AC voltage V(t) = 20 sin ω t of frequency 50 Hz is applied to a… | JEE Main 2021 (20 Jul, Shift 1)

Q58JEE Main 2021 · 20 JUL · SHIFT 1Medium

Pure R, L and C Circuits

AC voltage V(t) = 20 sin $\omega$ t of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m2. The amplitude of the oscillating displacement current for the applied AC voltage is _________. [Take $\varepsilon$ 0 = 8.85 $\times$ 10 $-$ 12 F/m]

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Answer: (c)

$C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 1}{2 \times 10^{-3}} = 4.425 \times 10^{-9}$ F
Amplitude of displacement current $= V_0\,\omega C$ with $\omega = 2\pi \times 50 = 100\pi$ rad/s.
$I_0 = 20 \times 100\pi \times 4.425 \times 10^{-9} \approx 2.78 \times 10^{-5}$ A
So the answer is (c) $27.79\ \mu$ A.

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