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Natural response RL circuit

  1. Apr 11, 2017 #1
    as you can see In the image I provided I have derived the equation for i(t) and v(t).

    On circuit A I use a passive sign convention(current flowing from positive to negative) on the inductor, hence the equation and their corresponding graph.
    i(t) is decaying exponentially, v(t) is decaying exponentially.

    On circuit B I did use what I learned in physics, i.e the voltage induced across an inductor is always opposite to the increase/decrease in current
    flowing throuhg it. Since the current i(t) is decreasing, the voltage should support that current hence the sign. when I perform kvl on circuit B I ended up having growth equation for both i(t) and v(t). Why is that?

    Also in circuit A I expect to have same graph for both i(t) and v(t) (Although they are both decaying).
    Please, kindly clear this up for me if you have time. Thanks! eda.png
     
    Last edited by a moderator: Apr 18, 2017
  2. jcsd
  3. Apr 11, 2017 #2

    Svein

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    Science Advisor

    What is the final value of the current through the inductor (when the current does not vary any longer)?
     
  4. Apr 18, 2017 #3

    eq1

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    The signs are not consistent between the two circuits therefore you can't use v=Ldi/dt (using the symbols from the schematics) in both algebraic solutions. i.e. circuit a has positive current entering the + terminal of L. Circuit B has positive current leaving the + terminal of L and this is why it shows current growth.

    Edit: I just realized you made the sign error on purpose. The issue is still a sign error. But you dropped it from the "physics" version of the equation. See the link below.
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/induct.html#c1
     
    Last edited: Apr 18, 2017
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