A capacitor is joined to an ac source. The charge on one of its plates varies as \(q=q_m\sin\omega t\), where \(q_m=200\ \mu\mathrm{C}\) and \(\omega=500\ \mathrm{rad\,s^{-1}}\). The amplitude of the current in the circuit is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(0.1\ \mathrm{A}\)
The current is the rate at which charge flows onto the plate, \( i=dq/dt \). For \( q=q_m\sin\omega t \), this gives \( i=\omega q_m\cos\omega t \). So the current amplitude is \( i_m=\omega q_m \). Since \( q_m=Cv_m \), this is the same as \( i_m=\omega Cv_m \).
Given:
\( q_m=200\ \mu\mathrm{C}=2\times10^{-4}\ \mathrm{C} \)
\( \omega=500\ \mathrm{rad\,s^{-1}} \)
\( \begin{aligned} i_m &= \omega q_m \\ &= 500\times2\times10^{-4} \\ &= 0.1\ \mathrm{A} \end{aligned} \)
NCERT (2026-27):
AC Voltage Applied to a Capacitor – “From the Kirchhoff’s loop rule, the voltage across the source and the capacitor are equal,”
\( \displaystyle v_m\sin\omega t=\frac{q}{C} \)
“To find the current, we use the relation”
\( \displaystyle i=\frac{dq}{dt} \)
\( \displaystyle i=\frac{d}{dt}(v_mC\sin\omega t)=\omega Cv_m\cos(\omega t) \)
“… where the amplitude of the oscillating current is im = ωCvm. …”