A bulb rated 300 W, 220 V is connected across a 110 V supply. Assuming its resistance stays the same, the percentage reduction in its power is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 75%
The power taken by a resistance \( R \) across a voltage \( V \) is \( P=V^{2}/R \). With the resistance of the bulb unchanged, the power is proportional to \( V^{2} \), so halving the voltage makes the power one-fourth.
\( \begin{aligned} \frac{P'}{P} &= \frac{V'^{2}}{V^{2}} \\ &= \left(\frac{110\,\mathrm{V}}{220\,\mathrm{V}}\right)^{2} \\ &= \frac{1}{4} \end{aligned} \)
\( \begin{aligned} P' &= \frac{P}{4} \\ &= \frac{300\,\mathrm{W}}{4} \\ &= 75\,\mathrm{W} \end{aligned} \)
The percentage reduction in power is:
\( \begin{aligned} \text{Reduction} &= \frac{P-P'}{P}\times100\% \\ &= \frac{300\,\mathrm{W}-75\,\mathrm{W}}{300\,\mathrm{W}}\times100\% \\ &= 75\% \end{aligned} \)