CBSE Class 12 Physics Alternating Current Objective Questions

CBSE Class 12 Physics Alternating Current Objective Questions. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

OBJECTIVE QUESTIONS

1. A 10 μF capacitor is connected across a 200 V, 50 Hz a.c. supply. The peak current through the circuit is
(a) 0.6 amp
(b) 0.6 √2 amp
(c) (0.6 / √2) amp
(d) (0.6 π/2) amp.

Answer: (b) 0.6 √2 amp

1. Capacitive reactance X_C = \( \frac{1}{2\pi f C} \) = \( \frac{1}{2\pi \times 50 \times 10 \times 10^{-6}} \) ≈ 318 Ω.
2. r.m.s. current I = V / X_C = 200 / 318 ≈ 0.63 A, which is about 0.6 A.
3. Peak current I₀ = √2 × I_r.m.s. ≈ 0.6 √2 A ≈ 0.85 A.
4. Hence, the correct option is (b).

Teacher's Note:
a) Peak value = √2 times the r.m.s. value; the supply voltage given is r.m.s.
b) Do not stop at 0.6 A; that is only the r.m.s. current.

2. In a L R circuit the a.c. source has voltage 220 V and the potential difference across the inductance is 176 volts. The potential difference across the resistance will be
(a) 44 V
(b) 396 V
(c) 132 V
(d) \( \sqrt{250 \times 176} \) V

Answer: (c) 132 V

1. In an L-R series circuit V² = V_R² + V_L².
2. V_R² = 220² - 176² = 48400 - 30976 = 17424
3. V_R = √17424 = 132 V
4. Hence, the correct option is (c).

Teacher's Note:
a) The voltages across R and L are 90° out of phase, so they add as vectors, not directly (44 V is the wrong subtraction).

3. An a.c. voltage source E = 200 √2 sin 100 t is connected across a circuit containing an a.c. ammeter and capacitor of capacity 1μF. the reading of ammeter is
(a) 10 mA
(b) 20 mA
(c) 40 mA
(d) 80 mA

Answer: (b) 20 mA

1. From E = E₀ sin ωt, E₀ = 200√2 V and ω = 100 rad/s.
2. r.m.s. voltage = E₀ / √2 = 200 V
3. X_C = \( \frac{1}{\omega C} \) = \( \frac{1}{100 \times 10^{-6}} \) = 10⁴ Ω
4. An a.c. ammeter reads the r.m.s. current = 200 / 10⁴ = 0.02 A = 20 mA.
5. Hence, the correct option is (b).

Teacher's Note:
a) An a.c. ammeter shows the r.m.s. value, not the peak value.
b) Here ω = 100 is given directly, so do not multiply it by 2π.
c) Convert amperes to milliamperes at the end.

4. A 12 Ω resistor and a 0.21 henry inductor are connected in series to an a.c. source operating at 20 volt, 50 cycle. The phase angle between the current and source voltage is
(a) tan⁻¹ √3
(b) tan⁻¹ (1/√3)
(c) tan⁻¹ (5.5)
(d) tan⁻¹ (1)

Answer: (c) tan⁻¹ (5.5)

1. X_L = 2πfL = 2 × 3.14 × 50 × 0.21 ≈ 66 Ω
2. tan φ = X_L / R = 66 / 12 = 5.5
3. So φ = tan⁻¹ (5.5).
4. Hence, the correct option is (c).

Teacher's Note:
a) The 20 V source voltage is not needed for the phase angle.
b) Phase angle in an L-R circuit is tan⁻¹ (X_L/R).

5. In order to obtain a time constant of 10 seconds in an R.C. circuit containing a resistance of 10³ Ω, the capacity of the condenser should be
(a) 10 μF
(b) 100 μF
(c) 1000 μF
(d) 10000 μF

Answer: (d) 10000 μF

1. Time constant τ = RC
2. C = τ / R = 10 / 10³ = 10⁻² F
3. 10⁻² F = 10⁻² × 10⁶ μF = 10000 μF
4. Hence, the correct option is (d).

Teacher's Note:
a) The printed key gives (c) 1000 μF, but 1000 μF with 10³ Ω gives only 1 second; the correct capacity is 10000 μF.
b) Always convert farad to microfarad by multiplying by 10⁶.

6. An A.C. series circuit contains 40 Ω of resistance and 30 Ω of inductive reactance. Then the impedance of circuit is
(a) 70 Ω
(b) 10 Ω
(c) 50 Ω
(d) 70 Ω

Answer: (c) 50 Ω

1. Impedance Z = √(R² + X_L²)
2. Z = √(40² + 30²) = √(1600 + 900)
3. Z = √2500 = 50 Ω
4. Hence, the correct option is (c).

Teacher's Note:
a) R and X_L are at right angles, so add them as vectors; 70 Ω is the trap of simple addition.

7. A group of electric lamps having a total power rating of 1000 watt is supplied by an a.c. voltage E = 200 sin (310 t + 60°). Then the r.m.s. value of the circuit current is
(a) 10 amp
(b) 10 √2 amp
(c) 20 amp
(d) 20 √2 amp.

Answer: I_r.m.s. = 5√2 A ≈ 7.07 A (none of the four options matches; the peak current is 10 A, option (a))

1. The peak voltage is E₀ = 200 V, so the r.m.s. voltage is E_r.m.s. = 200 / √2 = 100√2 V ≈ 141 V.
2. Lamps are purely resistive, so P = E_r.m.s. × I_r.m.s.
3. I_r.m.s. = 1000 / (100√2) = 5√2 A ≈ 7.07 A
4. The peak current is I₀ = √2 × 5√2 = 10 A, which is option (a), but the question asks for the r.m.s. value.

Teacher's Note:
a) The printed key gives (b) 10√2 amp; this does not follow from the data, since 10√2 A ≈ 14.1 A would need an r.m.s. voltage of about 71 V.
b) Do not confuse peak current (10 A) with r.m.s. current (7.07 A).
c) The data in the question gives 5√2 A ≈ 7.07 A, which is not among the four options.

8. In the alternating current circuit shown in following figure, the currents through the inductor and the capacitor are 1.2 amp and 1.0 amp. respectively. The current drawn from the generator is
(a) 0.4 amp.
(b) 0.2 amp.
(c) 1.0 amp.
(d) 1.2 amp.

[Figure: Parallel circuit. The inductor L is in the top branch, the capacitor C is in the middle branch, and the a.c. generator is in the bottom branch, all connected across the same two side wires.]

Answer: (b) 0.2 amp.

1. The inductor and the capacitor are connected in parallel across the generator.
2. Current in the inductor lags the voltage by 90° and current in the capacitor leads it by 90°, so they are opposite in phase.
3. Current drawn = 1.2 - 1.0 = 0.2 A
4. Hence, the correct option is (b).

Teacher's Note:
a) In parallel L-C branches the currents subtract; in series the voltages subtract.
b) Do not add the currents (2.2 A is the trap).

9. An alternating voltage E=200 √2 sin 100 t, where V is in volt and t in seconds, is connected to a series combination of 1 μF capacitor and a 10 kΩ resistor through an ac ammeter. The reading of the ammeter will be
(a) √2 mA
(b) 10 √2 mA
(c) 2 mA
(d) 20 mA

Answer: (b) 10 √2 mA

1. r.m.s. voltage = 200√2 / √2 = 200 V and ω = 100 rad/s.
2. X_C = \( \frac{1}{\omega C} \) = \( \frac{1}{100 \times 10^{-6}} \) = 10 kΩ
3. R = 10 kΩ, so Z = √(R² + X_C²) = 10√2 kΩ
4. I = V / Z = 200 / (10√2 × 10³) = \( \frac{20}{\sqrt{2}} \) × 10⁻³ A = 10√2 mA
5. Hence, the correct option is (b).

Teacher's Note:
a) An ammeter reads the r.m.s. current of the circuit.
b) Series R and C: use Z = √(R² + X_C²), not R + X_C.

10. An inductor of 1 henry is connected across a 220 V, 50 Hz supply. The peak value of the curent is approximately
(a) 0.5 A
(b) 0.7 A
(c) 1 A
(d) 1.4 A

Answer: (c) 1 A

1. X_L = 2πfL = 2 × 3.14 × 50 × 1 = 314 Ω
2. r.m.s. current = 220 / 314 ≈ 0.70 A
3. Peak current I₀ = √2 × 0.70 ≈ 0.99 A ≈ 1 A
4. Hence, the correct option is (c).

Teacher's Note:
a) 0.7 A is the r.m.s. value; the question asks for the peak.

11. An inductive coil has a resistance of 100 Ω. When an ac signal of frequency 1000 Hz is applied to the coil, the voltage leads the current by 45°. The inductance of the coil is
(a) \( \frac{1}{10\pi} \)
(b) \( \frac{1}{20\pi} \)
(c) \( \frac{1}{40\pi} \)
(d) \( \frac{1}{60\pi} \)

Answer: (b) \( \frac{1}{20\pi} \)

1. tan φ = X_L / R with φ = 45°, so X_L = R = 100 Ω
2. X_L = 2πfL, so L = X_L / (2πf) = 100 / (2π × 1000)
3. L = \( \frac{1}{20\pi} \) H
4. Hence, the correct option is (b).

Teacher's Note:
a) A phase angle of 45° means X_L and R are equal.

12. An LCR series circuit consists of a resistance of 10 Ω, a capacitance of reactance 60 Ω and an inductor coil. The circuit is found to resonate when put across a 300 V, 100 Hz supply. The inductance of the coil is (take π = 3).
(a) 0.1 H
(b) 0.01 H
(c) 0.2 H
(d) 0.02 H

Answer: (a) 0.1 H

1. At resonance X_L = X_C = 60 Ω
2. X_L = 2πfL, so L = 60 / (2 × 3 × 100) = 60 / 600
3. L = 0.1 H
4. Hence, the correct option is (a).

Teacher's Note:
a) At resonance the resistance 10 Ω and the 300 V supply do not affect L.
b) Use the given value π = 3 as instructed.

13. When 100 V dc is applied across a coil, a current of 1 A flows through it. When 100 V ac of 50 Hz is applied across the same coil, only 0.5 A flows. the resistance and inductance of the coil are (take π² = 10).
(a) 50Ω, 0.3 H
(b) 50 Ω, √0.3 H
(c) 100 Ω, 0.3 H
(d) 100 Ω, √0.3 H

Answer: (d) 100 Ω, √0.3 H

1. With dc only the resistance acts: R = 100 / 1 = 100 Ω
2. With ac, impedance Z = 100 / 0.5 = 200 Ω
3. X_L = √(Z² - R²) = √(200² - 100²) = 100√3 Ω
4. L = X_L / (2πf) = 100√3 / (2π × 50) = \( \frac{\sqrt{3}}{\pi} \) H
5. Using π² = 10, L = \( \frac{\sqrt{3}}{\sqrt{10}} \) = √0.3 H
6. Hence, the correct option is (d).

Teacher's Note:
a) A coil has no effect of inductance with dc, so dc data gives R directly.
b) Take care: the options differ in 0.3 and √0.3.

14. In an ac circuit V = 100 sin (100 t) volt and I = 100 sin (100 t + π/3) mA. The power dissipated in the circuit is
(a) 10⁴ W
(b) 10 W
(c) 2.5 W
(d) 5 W

Answer: (c) 2.5 W

1. V₀ = 100 V and I₀ = 100 mA = 0.1 A; the phase difference φ = π/3.
2. Average power P = \( \frac{V_0 I_0}{2} \) cos φ
3. P = \( \frac{100 \times 0.1}{2} \) × cos 60° = 5 × 0.5
4. P = 2.5 W
5. Hence, the correct option is (c).

Teacher's Note:
a) Convert milliampere to ampere first.
b) Use the peak values with a factor of half, or the r.m.s. values.

15. An electric bulb which runs at 80 V dc and consumes 10 A current is connected across a 100 V, 50 Hz ac supply. The inductance of the choke required so that it consumes rated power, is (take π = 3).
(a) 0.01 H
(b) 0.02 H
(c) 0.04 H
(d) 0.08 H

Answer: (b) 0.02 H

1. Resistance of the bulb R = 80 / 10 = 8 Ω
2. To consume rated power the current must stay 10 A, so Z = 100 / 10 = 10 Ω
3. X_L = √(Z² - R²) = √(100 - 64) = 6 Ω
4. L = X_L / (2πf) = 6 / (2 × 3 × 50) = 0.02 H
5. Hence, the correct option is (b).

Teacher's Note:
a) The bulb must carry the same 10 A, so the choke is added in series.
b) A choke (inductor) is used because it wastes no power, unlike a series resistor.

16. The tuning circuit of a radio receiver has a resistance of 50 Ω , an inductor of 10 mH and a variable capacitor. A 1 MHz radio wave produces a potential difference of 0.1 mV. The value of the capacitor to produce resonance is (take π²=10).
(a) 2.5 × 10⁻¹² F
(b) 5.0 × 10⁻¹² F
(c) 2.5 × 10⁻¹¹ F
(d) 5.0 × 10⁻¹¹ F

Answer: (a) 2.5 × 10⁻¹² F

1. At resonance f = \( \frac{1}{2\pi\sqrt{LC}} \), so C = \( \frac{1}{4\pi^2 f^2 L} \)
2. C = \( \frac{1}{4 \times 10 \times (10^6)^2 \times 10 \times 10^{-3}} \)
3. C = \( \frac{1}{4 \times 10^{11}} \) = 2.5 × 10⁻¹² F
4. Hence, the correct option is (a).

Teacher's Note:
a) Convert 1 MHz to 10⁶ Hz and 10 mH to 10⁻² H before substituting.
b) The resistance and the 0.1 mV signal do not affect the capacitor value at resonance.

17. The impedance of a circuit consists of 3 Ω resistance and 4 Ω reactance. The power factor of the circuit is
(a) 0.4
(b) 0.6
(c) 0.8
(d) 1.0

Answer: (b) 0.6

1. Impedance Z = √(3² + 4²) = 5 Ω
2. Power factor = cos φ = R / Z = 3 / 5 = 0.6
3. Hence, the correct option is (b).

Teacher's Note:
a) Power factor is resistance divided by impedance; 0.8 would be X / Z.

18. Two coils A and B are connected in series across a 240 V, 50 Hz supply. The resistance of A is 5 Ω and the inductance of B is 0.02 H. The power consumed is 3 kW and the power factor is 0.75. The impedance of the circuit is
(a) 0.144 Ω
(b) 1.44 Ω
(c) 14.4 Ω
(d) 144 Ω

Answer: (c) 14.4 Ω

1. P = VI cos φ, so I = 3000 / (240 × 0.75)
2. I = 16.67 A
3. Impedance Z = V / I = 240 / 16.67
4. Z = 14.4 Ω
5. Hence, the correct option is (c).

Teacher's Note:
a) Convert 3 kW to 3000 W.
b) The given resistance and inductance are not needed for finding Z.

19. A resistance R, an inductance L and a capacitance C are connected in series across an ac source of angular frequency ω. If the resonant frequency is ω₀, then the current will lag behind the voltage if
(a) ω < ω₀
(b) ω > ω₀
(c) ω = ω₀
(d) ω < 0

Answer: (b) ω > ω₀

1. Current lags the voltage when the circuit is inductive, that is X_L is greater than X_C.
2. X_L = ωL grows with ω and X_C = 1/(ωC) falls with ω.
3. At ω₀ they are equal, so for ω greater than ω₀, X_L is greater than X_C.
4. Hence, the correct option is (b).

Teacher's Note:
a) Below resonance the circuit is capacitive and the current leads.

20. An LCR series circuit containing a resistance of 120Ω has angular resonance frequency 4 × 10⁵ rad s⁻¹. At resonance the voltage across resistance and inductance are 60 V and 40 V respectively. The values of L and C are
(a) 0.2 m H, 1/32 μF
(b) 0.4 m H, 1/16 μF
(c) 0.2 mH, 1/16 μF
(d) 0.4 mH, 1/32 μF

Answer: (a) 0.2 m H, 1/32 μF

1. At resonance I = V_R / R = 60 / 120 = 0.5 A
2. X_L = V_L / I = 40 / 0.5 = 80 Ω
3. L = X_L / ω₀ = 80 / (4 × 10⁵) = 2 × 10⁻⁴ H = 0.2 mH
4. C = \( \frac{1}{\omega_0^2 L} \) = \( \frac{1}{16 \times 10^{10} \times 2 \times 10^{-4}} \) = \( \frac{1}{3.2 \times 10^{7}} \) F = \( \frac{1}{32} \) μF
5. Hence, the correct option is (a).

Teacher's Note:
a) At resonance X_L = X_C, so C can also be found from X_C = 80 Ω.
b) Check: C = 1/(ω₀ X_C) = 1/(4 × 10⁵ × 80) = 1/(3.2 × 10⁷) F.