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:

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

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

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

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

Step-by-Step Solution

The displacement current ( $I_d$ ) inside the capacitor is equal to the conduction current ( $I$ ) flowing in the connecting wires.

Charge on Capacitor: The charge $q$ on the capacitor at any time $t$ is given by $q = CV$ . Substituting the given voltage $V = V_0 \sin \omega t$ : $q = C(V_0 \sin \omega t)$
Calculate Current: The current is the rate of change of charge. $I_d = \frac{dq}{dt} = \frac{d}{dt}(C V_0 \sin \omega t)$ $I_d = C V_0 \frac{d}{dt}(\sin \omega t)$ $I_d = C V_0 (\omega \cos \omega t)$ $I_d = V_0 \omega C \cos \omega t$

Therefore, the displacement current is $V_0 \omega C \cos \omega t$ .

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