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R-L Circuit - Simple

  • Thread starter ttiger2k7
  • Start date
58
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[SOLVED] R-L Circuit - Simple

1. Homework Statement

In the figure below, suppose that the switch is initially open, and at time t=0, the switch is closed. Let t1=aL/R be the time that the current through the inductor L is 70.0 percent of its value when t is infinity. Find the dimensionless number a.

http://cse.unl.edu/~ejones/Images212/LR.gif [Broken]

2. Homework Equations

[tex]i=I_{0}e^{-(R/L)t)[/tex]

3. The Attempt at a Solution

I tried and no luck:

Given the problem, when t is 0 the current is initially 0. So I_{0} is 0.

And I want to find [tex].7i[/tex] (70% of i) and I already know that t1=aL/R

So plugging all the info in:

[tex].7i=0*e^{-(R/L)(aL/R)}[/tex]

[tex].7i=0[/tex]

[tex]i=0[/tex]

This is obviously wrong, and makes no sense to me. Help would be appreciated.
 
Last edited by a moderator:

Answers and Replies

10
0
Your equation is wrong. In infinity the current is not 0.

You must notice, that in infinity the voltage on inductor is 0. Hence you can calculate the current in infinity.

Than you should 'construct' right equation for the current and solve the problem (of course you can solve some differential equations instead). The one thing that is right is that the current will be exponential-like with time constant R/L
 
58
0
Okay, so is this right for finding current @ infinity?

Since voltage on inductor is 0,

[tex]L\frac{di}{dt}=0[/tex]

So, using Kirchoff's loop rule:

[tex]\epsilon-L\frac{di}{dt}-iR=0[/tex]

[tex]\epsilon-0-iR=0[/tex]

[tex]i=\frac{\epsilon}{R}[/tex]

**

EDIT: got it. Thanks!
 
Last edited:

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