Saved Bookmarks
| 1. |
An inductor is connected in series with a bulb to an a.c source. What happens to brighness of bulb when number of turns in the inductor is reduced ? |
|
Answer» Current through the bulb `I_(v) = (E_(v))/(Z) = (E_(V))/(sqrt(R^(2) + X_(L)^(2)))`, where `X_(L) = omega L` As `L prop N^(2)`, therefore, on decreasing number of turns, L decreases , Z decreases, `I_(v)` increases. Hence brightness of bulb increases. |
|