# Inductance in an RL circuit

1. Nov 13, 2009

### exitwound

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

1. The problem statement, all variables and given/known data

2. Relevant equations

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

3. 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.
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?

2. Nov 13, 2009

### 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

3. Nov 13, 2009

### exitwound

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