Answer
$i=20e^{-10t}$
Work Step by Step
Using RC Circuit we get:
$$R\frac{dQ}{dt}+\frac{Q}{C}=E$$
We are given:
$E=100$
$c=\frac{1}{50}F$
$R=5$
Substituting:
$$\frac{dQ}{dt}+\frac{Q}{\frac{5}{50}}=\frac{100}{5}$$
$$\frac{dQ}{dt}+10Q=20$$
The solution can be found by:
$$Q=e^{-\int10 dt}(c+\int e^{10t}20dt)$$
$$Q=ce^{-10t}+2$$
Since $t=0,Q=10$, we get:
$$0=c+2 \rightarrow c=-2$$
The charge at $t \geq 0$ is:
$$Q=-2e^{-10t}+2$$
Thus the current in the circuit for $t \geq 0$ is:
$$i=\frac{dQ}{dq}=20e^{-10t}$$
