A `100 W` bulb `B_1`, and two` 60 W` 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

(d) `P = (V^(2))/(R )` so `R = (V^(2))/(P) implies R_(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)`
and `W_(3) = ((250)^(2))/(R_(3))`
`W_(1) : W_(2) : W_(3) = 15 : 25 : 64` or `W_(1) lt W_(2) lt W_(3)`