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MHT CET 2025 (23 Apr, Shift 1) · Pure R, L and C Circuits

Pure R, L and C Circuits question from MHT CET 2025 (23 Apr, Shift 1)

Q18MHT CET 2025 · 23 APR · SHIFT 1Easy

Pure R, L and C Circuits

In an a.c. circuit with pure capacitance ' C ' and a.c. source $E = E_0 \sin \omega t$ , the equation of instantaneous current is given by

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Answer: (b)

$q = CE_0\sin\omega t$ , so $I = \frac{dq}{dt} = E_0\omega C\cos\omega t$ .
$I = E_0\omega C\sin\left(\omega t + \frac{\pi}{2}\right)$ : current leads by $\frac{\pi}{2}$ .
So the answer is (b).

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