Saved Bookmarks
| 1. |
A (current vs time) graph of the current passing through a solenoid is shown in Fig. For which time is the back electromotive force (u) a maximum? If the back emf t = 3 s is e, find the back emf at t = 7 s, 15 s and 40 s OA, AB and BC are straight line segments. |
|
Answer» As back e.m.f. `e = L (dI)/(dt)`, e will be maximum when `dI//dt` = max. The graph in Fig. shows that `dI//dt` is maximum for 5 x lt 6 lt 10 s. For `0 s lt t lt 5 s, (dI)/(dt) = (1)/(5)` `:.` At `t = 3 sec , e = L (dI)/(dt) = (L)/(5)` For `5 s lt t lt 10 s, (dI)/(dt) = (-2-1)/(5) = (-3)/(5) :.` At `7 s, e = L(dI)/(dt) = L (-(3)/(5)) = -3 e` For `10 s lt t lt 30 s, (dI)/(dt) = (2)/(20)=(1)/(10) :.` At `t = 15 s, e = L (dI)/(dt) = (L)/(10) = (e)/(2)` For `t gt 30 s, (dI)/(dt) = 0 :.` At `t = 40 s, e = 0` |
|