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Question about current decay in R-L circuit

  1. May 26, 2012 #1

    I am reading Young, Freedman's University physics and I have some questions about the
    sign conventions followed in getting the equations for the current decay in the R-L circuit.
    I have attached two snapshots which will help the discussions. Consider the series R-L circuit
    as shown in the diagram. Say switch S1 is closed and switch S2 is open for a long time. So that means steady current has been established in the upper part of
    the circuit. Now open S1 and close S2. Now authors give the following
    equation for the first case, when the current is building.

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

    which makes sense using the second diagram I attached. But the authors say that , for the
    second case , when the current is decaying through the bottom part of the circuit, the equation to be solved becomes,

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

    which doesn't make sense. Since the current is now decreasing, going around the loop
    in the direction of current (assuming that the direction of the current is still the same),
    and using the second diagram which I have uploaded, the Kirchhoff's rule says that

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

    But this doesn't give correct final result for the decay. So am I doing something wrong ?


    Attached Files:

  2. jcsd
  3. May 26, 2012 #2
    The minus sign in L di/dt is a consequence of Lenz's law (or the minus sign in Faraday's). Just because we know the current will decrease, we don't change the sign because that would give us an ever increasing current (contrary to our expectations)
  4. May 26, 2012 #3
    But Gordianus, but is there not a potential gain in such situation of we trace the loop in the same direction as the decaying current. I was trying to follow the rules of Kirchhoff using the second diagram given for the situation where [itex]\frac{di}{dt} <0[/itex].
  5. May 26, 2012 #4
    Exactly. When di/dt<0 you want a positive voltage drop. If you multiply by -L you obtain a positive number. On the contrary, if you multiply by L, you obtain a negative number (this isn't what you want).
    In a nut, you don't have to switch the sign from positive to negative. The same equation gives the right answer for both cases.
  6. May 27, 2012 #5
    Gordianus, I think it makes sense now.... So are the signs used the way they are used to get the end equations working ?
  7. May 27, 2012 #6
    The safe method is this:
    a) You assume the current flows in some direction (it doesn't matter which one).
    b) You apply KVL moving in the same direction current does.

    With these previous assumptions the voltage drop across the inductance is -L di/dt, whether the current grows or decays.
  8. May 27, 2012 #7
    ok makes sense
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