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Inductance in an RL circuit

  • Thread starter exitwound
  • Start date
  • #1
292
1
This is not a homework problem, but practice. I have the answer. I need to know why.

Homework Statement



Capture.JPG


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.
At [itex]t = 1.0\tau[/itex], the current in the circuit is:
[tex]i=\frac{E}{R}(1-e^{-1.0})[/tex]
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?
 

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
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
292
1
Oh. I see. I think I was just misunderstood what they were saying. Thanks.
 

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