Answer
$i=5-5e^{-40t}$
Work Step by Step
Using RC Circuit we get:
$$i+\frac{1}{RC}q=\frac{E}{R}$$
$$iR+L\frac{di}{dt}=E$$
We are given:
$E=20$
$L=0.1$
$R=4$
Substituting:
$$\frac{di}{dt}+\frac{4}{0.1}i=\frac{20}{0.1}$$
The solution can be found by:
$$i=e^{-\int 40 dt}(c+\int 200e^{\int 40 dt})$$
$$i=ce^{-40t}+5$$
Since $i(0)=0$, we get:
$$0=c+5 \rightarrow c=-5$$
Thus the current in the circuit for $t \geq 0$ is:
$$i=5-5e^{-40t}$$
