An inductor `(L =20 mH)`, a resistor `( R = 100 Omega)` and a battery `(epsilon = 10 V)` are connected in series. After along time the circuit is short- circuited and then the battery is disconnected. Find the current in the circuit 1ms after short- circuiting.

Explanation

The initial current is `I = i_0 =verepsilon/R =(10V)/(100 Omega) = `0.10 A`
The time constant is `tau = L/R=(20mH)/(100 Omega) =0`
`20 ms`
The current at `t = 1 ms is I = i_0 e^(-t/tau) = 0`
`10 Ae^(1ms/0
20ms) = (0
10A)e^(-5) = 6`
`7X10^(-4) A`.