In th circuit the key `(K)` is closed at `t=0`. Find the current in ampere through the key `(K)` at the instant `t=10^(2)ln2` sec.

Answer

2

Explanation

`i_(1)` is current in the first loop
`i_(1)=40/20(1-e^(t/(5xx10^(-4))))=3/2` amp
`i_(2)` is current in the second loop
`i_(2)=40/40e^(- t/(10^(-3)))=1/2` amp
So, total current through key `(K)`
`i=i_(1)+i_(2)`
`=3/2+1/2=2` amp