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

Explanation

A bulb is essentially a resistance `R=(V^(2))/(P)` where P
denotes the power of the bulb
`therefore` Resistance of `B_(1)(R_(1))=(V^(2))/(100)`
Resistance of `B_(2)(R_(2))=(V^(2))/(60)`
Resistance of `B_(3)(R_(3))=(V^(2))/(60)`
`thereforeI_(1)=` Current in `B_(1)=(250)/((R_(1)+R_(2)))=(250xx300)/(8V^(2))`
`I_(2)=` Current in `B_(2)=(250)/((R_(1)+R_(2)))=(250xx300)/(8V^(2))`
`I_(1)=` current in `B_(3)=I_(1) as B_(1),B_(2)` are in series
`thereforeW_(1)` output power of `B_(1)=I_(1)^(2)R_(1)`
`thereforeW_(1)=((250xx300)/(8V^(2)))^(2)xx(V^(2))/(100)`
`W_(2)=I_(2)^(2)R_(2) or W_(2)=((250xx300)/(8V^(2)))^(2)xx(V^(2))/(60)`
`W_(3)=I_(2)^(2)R_(3) or W_(3)=((250xx300)/(8V^(2)))^(2)xx(V^(2))/(60)`
`thereforeW_(1):W_(2):W_(3)=15:25:64` or `W_(1)ltW_(2)ltW_(3)`