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JEE Main 2024 (1 Feb, Shift 1) · Pure R, L and C Circuits

Pure R, L and C Circuits question from JEE Main 2024 (1 Feb, Shift 1)

Q32JEE Main 2024 · 1 FEB · SHIFT 1Medium

Pure R, L and C Circuits

A parallel plate capacitor has a capacitance $C = 200\text{ pF}$ . It is connected to $230\text{ V}$ ac supply with an angular frequency $300\text{ rad/s}$ . The rms value of conduction current in the circuit and displacement current in the capacitor respectively are:

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Answer: (b)

$I_{rms} = V_{rms}\,\omega C = 230 \times 300 \times 200 \times 10^{-12} = 1.38 \times 10^{-5}$ A $= 13.8\ \mu$ A
Displacement current between the plates always equals the conduction current in the wires.
So the answer is (b) $13.8\ \mu$ A and $13.8\ \mu$ A.

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