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

As resistance of a bulb, `R = (V^(2))/(P)`,
Hence, `R_(1):R_(2):R_(3)=(1)/(100):(1)/(60):(1)/(60)`
Now the combined potential difference across `B_(1)` and `B_(2)` is same as the potential difference across `B_(3)`. Hence, `W_(3)` is more than `W_(1)` and `W_(2)` . Now, `B_(1)` and `B_(2)`, being in series, carry same current and `R_(1)lt R_(2)`, therefore `W_(1) lt W_(2)`,
`W_(1) lt W_(2) lt W_(3)`