physics : class12 Current in Pure Capacitive Circuit : Wattless Current : Electromagnetic Induction and Alternating Currents MCQ Questions and Answers
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Current in Pure Capacitive Circuit
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MCQ Questions
1. A pure capacitor of capacitance C is connected to an alternating voltage v(t) = V₀ sin(ωt). Which expression correctly gives the current through the capacitor?
A.
i(t) = (V₀/ωC) sin(ωt)
B.
i(t) = ωCV₀ cos(ωt)
C.
i(t) = ωCV₀ sin(ωt)
D.
i(t) = (V₀/C) cos(ωt)
Correct Answer:
i(t) = ωCV₀ cos(ωt)
Explanation:
For a capacitor, i(t) = C dv/dt. Differentiating v(t) = V₀ sin(ωt) gives i(t) = CωV₀ cos(ωt), showing that the current leads the voltage by 90°.
2. A pure capacitor of capacitance 20 μF is connected to a 200 V rms, 50 Hz AC supply. What is the rms current through the capacitor?
A.
0.314 A
B.
0.628 A
C.
1.257 A
D.
2.513 A
Correct Answer:
1.257 A
Explanation:
For a pure capacitive circuit, Iᵣₘₛ = Vᵣₘₛ/XC = 2πfCVᵣₘₛ. Substituting f = 50 Hz, C = 20 μF, and Vᵣₘₛ = 200 V gives Iᵣₘₛ ≈ 1.257 A.
3. The voltage across a 10 μF capacitor is v(t) = 100 sin(1000t) volts. Which expression represents the current through it?
A.
i(t) = 0.1 sin(1000t) A
B.
i(t) = 1.0 sin(1000t) A
C.
i(t) = 1.0 cos(1000t) A
D.
i(t) = 10 cos(1000t) A
Correct Answer:
i(t) = 1.0 cos(1000t) A
Explanation:
Using i = C dv/dt, i(t) = 10 × 10⁻⁶ × 100 × 1000 cos(1000t) = 1.0 cos(1000t) A. The cosine form confirms a 90° current lead.
4. In a pure capacitive AC circuit, the rms voltage is 120 V and the rms current is 2 A. What is the average power consumed by the capacitor?
A.
0 W
B.
120 W
C.
240 W
D.
480 W
Correct Answer:
0 W
Explanation:
In a pure capacitive circuit, current leads voltage by 90°, so the power factor is cos 90° = 0. Therefore, average power P = VI cos φ = 0, although current and voltage are both nonzero.
5. A capacitor has capacitive reactance 100 Ω at a frequency of 50 Hz. If the frequency is increased to 200 Hz without changing the capacitance, what will be its new capacitive reactance?
A.
25 Ω
B.
50 Ω
C.
200 Ω
D.
400 Ω
Correct Answer:
25 Ω
Explanation:
Capacitive reactance is XC = 1/(2πfC), so it is inversely proportional to frequency. Increasing the frequency fourfold reduces the reactance from 100 Ω to 25 Ω.
6. In a pure capacitive circuit, the current is i(t) = I₀ sin(ωt). Which expression gives the corresponding voltage, assuming zero average voltage?
A.
v(t) = ωCI₀ cos(ωt)
B.
v(t) = (I₀/ωC) cos(ωt)
C.
v(t) = -(I₀/ωC) cos(ωt)
D.
v(t) = (I₀/C) sin(ωt)
Correct Answer:
v(t) = -(I₀/ωC) cos(ωt)
Explanation:
Since i = C dv/dt, integration gives v(t) = -(I₀/ωC) cos(ωt), apart from a constant. Thus, the voltage lags the current by 90°.
7. A pure capacitor is connected successively to AC sources of the same rms voltage but frequencies f and 3f. What is the ratio of the rms currents in the two cases, I₍₃f₎/I₍f₎?
A.
1/3
B.
1
C.
3
D.
9
Correct Answer:
3
Explanation:
For a fixed capacitance and voltage, Iᵣₘₛ = 2πfCVᵣₘₛ. Therefore, tripling the frequency triples the rms current, giving I₍₃f₎/I₍f₎ = 3.
8. The charge on a capacitor varies as q(t) = Q₀ cos(ωt). Which statement correctly describes the current and its phase relation to the charge?
A.
i(t) = ωQ₀ cos(ωt), in phase with charge
B.
i(t) = -ωQ₀ sin(ωt), and current leads charge by 90°
C.
i(t) = (Q₀/ω) sin(ωt), and current lags charge by 90°
D.
i(t) = -Q₀ sin(ωt), in phase opposition with charge
Correct Answer:
i(t) = -ωQ₀ sin(ωt), and current leads charge by 90°
Explanation:
Current is the time derivative of charge: i = dq/dt = -ωQ₀ sin(ωt). Relative to cos(ωt), this current is shifted by -90° in its expression, meaning the current reaches its extrema 90° ahead of the charge waveform.
9. A 5 μF capacitor is connected to a 100 V rms supply. The rms current is 0.314 A. What is the approximate frequency of the supply?
A.
50 Hz
B.
100 Hz
C.
200 Hz
D.
314 Hz
Correct Answer:
100 Hz
Explanation:
Using Iᵣₘₛ = 2πfCVᵣₘₛ, f = Iᵣₘₛ/(2πCVᵣₘₛ). Substituting 0.314 A, 5 μF, and 100 V gives f ≈ 100 Hz.
10. For a pure capacitive circuit supplied by v(t) = V₀ sin(ωt), which expression correctly represents the instantaneous power and its average over one complete cycle?
A.
p(t) = (ωCV₀²/2) sin(2ωt), and average power is zero
B.
p(t) = (ωCV₀²/2) cos(2ωt), and average power is V₀²ωC/2
C.
p(t) = ωCV₀² sin²(ωt), and average power is V₀²ωC/2
D.
p(t) = V₀²/(ωC), constant and nonzero
Correct Answer:
p(t) = (ωCV₀²/2) sin(2ωt), and average power is zero
Explanation:
Here i(t) = ωCV₀ cos(ωt), so p(t) = vi = ωCV₀² sin(ωt)cos(ωt) = (ωCV₀²/2) sin(2ωt). Its positive and negative values cancel over a complete cycle, giving zero average power.
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