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 between the plates of a capacitor, we need to understand the relationship between the voltage applied across the capacitor, the resulting electric field, and the current. The given problem involves an alternating current (AC) voltage source applied across a capacitor, with the voltage described as \( V = V_0 \sin \omega t \).

The capacitance of a capacitor, \( C \), is defined by the equation:

\(Q = CV\)

where \( Q \) is the charge on the capacitor plates. Differentiating this with respect to time gives us the current:

\(I = \frac{dQ}{dt} = C \frac{dV}{dt}\)

Given the voltage \( V = V_0 \sin \omega t \), the derivative with respect to time is:

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

Hence, the current \( I \) through the capacitor is:

\(I = C \times V_0 \omega \cos \omega t = V_0 \omega C \cos \omega t\)

The displacement current, which is equivalent to the conduction current in terms of magnitude in this setup, is given by:

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

Thus, the displacement current between the plates of the capacitor is correctly described by the expression \( I_d = V_0 \omega C \cos \omega t \).

The correct answer is therefore:

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