InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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» 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)`. |
|