Question

An electric bulb of resistance $500\Omega$, draws a current of $0.4$A. Calculate the power of the bulb and the potential difference at its end.

Answer 1

Resistance of bulb $R=500Omega$

Current flowing through it $I=0.4A$

So, potential difference across its ends $V=IR$

$\therefore$ $V=0.4\times 500=200$ volts

Power of the bulb $P=I^{2}R$

$\therefore$ $P=(0.4{)}^2\times 500=80W$