Question: 26. A capacitor of capacitance ‘ C ‘, is connected across an ac source of voltage V, given by
\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}The displacement current between the plates of the capacitor, would then be given by :
(1) I_{d}=V_{0} \omega C \cos \omega t
(2) I_{d}=\frac{V_{0}}{\omega C} \cos \omega t
(3) I_{d}=\frac{V_{0}}{\omega C} \sin \omega t
(4) I_{d}=V_{0} \omega C \sin \omega t
Answer: Option (1)
Explanation:
The charge on a capacitor is related to the potential difference across it by the relation Q = CV.
Given the applied alternating voltage V = V_{0}\sin \omega t, the charge on the capacitor at any instant is
Q = C V_{0}\sin \omega t.
The displacement current is defined as the rate of change of charge with respect to time
and is given by I_{d}=\frac{dQ}{dt}.
Differentiating the expression for charge,
I_{d} = \frac{d}{dt}\left(CV_{0}\sin \omega t\right).
This gives
I_{d} = C V_{0}\omega \cos \omega t.
Hence, the displacement current between the plates of the capacitor is
I_{d}=V_{0}\omega C \cos \omega t.
Therefore, the correct answer is Option (1).