Question

A capacitor of capacitance ‘C’, is connected across an ac source of voltage V, given by V=V₀sinωt The displacement current between the plates of the capacitor, would then be given by

• A. \(I_d=V_0\,\omega Csin\omega t\)
• B. \(I_d=V_0\,\omega Ccos\omega t\)
• C. \(I_d=\frac{V_0}{\omega C}cos\omega t\)
• D. \(I_d=\frac{V_0}{\omega C}sin\omega t\)

Answer 1

The Correct Option is B

To determine the displacement current \(I_d\) for a capacitor connected to an AC voltage source, we start by examining the given voltage expression: \(V = V_0 \sin \omega t\), where:

• \(V_0\) is the peak voltage, 
• \(\omega\) is the angular frequency,
• \(t\) is the time.

In an AC circuit, the relationship between current and voltage in a capacitor is different from that in a resistor. For a capacitor, the current (or displacement current in this case) leads the voltage by a phase angle of \(\frac{\pi}{2}\) radians.

The displacement current through the capacitor is given by:

• The capacitor equation: \(Q = CV\), where \(Q\) is the charge on the capacitor.
• Differentiate with respect to time to find the current: \(I_d = \frac{dQ}{dt} = C \frac{dV}{dt}\).

Now, differentiate the voltage \(V = V_0\sin\omega t\) with respect to time:

• \(\frac{dV}{dt} = V_0 \omega \cos \omega t\).

Substitute into the expression for current:

• \(I_d = C \left( V_0 \omega \cos \omega t \right)\).

Simplifying, we get:

• \(I_d = V_0 \omega C \cos \omega t\).

Thus, the displacement current is given by the expression: \(I_d = V_0 \omega C \cos \omega t\). This matches the provided correct answer.

The other options can be ruled out since they do not incorporate the correct behavior of current leading the voltage by \(\frac{\pi}{2}\) radians, as seen in the use of \(\cos\) instead of \(\sin\) for current.

Therefore, the correct option is:

• \(I_d = V_0 \omega C \cos \omega t\)