Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

$I_d = V_0 \omega C \cos \omega t$

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

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

$I_d = V_0 \omega C \sin \omega t$

Step-by-Step Solution

The charge on the capacitor is $q = CV = C V_0 \sin \omega t$ . The displacement current $I_d = \frac{dq}{dt} = \frac{d}{dt}(C V_0 \sin \omega t) = C V_0 \omega \cos \omega t$ .

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