Why is the Time Constant Used in Solving Inductance Problems?

[exitwound]

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

Homework Statement

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?

 

[Delphi51]

In the problem and the example you have "at time t = 1.0*τ" so that is why
e^(t/τ) = e^(1.0*τ/τ) = e^1

 

[exitwound]

Oh. I see. I think I was just misunderstood what they were saying. Thanks.