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RL Circuit Time Constant

  1. Mar 5, 2006 #1
    Can anyone tell me what formula I would use to find the time constant for a circuit that reaches 85% of its final value 1.86 seconds after the switch is closed.
     
  2. jcsd
  3. Mar 5, 2006 #2
    I am assuming that this circuit is in series with a constant voltage source, and initially there is no current when the switch is first thrown, for which case we have the first order ode arrising from taking Kirchhoff's voltage law around the loop.
    [tex] \dot{I}+\frac{R}{L} I-\frac{\varepsilon_{0}}{R} = 0 , I(0) = 0 \cr
    I(t) = \frac{\varepsilon_{0}}{R} \left( 1- \exp(-t \tau) \right) , \tau = R/L [/tex]
    where tau is the time constant you seek. Now use the information given, by taking the ratio @ t=0, and solve. Hope this helps, sincerely, x.
     
  4. Oct 31, 2009 #3
    85% completion means 15% incomplete.

    take the natural log of .15 (the decimal form of 15%). This is -1.897. toss out the negative as this is what you get when working with logs of numbers less than 1. what you are left with is 1.897. This is the number of time constants needed to reach 85% of completion.

    divide you 1.86 seconds by 1.897, and you get about 0.980 seconds per time constant.

    This is awfully close to 1, which might be the round answer you are expected to get. use this formula to plot 1,2,3,4,and 5 time constants, and you'll begin to see the pattern:

    1-e-tc#.

    that is: one minus (eulers number raised to the negative tc#). You'll get approximately .63, .87, .95, .98 and .99. But do it for yourself, with and without the "one minus"
     
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