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

Answer

D

Explanation

`I= I_(0e)^((1)/(tau)), tau = (L)/(R )`
`= (15)/(150) e^((-1 xx 10^(-3))/(1 //5 xx10^(3))) = 0.67 mA`