The switch in figure is closed at time `t = 0`. Find the current in the inductor and the current through the switch as functions of time thereafter.

Answer

A, B

Explanation

At `t=0`, inductor offers infinite resistance. Hence, current through inductor wire is zero. Whole currennt passes through two resistors of `4Omega` each.
`i_1=10/(4+4)=1.25A`
A`t=oo`, inductor offers zero resistance.
`R_("net")=4+(8xx4)/(8+4)`
`=6.64Omega`
so,main current
`i_2=10/R_("net")=1.5A`
This distributes in `4Omega` and `3Omega` in inverse ratio of resitance. Hence, current through `4Omega` is `1A` and thorugh `8Omega` is `0.5A`.
for equivalent `tau_L` of the circuit `R_("net")` across inductor after short circuting the battery is `10Omega`.
`:. tau_L=1/R_("net")=1/10=0.1s`

`i_L=0.5(1-e^(t/0.1))`
`=0.5(1-e^(-10t))`

`i=1.25+0.25(1-e^(-t//0.1))`
`=15 - 0.25 e^(-10t)`