A 100 W bulb `B_(1)` and two 60 W bulbs `B_(2)` and `B_(3)`, are connected to a 250V 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)gtW_(2)=W_(3)`
B. `W_(1)gtW_(2)gtW_(3)`
C. `W_(1)ltW_(2)=W_(3)`
D. `W_(1)ltW_(2)ltW_(3)`

Answer

D

Explanation

`P=(V^(2))/(R)` so `R=(V^(2))/(P)`
`thereforeR_(1)=(V^(2))/(100)` and `R_(2)=R_(3)=(V^(2))/(60)`
Now, `W_(1)=((250)^(2))/((R_(1)+R_(2))^(2))R_(1)`
`W_(2)=((250)^(2))/((R_(1)+R_(2))^(2))R_(2)` ad `W_(3)=((250)^(2))/(R_(3))`
`W_(1):W_(2):W_(3)=15:25:64` or `W_(1)ltW_(2)ltW_(3)`