Question

An electric bulb rated $220V$, $60W$ is working at full efficiency. State the resistance of the coil of the bulb.

Answer 1

Step:1 Given data

Power of bulb (P)$=60W$

Potential difference (V) $=220V$

Step:2 Finding the resistance of the coil of the bulb.

As we know, the power of the bulb $\left(P\right)$

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

The resistance of the bulb $\left(R\right)$

$R=\frac{V_2}{P}$

$R=\frac{220\times 220}{60}$

$R=806.7\Omega$

Hence, the resistance of the coil is $806.7\Omega$