In an ac circuit with voltage V and current I, the average power diss…

The correct answer is Option (4) → dependent on the phase angle between V and I

For \( v=v_m\sin\omega t \) and \( i=i_m\sin(\omega t+\phi) \), the instantaneous power has a steady term and a term that changes with time. Only the steady term has a non-zero average over one cycle. So the average power is:

    \( \begin{aligned} P &= \frac{v_mi_m}{2}\cos\phi \\ &= VI\cos\phi \end{aligned} \)

Here \( V \) and \( I \) are rms values, and \( \cos\phi \) is the power factor.

Option (1) \( VI \) — Incorrect. This holds only when the voltage and current are in phase (\( \phi=0 \), a purely resistive circuit). With any phase difference, the average power is smaller.

Option (2) \( \frac{1}{2}VI \) — Incorrect. The factor \( 1/2 \) goes with peak values, as in \( \frac{1}{2}v_mi_m\cos\phi \). Even this equals \( \frac{1}{2}v_mi_m \) only for \( \phi=0 \). No fixed fraction of VI gives the power in every circuit.

Option (3) \( \frac{1}{\sqrt{2}}VI \) — Incorrect. This equals \( VI\cos\phi \) only for \( \phi=45^\circ \). For any other phase angle, the average power is different.

Option (4) dependent on the phase angle between V and I — Correct. Since \( P=VI\cos\phi \), the same V and I give anything from the maximum VI (\( \phi=0 \)) down to zero (\( \phi=\pi/2 \), a pure inductor or capacitor).

NCERT (2026-27):
Power in AC Circuit: the Power Factor – “… Therefore,

    \( \displaystyle P=\frac{v_mi_m}{2}\cos\phi=\frac{v_m}{\sqrt{2}}\,\frac{i_m}{\sqrt{2}}\cos\phi=VI\cos\phi \qquad [7.30(a)] \)

… So, the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle φ between them. The quantity cosφ is called the power factor.”