An inductor of 10mH is connected to a 18V battery through a resistor of 10k`Omega` and a switch After a long time, when the maximum current is set up in the circuit, the current is switched off. Calculate the cur- rent in the circuit after 2 `mu`s.

Explanation

Given that L=10 mH = `10^(2)H, R=10k Omega= 10^(4)Omega` and E=18V , We know that
`I_(0)=(E)/(R)=(18)/(10^(4))A`
Time constant, `tau_(L)=(L)/(R)=(10xx10^(-3))/(10xx10^(3))=10^(-6)` sec
We know that, `I=I_(0)e^(-(R//L)t)`
Here `t = 2 mu s = 2xx10^(-6) s, I_(0) = 18 xx 10^(-4)A`
`I=18 xx 10^(-4)e^(-2)=2.48 xx 10^(-4)A`