# Inductance in an RL circuit

This is not a homework problem, but practice. I have the answer. I need to know why.

## Homework Equations

$$i=\frac{E}{R}(1-e^{\frac{-t}{\tau}})$$ "charging" an Inductor

## The Attempt at a Solution

I have solved part a) correctly.

$$i=\frac{E}{R}(1-e^{\frac{-t}{\tau}})$$
$$.8\frac{E}{R}=\frac{E}{R}(1-e^{\frac{-t}{\tau}})$$
$$.2 = e^{\frac{-t}{\tau}}$$
$$ln .2 = \frac{-t}{\tau}$$
$$t= 8.45\mu s$$

Part B is stumping me. I have a walkthrough of the problem but I don't understand why they do what they do. Here's the walkthrough.
At $t = 1.0\tau$, the current in the circuit is:
$$i=\frac{E}{R}(1-e^{-1.0})$$
Maybe I'm just missing something stupid, but how do they end up with $-t/\tau$= -1? If $\tau$ is the time constant of the inductance (L/R), why do they "assume" or how do they calculate that t=1 and $\tau$=1?