A voltage \(v=100\sin(100\pi t)\) volt is applied across a pure capacitor. The capacitive reactance is 50 Ω. The current in the circuit at \(t=\frac{1}{300}\ \mathrm{s}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1 A
In a pure capacitor the current leads the voltage by \( \pi/2 \). For \( v=v_m\sin\omega t \), the current is \( i=i_m\sin(\omega t+\pi/2) \). This is the same as \( i=i_m\cos\omega t \). Its amplitude is \( i_m=v_m/X_C \).
\( \begin{aligned} i_m &= \frac{v_m}{X_C} \\ &= \frac{100}{50} \\ &= 2\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} \omega t &= 100\pi\times\frac{1}{300} \\ &= \frac{\pi}{3} \end{aligned} \)
\( \begin{aligned} i &= i_m\cos\omega t \\ &= 2\cos\frac{\pi}{3} \\ &= 2\times0.5 \\ &= 1\ \mathrm{A} \end{aligned} \)
NCERT (2026-27):
AC Voltage Applied to a Capacitor – “… Using the relation, cos(ωt) = sin(ωt + π/2), we have”
\( \displaystyle i=i_m\sin\left(\omega t+\frac{\pi}{2}\right) \)\( \qquad (7.16) \)
AC Voltage Applied to a Capacitor – “… so that the amplitude of the current is”
\( \displaystyle i_m=\frac{v_m}{X_C} \)\( \qquad (7.18) \)