A 100 W bulb `B_(1)` and two 60W bulbs `B_(2) and B_(3)` , are connected to a 250 V source as shown in the figure . Now `W_(1) , W_(2) and W_(3)` are the output powers of the bulbs `B_(1) , B_(2) and B_(3)` respectively, then

Options

A. `W_(1) gt W_(2) = W_(3)`
B. `W_(1) gt W_(2) gt W_(3)`
C. `W_(1) lt W_(2) = W_(3)`
D. `W_(1) lt W_(2) lt W_(3)`

Answer

D

Explanation

`W_(1) lt W_(2) lt W_(2)`
Resistance `R_(1)` of sphere `B_(1) = (V^(2))/(P_(1)) `
`= ((250)^(2))/(100)`
= `625 Omega `
` = 1042 Omega`
Resistance`R_(3) ` of sphere `B_(3) = (V^(2))/(P_(3)) = ((250)^(2))/(60)`
= ` 1042 Omega`

Now
Output power of `B_(1) , W_(1) = (V^(2))/((R_(1) + R_(2))^(2)) . R_(1)`
`= ((250)^(2))/((625 + 1042)^(2)) xx 625 `
= 14.1 W
Output power of `B_(2) , W_(2)= (V^(2))/((R_(1) + R_(2))^(2)) xx R_(2)`
`= ((250)^(2))/((625 + 1042)^(2)) xx 1042`
23.4 W
Output power of `B_(3) , W_(3) = (V^(2))/(R_(3)^(2)) xx R_(3)`
` = ((250)^(2))/( 1042) = 60 ` W
`therefore W_(1) lt W_(2) lt W_(3)`