A voltage \( V=25\sin(50\pi t) \) volt is applied across a pure inductor of \( 20\,\mathrm{mH} \). The average power dissipated over a complete cycle is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → zero
In a pure inductor the current lags the voltage by \( \pi/2 \). The instantaneous power \( p=-\frac{1}{2}i_mv_m\sin(2\omega t) \) is positive for one quarter cycle and negative for the next. The energy stored in the magnetic field is returned to the source. So the average power over a complete cycle is zero.
With the phase angle \( \phi=90^\circ \), the average power is:
\( \begin{aligned} P &= VI\cos\phi \\ &= VI\cos90^\circ \\ &= 0 \end{aligned} \)
The values \( 25\,\mathrm{V} \), \( 50\pi\,\mathrm{rad\,s^{-1}} \) and \( 20\,\mathrm{mH} \) fix the size of the current. They do not change the average power, which is zero for any pure inductor.
NCERT (2026-27):
Summary – “3. … The current in the inductor lags the voltage by π/2. The average power supplied to an inductor over one complete cycle is zero.”