- #1
exitwound
- 292
- 1
This is not a homework problem, but practice. I have the answer. I need to know why.
[tex]i=\frac{E}{R}(1-e^{\frac{-t}{\tau}})[/tex] "charging" an Inductor
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?
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.
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?