Question

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:

• A. $I _{ d }= V _{0} \omega C \cos \omega t$
• B. $I_{d}=\frac{V_{0}}{\omega C} \cos \omega t$
• C. $I _{ d }=\frac{ V _{0}}{\omega C } \sin \omega t$
• D. $I _{ d }= V _{0} \omega C \sin \omega t$

Answer 1

Answer: (A)

$I_d=V_o \omega C \cos \omega t$
Given $V=V_0 \sin \omega t$
Now, displacement current $i_d$ is given by,
$I_d=C \frac{d V}{d t}=C \frac{d}{d t}\left(V_0 \sin \omega t\right)$
$= C \left( V _0 \omega\right) \cos \omega t$
$=V_0 \omega C \cos \omega t$