An alternating voltage of amplitude 40 V and frequency 4 kHz is applied directly across the capacitor of 12. The maximum displacement current between the plates of the capacitor is nearly: (a) 10 A (b) 8 A (c) 13 A (d) 12 A

Answer: (d). Id = CdV/dt; for V = V₀ sin ω t, Id,max = Cω V₀. ω = 2π × 4000 = 2.51 × 10⁴ rad/s. Id,max = 12 × 10⁻⁶ × 2.51 × 10⁴ × 40 ≈ 12.06 A. So the answer is (d) 12 A.

An alternating voltage of amplitude 40 V and frequency 4 kHz is… | JEE Main 2024 (5 Apr, Shift 1)

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Displacement Current, Maxwell's Equations & Production

An alternating voltage of amplitude $40\text{ V}$ and frequency $4\text{ kHz}$ is applied directly across the capacitor of $12\ \mu\text{F}$ . The maximum displacement current between the plates of the capacitor is nearly:

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Answer: (d)

$I_d = C\dfrac{dV}{dt}$ ; for $V = V_0 \sin \omega t$ , $I_{d,max} = C\omega V_0$ .
$\omega = 2\pi \times 4000 = 2.51 \times 10^{4}$ rad/s.
$I_{d,max} = 12 \times 10^{-6} \times 2.51 \times 10^{4} \times 40 \approx 12.06$ A.
So the answer is (d) 12 A.

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