Answer to Question #11377 in Electric Circuits for saksham
Two bulbs B1(40W,220V) and B2(100W,220V) are connected in series to an emf of 220 V. Which bulb will glow brighter?
Let's find out the resistances of each bulb from the data given: R1 = U1^2 / P1 = 220^2 / 40 = 1210 Ohm R2 = U2^2 / P2 = 220^2 / 100 = 484 Ohm Let's use P = I^2 * R for each bulb and find out if it is greater than its power rating. The one with higher power rating will get fused. P1 = (220 / (1210 + 484))^2 * 1210 = 20.408 W P2 = (220 / (1210 + 484))^2 * 484 = 8.163 W. Answer: first bulb will glow brighter.