A pure inductor is joined to an ac source \(v=v_m\sin\omega t\). Later, a pure capacitor is joined to the same source. Consider the following statements:
(A) The instantaneous power supplied to the inductor varies as \(\sin 2\omega t\).
(B) The average power supplied to the capacitor over a complete cycle is \(i_mv_m/2\).
(C) In the inductor circuit, the current has no constant part, because the source emf oscillates symmetrically about zero.
(D) In the capacitor, the current reaches its maximum value later than the voltage by one-fourth of a period.
Which of the above statements are incorrect?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (B) and (D) only
In a pure inductor or a pure capacitor, the current and the voltage differ in phase by \( \pi/2 \). The power \( p=vi \) then goes as \( \sin\omega t\cos\omega t \), which is half of \( \sin 2\omega t \). Its average over a full cycle is zero.
(A) The instantaneous power supplied to the inductor varies as \( \sin 2\omega t \). — Correct statement. For the inductor, \( p_L=-(i_mv_m/2)\sin 2\omega t \). So the power changes at twice the angular frequency of the source.
(B) The average power supplied to the capacitor over a complete cycle is \( i_mv_m/2 \). — Incorrect statement. The average of \( \sin 2\omega t \) over a cycle is zero, so the capacitor takes zero average power. The value \( i_mv_m/2 \) is the average power in a pure resistor.
(C) In the inductor circuit, the current has no constant part, because the source emf oscillates symmetrically about zero. — Correct statement. A current that swings evenly about zero has no steady part. So the integration constant in the current is zero.
(D) In the capacitor, the current reaches its maximum value later than the voltage by one-fourth of a period. — Incorrect statement. In a capacitor the current leads the voltage by \( \pi/2 \), so it reaches its maximum one-fourth of a period earlier. A later maximum is seen in the inductor.
NCERT (2026-27):
AC Voltage Applied to an Inductor – “The instantaneous power supplied to the inductor is”
\( \displaystyle p_L=i\,v=i_m\sin\left(\omega t-\frac{\pi}{2}\right)\times v_m\sin(\omega t)=-i_mv_m\cos(\omega t)\sin(\omega t) \)\( \displaystyle =-\frac{i_mv_m}{2}\sin(2\omega t) \)
Summary – “… As in the case of inductor, the average power supplied to a capacitor over one complete cycle is zero.”
AC Voltage Applied to an Inductor – “… Since the source has an emf which oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about zero, so that no constant or time-independent component of the current exists. Therefore, the integration constant is zero. …”
AC Voltage Applied to a Capacitor – “… We see that the current reaches its maximum value earlier than the voltage by one-fourth of a period.”