Question

A bulb rated $220V$ $-$ $300W$ is connected across $110V$ supply. The percentage reduction in the power is

• A. $25\%$
• B. $50\%$
• C. $60\%$
• D. $75\%$

Answer 1

Answer: (D)

$P=300W$ and $V=220V$
Resistance of the bulb
$R=\frac{V^{2}}{P}=\frac{220 \times 220}{300}=\frac{484}{3}\Omega$
When the bulb is connected across $110V$ , then power delivered
$P=\frac{V^{2}}{R}=\frac{110 \times 110 \times 3}{484}=75W$
Reduction in power $=300-75$
$=225W$
Percentage reduction in power
$=\frac{225}{300}\times 100=75\%$