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MHT CET 2026 (17 Apr, Shift 2) · Pure R, L and C Circuits

Pure R, L and C Circuits question from MHT CET 2026 (17 Apr, Shift 2)

Q3MHT CET 2026 · 17 APR · SHIFT 2Easy

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)

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

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