In a series LCR circuit, the source voltage is \(v=v_m\sin\omega t\) and the current is \(i=i_m\sin(\omega t+\phi)\). The instantaneous power supplied by the source is
\(p=\frac{v_mi_m}{2}\left[\cos\phi-\cos(2\omega t+\phi)\right]\)
If the source is joined to a pure inductor only, the instantaneous power \(p\) becomes
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-\frac{v_mi_m}{2}\sin2\omega t\)
In a pure inductor the current lags the voltage by \( \pi/2 \). The current is written as \( i=i_m\sin(\omega t+\phi) \), so \( \phi=-\pi/2 \). Put this value in the given expression.
\( \begin{aligned} \cos\phi &= \cos(-\pi/2) \\ &= 0 \end{aligned} \)
\( \begin{aligned} \cos(2\omega t+\phi) &= \cos(2\omega t-\pi/2) \\ &= \sin2\omega t \end{aligned} \)
\( \begin{aligned} p &= \frac{v_mi_m}{2}\left[0-\sin2\omega t\right] \\ &= -\frac{v_mi_m}{2}\sin2\omega t \end{aligned} \)
Option (1) \(\frac{v_mi_m}{2}\sin2\omega t\) — Incorrect. This comes from \( \phi=+\pi/2 \), where the current leads the voltage. It is the power supplied to a pure capacitor.
Option (2) \(-\frac{v_mi_m}{2}\sin2\omega t\) — Correct. It follows from \( \phi=-\pi/2 \) for the lagging current. This power is positive for half of the time and negative for the other half.
Option (3) \(\frac{v_mi_m}{2}(1-\cos2\omega t)\) — Incorrect. This is the result for \( \phi=0 \), that is, for a pure resistor. Such a power is never negative.
Option (4) zero — Incorrect. For a pure inductor the average power over a cycle is zero. The instantaneous power is not zero. It changes with time at the angular frequency \( 2\omega \).
NCERT (2026-27):
AC Voltage Applied to an Inductor – “… Using −cos(ωt) = sin(ωt − π/2), we have”
\( i=i_m\sin\left(\omega t-\frac{\pi}{2}\right) \)\( \qquad (7.12) \)
AC Voltage Applied to an Inductor – “The instantaneous power supplied to the inductor is”
\( p_L=i\,v=i_m\sin\left(\omega t-\frac{\pi}{2}\right)\times v_m\sin(\omega t)=-i_mv_m\cos(\omega t)\sin(\omega t)= \)\( -\frac{i_mv_m}{2}\sin(2\omega t) \)