An electric bulb of resistance `400 Omega`, draws a current of `0.5 A`. Calculate the power of the bulb and the potential difference at its ends.

Answer

100 W, 200 V

Explanation

`P = I^2 R = (0.5 A)^2 (400 Omega) = 100 W`
As `P = VI, V = (P)/(I) = (100 W)/(0.5 A) = 200 V`.