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.

Explanation

`10 - 4I - 8I_(1) - (dI_(1))/(dt) = 0.0` (i)
`I_(1) + I_(2) = I`
`10 - 4I - 4I_(2) = 0` (ii)
`rArr I_(2) = ((10 - 4I)/(4))`
`I_(1) + (10)/(4) = 2I rArr 4I = 2I_(1) + 5`

put in equation (i), `10 - 2I_(1) - 5 - 8I_(1) = (dI_(1))/(dt)`
`dt = (dI_(1))/(5 - 10I_(1))`
`-t xx 10 = In ((5 - 10I_(1))/(5))`
Current through inductor `I_(1) = (1)/(2) (I - e^(-10t))`
Current through switch `I = (5)/(4) + (I_(1))/(2) = (5)/(4) + (1)/(4) (I - e^(-10t))`