Saved Bookmarks
| 1. |
An inductor (L =0.03 H) and a resistor `(R = 0.15k(Omega))` are connected in series to a battery of 15 V EMF in a circuit shown below. The key `K_(1)` is opened and Key `K_(2)` is closed simultaneously. At t =1 ms, the current in the circuit will be `(e^(5) = 150)` A. 6.7mAB. 0.67mAC. 100mAD. 67mA |
|
Answer» Correct Answer - B (b) I_(0)=(15xx100)/(0.15xx10^(3))=0.1A` ltbr. `I_(oo)=0` I(t)=[I(0)-I(oo)]e^((-t)/(L//R))+i(oo)` I(t) 0.1 e^((-t)/(L//R))=0.1e^(R/L)` I(t)=0.1 e^((0.15xx1000)/(0.03)=0.67mA`. |
|