# Current using kirchhoff's loop rule

## Homework Statement

By applying kirchhoffs rules, show that the in the RL circuit is given by: $$I=\frac{\epsilon}{R}\left(1-e^{-\frac{Rt}{L}}\right)$$

## Homework Equations

kirchhoff's second rule: $$\epsilon-IR-L\frac{dI}{dt}=0$$

## The Attempt at a Solution

$$\epsilon=IR+L\frac{dI}{dt}$$
$${\epsilon}dt=IRdt+LdI$$
$$\frac{\epsilon}{I}dt-Rdt=L\frac{1}{I}dI$$
then i thought to integrate both sides, LHS w.r.t t and RHS w.r.t I, but doing this on paper doesnt get to the right answer, am i going wrong somewhere with the rearranging??