Two bulbs of 220 V and 100 W are firs connected in series and then in parallel with a supply of 220 V. Total power in buth the cases will be ........

Options

A. 50 W, 100 W
B. 100 W, 50 W
C. 200 W, 150 W
D. 50 W, 200 W

Answer

D

Explanation

50 W, 200 W
Power in series combination of resistances
`P_(s) = (P_(1) xx P_(2))/(P_(1) + P_(2))`
but `P_(1) = P_(2)`
= 100 W
`therefore P_(S) = (100 xx 100)/(100 + 100)`
50 W
Power in parallel combination,
`P_(p) = P_(1) + p_(2)`
but `P_(1) = P_(2) `
= 100 W
`P_(p) = 100 + 100 `
= 200 W