capacitor voltage lags the current by 90°, so V₁ is behind V₂, not ahead, and V₁ is not in phase with the current.
3. Only (I) is correct.

Teacher's Note:
1. Resistor voltage is in phase with the current.

Question 14: In a purely inductive AC circuit, how does the amplitude of the current \( I_0 \) vary with the angular frequency ω of the source (voltage amplitude kept fixed)?
(a) It decreases inversely with ω
(b) It oscillates with ω
(c) It is independent of ω
(d) It decreases linearly with ω
Show Answer & Explanation

Answer: (a) It decreases inversely with ω

Explanation:
1. \( I_0 = V_0/\omega L \).
2. So it is inversely proportional to ω.

Teacher's Note:
1. The graph is a hyperbola.

Question 15: The phase difference between current and voltage in an AC circuit is π/4 radian. If the frequency of AC is 50 Hz, then the phase difference is equivalent to the time difference
(a) 0.78 s
(b) 15.7 ms
(c) 2.5 s
(d) 2.5 ms
Show Answer & Explanation

Answer: (d) 2.5 ms

Explanation:
1. \( \Delta t = \phi/\omega = (\pi/4)/(100\pi) = 1/400 \) s.
2. \( = 2.5 \) ms.

Teacher's Note:
1. Time difference equals phase divided by ω.

Question 16: A resistance of 300 Ω and an inductance of \( 1/\pi \) henry are connected in series to a A.C. voltage of 20 volts and 200 Hz frequency. The phase angle between the voltage and current is
(a) \( \tan^{-1}(4/3) \)
(b) \( \tan^{-1}(3/4) \)
(c) \( \tan^{-1}(3/2) \)
(d) \( \tan^{-1}(2/3) \)
Show Answer & Explanation

Answer: (a) \( \tan^{-1}(4/3) \)

Explanation:
1. \( X_L = 2\pi(200)(1/\pi) = 400\ \Omega \).
2. \( \tan\phi = X_L/R = 400/300 = 4/3 \).

Teacher's Note:
1. The voltage does not affect the phase.

Question 17: The reactance X₁ of a component decreases with frequency, and the reactance X₂ of another component increases linearly with frequency. We can then say that
(a) X₁ is an inductor and X₂ is a capacitor
(b) X₁ is a resistor and X₂ is a capacitor
(c) X₁ is a capacitor and X₂ is an inductor
(d) X₁ is an inductor and X₂ is a resistor
Show Answer & Explanation

Answer: (c) X₁ is a capacitor and X₂ is an inductor

Explanation:
1. \( X_C = 1/\omega C \) decreases with frequency.
2. \( X_L = \omega L \) increases linearly.
3. So X₁ is a capacitor and X₂ is an inductor.

Teacher's Note:
1. A resistor's opposition does not depend on frequency.

Question 18: In the L–R circuit R = 10 Ω and L = 2 H. If 120 V, 60 Hz alternating voltage is applied then the flowing current in this circuit will be
(a) 0.32 A
(b) 0.16 A
(c) 0.48 A
(d) 0.80 A
Show Answer & Explanation

Answer: (b) 0.16 A

Explanation:
1. \( X_L = 2\pi(60)(2) = 754 \) Ω.
2. \( Z = \sqrt{10^2 + 754^2}\approx754 \) Ω.
3. \( I = 120/754\approx0.16 \) A.

Teacher's Note:
1. The resistance is negligible compared with the reactance.

Question 19: A capacitor of capacitance 100 µF and a resistance of 100 Ω are connected in series with AC supply of 220 V, 50 Hz. The current leads the voltage by
(a) \( \tan^{-1}(1/2\pi) \)
(b) \( \tan^{-1}(1/\pi) \)
(c) \( \tan^{-1}(2/\pi) \)
(d) \( \tan^{-1}(4/\pi) \)
Show Answer & Explanation

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

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

Teacher's Note:
1. A capacitive circuit has the current leading.

Question 20: A 50 Hz a.c. source of 20 volts is connected across R and C in series. The voltage across R is 12 volts. 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. \( V^2 = V_R^2 + V_C^2 \).
2. \( V_C = \sqrt{20^2 - 12^2} = 16 \) V.

Teacher's Note:
1. R and C voltages are 90° out of phase.

Question 21: A 200 Ω resistance and 1 H inductance are connected in series with an A.C. circuit. The frequency of the source is \( 200/2\pi \) Hz. Then phase difference between V and I will be
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Show Answer & Explanation

Answer: (c) 45°

Explanation:
1. \( \omega = 200 \) rad/s, so \( X_L = 200\ \Omega = R \).
2. \( \tan\phi = 1 \), so \( \phi = 45^\circ \).

Teacher's Note:
1. Equal reactance and resistance give 45°.

Question 22: An alternating voltage is connected in series with a resistance r and an inductance L. If the potential drop across the resistance is 200 volt and across the inductance is 150 volt, the applied voltage is
(a) 350 volt
(b) 250 volt
(c) 500 volt
(d) 300 volt
Show Answer & Explanation

Answer: (b) 250 volt

Explanation:
1. \( V = \sqrt{200^2 + 150^2} = 250 \) V.

Teacher's Note:
1. Add R and L voltages as perpendicular vectors.

Question 23: The time constant of the given circuit (L and R in series) is τ. If the battery is replaced by an ac source having voltage \( V = V_0\cos\omega t \), the power factor of the circuit will be
(a) ωτ
(b) \( \dfrac1{\sqrt{1+(\omega\tau)^2}} \)
(c) \( \sqrt{1+(\omega\tau)^2} \)
(d) none
Show Answer & Explanation

Answer: (b) \( \dfrac1{\sqrt{1+(\omega\tau)^2}} \)

Explanation:
1. \( \cos\phi = R/\sqrt{R^2+\omega^2L^2} = 1/\sqrt{1+(\omega L/R)^2} \).
2. With \( L/R = \tau \): \( \cos\phi = 1/\sqrt{1+(\omega\tau)^2} \).

Teacher's Note:
1. Divide the numerator and denominator by R.

Question 24: A coil has an inductance of 0.7 henry and is joined in series with a resistance of 220 Ω. When the alternating emf of 220 V at 50 Hz is applied to it then the phase through which the current lags behind the applied emf and the wattless component of current in the circuit will be respectively
(a) 30°, 1 A
(b) 45°, 0.5 A
(c) 45°, 1.5 A
(d) 60°, 1.5 A
Show Answer & Explanation

Answer: (b) 45°, 0.5 A

Explanation:
1. \( X_L = 2\pi(50)(0.7) = 220\ \Omega = R \), so \( \phi = 45^\circ \).
2. \( I = 220/(220\sqrt2) = 1/\sqrt2 \) A.
3. Wattless component \( = I\sin45^\circ = 0.5 \) A.

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

Question 25: A small signal voltage \( V(t) = V_0\sin\omega t \) is applied across an ideal capacitor C:
(a) Current I(t) lags voltage V(t) by 90°.
(b) Over a full cycle the capacitor C does not consume any energy from the voltage source.
(c) Current I(t) is in phase with voltage V(t).
(d) Current I(t) leads voltage V(t) by 180°.
Show Answer & Explanation

Answer: (b) Over a full cycle the capacitor C does not consume any energy from the voltage source.

Explanation:
1. In an ideal capacitor the current leads the voltage by 90°, so (1), (3) and (4) are wrong.
2. The average power over a cycle is zero, so no energy is consumed.

Teacher's Note:
1. Pure reactance means no net energy loss.

Question 26: An inductor of reactance 40 Ω, a capacitor of reactance 40 Ω and a resistor R are connected in series to a supply. If the value of R is changed, then
(a) voltage across L remains same
(b) voltage across C remains same
(c) voltage across LC combination remains same
(d) voltage across LC combination changes
Show Answer & Explanation

Answer: (c) voltage across LC combination remains same

Explanation:
1. Since \( X_L = X_C \), the LC combination has zero net reactance and zero voltage drop.
2. This stays true for any R, so the voltage across the LC combination remains the same (zero), while the voltage across L or C alone changes with the current.

Teacher's Note:
1. At resonance the LC voltages cancel.

Question 27: In the given LCR circuit with R = 25 Ω, \( X_L = 400\ \Omega \), \( X_C = 400\ \Omega \) and a 100 V, 50 Hz source, the voltage across the terminals of the resistance and the current will be
(a) 400 V, 2 A
(b) 800 V, 2 A
(c) 100 V, 2 A
(d) 100 V, 4 A
Show Answer & Explanation

Answer: (d) 100 V, 4 A

Explanation:
1. \( X_L = X_C \), so the circuit is at resonance and \( Z = R = 25 \) Ω.
2. \( I = 100/25 = 4 \) A, and the whole 100 V appears across R.

Teacher's Note:
1. At resonance the source voltage falls entirely across R.

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