Two electric bulbs marked 40W, 220V and 60W, 220V when connected in series, across same voltage supply of 220V, the effective power is `P_1` and when connected in parallel, the effective power is `P_(2)` . Then `P_(1)/P_(2)` is-

Options

A. 0.5
B. 0.48
C. 0.24
D. 0.16

Answer

C

Explanation

Resistance of bulb `=("Rated voltage")^(2)/("Rated power")`
`R_(B_(1)) = (220)^(2)/40 Omega` and `R_(B_(2)) =(220)^(2)/60 Omega`
`R_(S) = R_(B1)) + R_(B_(2))`
`=(220^(2))/40 + (220^(2)/60) = (220^(2)) [1/40 + 1/60]= 24 W`
When the bulbs are connected in parallel
`1/R_(P) = 1/R_(B_(1)) + 1/R_(B_(2)) rArr 1/R_(P) = 40/(220)^(2) + 60/(220)^(2)`
`1/R(P) = 100/(220)^(2)`
`therefore P_(1)/P_(2) =(24 W)/(100 W) = 0.24`