An electric appliance is joined to the 220 V ac mains. It draws an rms current of 5 A, and the average power it uses is 550 W. The power factor of the appliance is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(0.5\)
The average power in an ac circuit is \( P=VI\cos\phi \), where \( V \) and \( I \) are rms values. The factor \( \cos\phi \) is the power factor.
Given:
\( P=550\ \mathrm{W} \)
\( V=220\ \mathrm{V} \)
\( I=5\ \mathrm{A} \)
\( \begin{aligned} \cos\phi &= \frac{P}{VI} \\ &= \frac{550}{220\times5} \\ &= 0.5 \end{aligned} \)
So the current and the voltage of this appliance differ in phase by \( 60^\circ \).
NCERT (2026-27):
Power in AC Circuit: the Power Factor – “… Therefore,”
\( \displaystyle P=\frac{v_mi_m}{2}\cos\phi=\frac{v_m}{\sqrt{2}}\frac{i_m}{\sqrt{2}}\cos\phi \)\( \displaystyle =VI\cos\phi \)\( \qquad [7.30(a)] \)
“… 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. …”