The switch `S` is closed at time `t=0` . Find the current through the inductor as a function of time `t`
(a)
(b) 1

Explanation

`E'=((6E)/(3r)+(12E)/(6r))/((1)/(3r)+(1)/(6r))=((12E+12E)/(6r))/((1)/(2r))=8E`
`r'=(3rxx6r)/(3r+6r)=2r`
`i=i_(0)(1-e^(-t//tau))`
`i_(0)=(E')/(r')=(8E)/(2r)=(4E)/(r)`
`tau=(L)/(r')=(L)/(2r)`
`i=(4E)/(r)(1-e^(-(2rt)/(L)))`

`E'=((6)/(3)+(4)/(2))/((1)/(3)+(1)/(2))=(2+2)/((5)/(6))=(24)/(5)V`
`r'=(3x2)/(3+2)=(6)/(5)Omega`
`i=i_(0)(1-e^(-t//tau))`
`i_(0)=(E')/(r')=(24//5)/(6//5)=4A`
`tau=(L)/(r')=(2)/(6//5)=(5)/(3)sec`
`i=4(1e^(-(3t)/(5)))=4(1-e^(-0.6t))`