Saved Bookmarks
| 1. |
Current is switched on in a circuit contaning a resistance of `10 Omega` and an inductance of 0.8 H. Calculate the time taken by the current to grow to half its maximum value. |
|
Answer» Here, `R = 10 Omega, L = 0.8 H`, `t = ?, I = I_(0) //2` As `I = I_(0) (1 - e^(-R//Lt)) = I_(0) //2` `:. (1)/(2) = 1 - e^(-Rt//L)` or `e^(-Rt//L) = 1 - (1)/(2) = (1)/(2)` `(Rt)/(L) log_(e) e = log_(e) 1 - log_(e) 2 = 0 - 0.693` `t = + 0.693 (L)/(R ) = (0.693 xx 0.8)/(10) = 0.055 s` |
|