Consider the following statements about power in ac circuits:
(A) The power rating of an element used in ac circuits refers to its average power rating.
(B) When the voltage and the current differ in phase by \(60^\circ\), the average power is half the product of the rms voltage and the rms current.
(C) The wattless component of the current is in phase with the applied voltage.
(D) The power factor of a circuit in which the current lags the voltage can be improved by joining an inductor of suitable value in parallel.
Which of the above statements are incorrect?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C) and (D) only
The average power is \( P=VI\cos\phi \). The current can be split into two parts. The part in phase with the voltage carries the power. The part at \( 90^\circ \) to the voltage carries none.
(A) The power rating of an element used in ac circuits refers to its average power rating. — Correct statement. A rating such as 100 W on a bulb is the power averaged over a cycle.
(B) When the voltage and the current differ in phase by \( 60^\circ \), the average power is half the product of the rms voltage and the rms current. — Correct statement. For \( \phi=60^\circ \), the power factor is \( \cos60^\circ=0.5 \). So the average power is \( 0.5\,VI \).
(C) The wattless component of the current is in phase with the applied voltage. — Incorrect statement. The wattless component \( I_q \) is perpendicular to the voltage, which is why it uses no power. The part in phase with the voltage is the power component \( I_p \).
(D) The power factor of a circuit in which the current lags the voltage can be improved by joining an inductor of suitable value in parallel. — Incorrect statement. The lagging wattless current has to be cancelled by an equal leading current. A capacitor joined in parallel gives this leading current. An inductor would add to the lag.
NCERT (2026-27):
Points to Ponder – “2. The power rating of an element used in ac circuits refers to its average power rating.”
Power in AC Circuit: the Power Factor – “… Therefore,”
\( \displaystyle P=\frac{v_mi_m}{2}\cos\phi=\frac{v_m}{\sqrt{2}}\frac{i_m}{\sqrt{2}}\cos\phi \)\( \displaystyle =VI\cos\phi \)\( \qquad [7.30(a)] \)
Example 7.7 – “… Let us resolve I into two components. Ip along the applied voltage V and Iq perpendicular to the applied voltage. Iq as you have learnt in Section 7.7, is called the wattless component since corresponding to this component of current, there is no power loss. …”
Example 7.7 – “… This can be done by connecting a capacitor of appropriate value in parallel so that Iq and I′q cancel each other and P is effectively Ip V.”