Question

A bulb of $40W$ is designed to operate at $240V$. What is the resistance of the filament of the bulb.

• A. $9600\Omega$
• B. $60\Omega$
• C. $1400\Omega$
• D. $1440\Omega$

Answer 1

The correct option is D

$1440\Omega$

Given:
Voltage $\left(V\right)=240V$
Power $\left(P\right)=40W$

Electric Power, $P=\frac{V^2}{R}$
$\Rightarrow R=\frac{V^2}{P}$

$\therefore$ Resistance of filament of bulb = $\frac{(240V{)}^2}{40W}=1440\Omega$