Verified Solution • 2025–26 Session moderate

QNEET-2021-Ph-637: NEET 2021 Physics - Class 12 NEET-2021-Ph-637 Solution: A capacitor of capacitance 'C' is connected across an ac source …

A capacitor of capacitance 'C' is connected across an ac source of voltage V given by

V = V_0\sin\omega t

The displacement current between the plates of the capacitor would then be given by:

I_d = V_0\omega C\sin\omega t

I_d = V_0\omega C\cos\omega t

I_d = \frac{V_0}{\omega C}\cos\omega t

I_d = \frac{V_0}{\omega C}\sin\omega t

Multiple Choice Options

A ${I_d} = {V_0}\omega C\sin \omega t$ I d _ = V 0 _ _ C sin _ t
B ${I_d} = {V_0}\omega C\cos \omega t$ I d _ = V 0 _ _ C cos _ t
C ${I_d} = {{{V_0}} \over {\omega C}}\cos \omega t$ I d _ = _ C V 0 _ _ cos _ t
D ${I_d} = {{{V_0}} \over {\omega C}}\sin \omega t$ I d _ = _ C V 0 _ _ sin _ t

Step-by-Step Verified Solution

Given, V = V 0 sin _ t We know, q = C V Now displacement current I d is given by, I d = d q d t = C d V d t I d = C ( V 0 _ cos _ _ t ) = V 0 _ C cos _ _ t

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