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

When switch `K_(1)` closed for long time current through inductor, `I=(E)/(R)=(15)/(0.15xx10^(3))=0.1A`
When `K_(1)` opened and `K_(2)` is closed, then
`i=Ie^(-t//t)`
`t=1ms=1xx10^(-3)s`

`therefore" "tau_(L)=(L)/(R)=(0.03)/(0.15xx10^(3))=(10^(-3))/(5)`
`thereforei=0.1e^(((((-1xx10^(-3))/(10^(-3))))/(5)))=0.1e^(-5)`
`i=(0.1)/(e^(5))=(0.1)/(150)=(1)/(15)xx10^(-2)xx10^(3)mA`
`i=0.67mA`