Question

An electric bulb is rated 120V-100W. Find a) Resistance of the filament
b)Maximum current flowing through it.

Answer 1

Answer:

(a) Resistance,  $R=\frac{V^{2}}{P}$ .     ; V = potential difference

                                               P = power.

             => $R=\frac{120^{2}}{100}$ = 144Ω.

(b)   Maximum current, $I_{max}=\frac{V}{R}$

                                => $I_{max}=\frac{120}{144}$ =0.833 ampheres.