A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 200 V, 50 Hz supply. The resistance of the circuit is negligible. The average power (average over one cycle) transferred to capacitor is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0
The average power in an ac circuit is \( P = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi \). The current through an ideal capacitor leads its voltage by \( \phi = 90^\circ \), so \( \cos\phi = 0 \).
\( P = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos90^\circ = 0 \)
Physically, the capacitor takes energy from the source during one quarter of the cycle and returns exactly the same energy in the next quarter. The values of L, C and the supply are not needed. With negligible resistance nothing is dissipated in the circuit.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
AC Voltage Applied to a Capacitor – “… The instantaneous power supplied to the capacitor is … So, as in the case of an inductor, the average power
\( \displaystyle P_C = \left\langle \frac{i_m v_m}{2}\sin(2\omega t) \right\rangle = \frac{i_m v_m}{2}\left\langle \sin(2\omega t) \right\rangle = 0 \)
since \( \langle \sin(2\omega t) \rangle = 0 \) over a complete cycle.”