Question

The switch in figure is open for $t<0$ then closed at time $t=0$. Find the current through the switch as function of time there after.

Answer 1

$i_2=\frac{10}{R_{net}}=1.5A$

The distributes in $4\Omega$ and $3\Omega$ in invers ration of resistance. Hence, current through $4\Omega$ is $1A$ and through $8\Omega$ is $0.5A$.
For equivalent ${\tau }_L$ of the circuit $R_{net}$ across inductor after short-circuiting the battery is $10\Omega$.
${\tau }_L=\frac{1}{R_{net}}=\frac{1}{10}\simeq 0.1s$

$i_L=(1-e^{\frac{t}{0.1}})=0.5(1-e^{-10t})$

$i=1.25+0.25(1-e^{-t/0.1})$
$=1.5-0.25e^{-10t}$