Question

$15V$ electric bulb is connected to $15V$ battery of negligible resistance. The resistance offered by the bulb is $5\Omega$. Find power of bulb.

• A. $3W$
• B. $15W$
• C. $75W$
• D. $45W$

Answer 1

The correct option is D $45W$

Given : $V=15$ volts $R=5\Omega$

Power of bulb $P=\frac{V^2}{R}=\frac{{15}^2}{5}=45$ W