5. 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: (2021)

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

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

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

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

Solution:

The displacement current between the capacitor plates is equal to the conduction current, $I_d = \frac{dq}{dt}$.
Since charge on the capacitor is $q = C V = C V_0 \sin \omega t$, we differentiate with respect to time.
$I_d = \frac{d}{dt}(C V_0 \sin \omega t) = V_0 \omega C \cos \omega t$.

Rankers Physics

5. 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: (2021)

Solution:

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