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)`

Options

A. 6.7mA
B. 0.67mA
C. 100mA
D. 67mA

Answer

B

Explanation

(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`.