An inductor (L = 0.03H) and a resistor (R = 0.15`Omega`) are connected in series to a battery of 15 V emf in a circuit shown below. The key `K_(1)` has been kept closed for a long time. Then, at t = 0, `K_(1)` is opened and key `K_(2)` is closed simultaneously. At t = 1 m/s 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

Current at `t=0 I_(0)=E_(0)/R`
`t=0 I_(0)=E_(0)/R`
For decay circuit `I=I_(0)e^((tR)/L)`
`I=E_(0)/R e^((tR)/L) rArr I=0.67 mA`