Question

An electric bulb of 15 V is connected to a battery of 15 V which has negligible resistance. If the resistance offered by the bulb is $5\Omega$ the power of the bulb is ..... W.

• A. 225
• B. 75
• C. 30
• D. 45

Answer 1

The correct option is D 45

$V=15V$ which is given
$R=5\Omega$ which is given
$P=\frac{V^2}{R}=\frac{15\times 15}{5}$

by putting all the values given we get ,
$P=45W$