Question

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

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 1

Correct Answer - D
`R_(1) = ((250)^(2))/(100) , R_(2) = R_(3) = ((20)^(2))/(60)`
`(R_(2)=R_(3)) gt R_(1)`
In upper branch containing `B_(1)` and `B_(2)` current is same
`W_(1) = i^(2) R_(1), W_(2) = i^(2) R_(2)`
`W_(2) gt W_(1)`
Total power consumed in upper branch = `((250)^2)/(R_1+R_2)`
Total power consumed in lower branch = `(250^2)/(R_2) = W_(3)`
`W_(1) lt W_(2) lt W_(3)`.