Answer to Question #282610 in Electric Circuits for Anonymous005

Question

A light bulb rated as 60 W at 220 V has a potential difference of 110 V across its ends. Find the power dissipated in the light bulb, assuming that the light bulb is ohmic (which it likely isn't). Ans: 15 W

Answer 1

$P=\frac{V^2}{R}$

Where $P=$ Power

$V=$ Voltage

$R=$ Resistance

Resistance of bulb $R=\frac{V^2}{P}=\frac{220^2}{60}=\frac{2420}{3}\Omega$

Power dissipated in the bulb

$P=\frac{110^2}{\frac{2420}{3}}=\frac{110^2}{2420}×3=15W$