An ac voltage of peak value 100 V is applied across a pure inductor. The peak value of the current is 2 A. The largest value of the instantaneous power supplied to the inductor is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 100 W
In a pure inductor the current lags the voltage by \( \pi/2 \). The instantaneous power supplied to it is \( p_L=-(i_mv_m/2)\sin2\omega t \). Its average over a cycle is zero. At a given instant, though, it is not zero. Its largest value comes when \( \sin2\omega t=-1 \).
Given:
\( v_m=100\ \mathrm{V} \)
\( i_m=2\ \mathrm{A} \)
\( \begin{aligned} p_{L,\max} &= \frac{i_mv_m}{2} \\ &= \frac{2\times100}{2} \\ &= 100\ \mathrm{W} \end{aligned} \)
The inductor takes 100 W at this instant. It gives the energy back to the source in the next quarter cycle.
NCERT (2026-27):
AC Voltage Applied to an Inductor – “The instantaneous power supplied to the inductor is”
\( \displaystyle p_L=i\,v=i_m\sin\left(\omega t-\frac{\pi}{2}\right)\times v_m\sin(\omega t)=-i_mv_m\cos(\omega t)\sin(\omega t) \)
\( \displaystyle =-\frac{i_mv_m}{2}\sin(2\omega t) \)