In a purely capacitive ac circuit, what is the amplitude of the curre…

The correct answer is Option (1) → \( V/X_C \)

In a purely capacitive circuit the capacitive reactance \( X_C=1/\omega C \) plays the role of resistance. So the current amplitude is the peak voltage divided by \( X_C \).

With \( v=V\sin\omega t \) across the capacitor, the charge on it is \( q=CV\sin\omega t \), and the current is:

    \( \begin{aligned} i &= \frac{dq}{dt} \\ &= \omega CV\cos\omega t \end{aligned} \)

The amplitude of this current is:

    \( \begin{aligned} i_m &= \omega CV \\ &= \frac{V}{1/\omega C} \\ &= \frac{V}{X_C} \end{aligned} \)

Option (1) \( V/X_C \) — Correct. This is \( i_m=v_m/R \) with \( X_C \) in place of \( R \). Its unit is volt per ohm, which is the ampere.

Option (2) \( X_C/V \) — Incorrect. Its unit is ohm per volt, that is, per ampere. So it cannot be a current. It would also make the current fall when the voltage rises.

Option (3) \( VX_C \) — Incorrect. Its unit is volt × ohm, which is not the unit of current. It would also give a larger current for a larger reactance, but the reactance opposes the current.

Option (4) \( V/X_C^2 \) — Incorrect. The working gives \( i_m=\omega CV \), which has the first power of \( 1/X_C \). An extra \( X_C \) in the denominator leaves a unit of ampere per ohm.

NCERT (2026-27):
AC Voltage Applied to a Capacitor – “… Comparing it to i_m = v_m/R for a purely resistive circuit, we find that (1/ωC) plays the role of resistance. It is called capacitive reactance and is denoted by X_c,

    \( X_c=1/\omega C \qquad (7.17) \)

so that the amplitude of the current is

    \( \displaystyle i_m=\frac{v_m}{X_C} \)\( \qquad (7.18) \)”