An electric bulb is rated '220 V, 100 W'. (a) What is it's resistance? (b) What safe current can be passed through it?

(a) Given,

P = 100 W

V = 220 volt

We know that,

Power P = $\dfrac{V^2}{R}$

Substituting the values in the formula above, we get,

$100 = \dfrac{220^2}{R} \\[0.5em] \Rightarrow R = \dfrac{220^2}{100} \\[0.5em] \Rightarrow R = 484 Ω \\[0.5em]$

Hence, resistance of electric bulb = 484 Ω

(b) From relation P = VI

Safe current I = $\dfrac{P}{V}$

Substituting the value we get,

$I = \dfrac{100}{220} \\[0.5em] \Rightarrow I = 0.45 A$

Hence, safe current limit = = 0.45 A