- #1

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## Homework Statement

## Homework Equations

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

## The Attempt at a Solution

I have solved part a) correctly.

[tex]i=\frac{E}{R}(1-e^{\frac{-t}{\tau}})[/tex]

[tex].8\frac{E}{R}=\frac{E}{R}(1-e^{\frac{-t}{\tau}})[/tex]

[tex].2 = e^{\frac{-t}{\tau}}[/tex]

[tex]ln .2 = \frac{-t}{\tau}[/tex]

[tex]t= 8.45\mu s[/tex]

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 [itex]-t/\tau[/itex]= -1? If [itex]\tau[/itex] is the time constant of the inductance (L/R), why do they "assume" or how do they calculate that t=1 and [itex]\tau[/itex]=1?At [itex]t = 1.0\tau[/itex], the current in the circuit is:

[tex]i=\frac{E}{R}(1-e^{-1.0})[/tex]