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

Here,` L = 0.3 Hm R = 0.15 K Omega = 0.15 xx 10^(3) Omega`,
`t = 1 ms = 1 xx 10^(-3) sec`.
`I_(0) = (V)/(R ) = (15)/(150) = (1)/(10) = 0.1`
`tau = (L)/(R ) = (0.03)/(150) = 2 xx 10^(-4) s`
During decay. `I = I_(0) e^(-t//tau) = 0.1 e^(-(10^(-3))/(2 xx 10^(-4))`
`I = 0.1 e^(-5) = (0.1)/(e^(5)) = (0.1)/(150) A = 0.67 mA`