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

Answer

B

Explanation

`I (0) = (15 xx 100)/(0.15 xx 10^(3)) = 0.1 A`
`I (infty) = 0 Rightarrow I (t) = [I(0) - I(infty)]e^(-t)/(L/R) + i(infty)`
`I(t) = 0.1 e^((-t)/(L/R)) = 0.1 e^((R)/(L))`
`I(t) = 0.1 e^((0.15 xx 1000)/(0.03)) = 0.67 mA`