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Steady state-circuits

  1. Mar 21, 2010 #1
    This is a simple question. I know the equation I am supposed to use but I can not get my head around one part. The part in brackets! I keep getting wild numbers as my answer

    The equation is :-

    I = V/R * (1 - e^t*R/L)

    can I ask someone to give me a step by step example of how you place the relevent numbers into it.

    L= 1 x 10 -4 H
    t=1.0 x 10-5 s
    e=2.72 (3sf)
    Last edited: Mar 21, 2010
  2. jcsd
  3. Mar 21, 2010 #2
    The equation describes how current changes with time in a RL series circuit fed by a battery V.
    However, there is a mistake, the equation should read:

    I(t)=V/R * [1-e^(-t*R/L)] (note the sign change)

    My guess is that you don't understand the meaning of e^. In computer jargon ^ stands for exponentiation. Now you shouldn't have problems with this equation.
  4. Mar 22, 2010 #3

    So are you saying that [1-e^(-t*R/L)] just means (-e) -2.72 to the power of the total -t*R/L
  5. Mar 22, 2010 #4


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    Staff: Mentor

    [tex]I(t) = \frac V R (1-e^{-\frac{tR}{L}})[/tex]
  6. Mar 22, 2010 #5
    Hi Borak

    I know thats the equation I need, but the part in brackets is confusing me. I am not sure I am utilising it correctly.

    My book has the equation but no examples of it workings, so I can not see how the bracket part can be substituted for the numbers I have worked out.
  7. Mar 22, 2010 #6


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    Staff: Mentor


    is an exponential function, e to


    power. If you don't know what that means and how to calculate it, you should quickly get back to your math books.
  8. Mar 22, 2010 #7
    Yeah, I guess.....my physics teacher just says read the books instead of him teaching it.

    Thanks anyway.
  9. Mar 22, 2010 #8


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    Staff: Mentor

    Don't get me wrong. What I am aiming at is that it is a simple calculation - if you have a problem with it it means your math skills are way too low for things you are expected to do. We can help you checking if your result is OK, we can point you to the fact exponential function exists, but it is up to you to learn about it.
  10. Mar 27, 2010 #9

    Honestly thanks, I am struggling slightly because I am playing catch up. Be away from school for a long while due to injury. I have re-read my books and eventually figured it out, some of the numbers I was getting originally as the answer were just not looking right, I really started to confuse myself.

    Thanks again
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