Question

An electric bulb rated 50 W – 200 V is connected across a 100 V supply. The power dissipation of the bulb is : 

(1) 12.5 W 

(2) 25 W 

(3) 50 W 

(4) 100 W 

Answer 1

Correct option is : (1) 12.5 W

Rated power & voltage gives resistance

\(\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(200)^{2}}{50}=\frac{40000}{50}\)

R = 800

\(\mathrm{P}=\frac{\left(\mathrm{V}_{\text {applied }}\right)^{2}}{\mathrm{R}}=\frac{(100)^{2}}{800}\)

P = 12.5 watt