A bulb of 220 V and 300 W is connected across 110 V circuit. The percentage reduction in power is:
Answer & explanation
Correct answer: option 4
P = 300 W, V = 220 V
$R = \frac{V^2}{P}=\frac{220 \times 220}{300}=\frac{484}{3} \Omega$
If this bulb is connected across 110 V, then the power delivered is
$P' = \frac{V^2}{R}=\frac{(110)^2}{484 / 3}=\frac{110 \times 110 \times 3}{484}$ = 75 W
∴ Percentage reduction in power = $\frac{300-75}{300}$ × 100% = 75%