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Capacitor/inductor charging

  1. Jun 2, 2009 #1
    This isn't homework, but it feels like it...it's from the Dorf & Svoboda Introduction to Electric Circuits, 6th ed. on p 298, if you happen to have that.

    I'm following the derivation of the formula for capacitor voltage for an RC circuit with a single resistor and single cap, or the current in the simple inductor circuit. Arriving at the general form of the differential equation, with time constant T:

    dx(t)/dt + x(t)/T = K

    rewritten:

    dx/dt = (KT - x)/T

    now, in the next step, the authors separate the variables and multiply each side by -1, yielding:

    dx/(x - KT) = -dt/T

    and go on to integrate both sides and arrive at:

    ln(x - KT) = -t/T + D

    D being the constant of integration. Raising e to both sides you get:

    x(t) = KT + Ae-t/T

    where A is eD, and you can go on using initial conditions to solve for the constant.



    I can't get the same result when I don't multiply both sides by -1 when separating variables before solving the equation. My steps are as follows:

    dx/dt = (KT - x)/T

    dx/(KT - x) = dt/T having not multiplied by -1

    ln(KT - x) = t/T + D

    KT - x = Aet/T

    x(t) = KT - Aet/T

    This doesn't appear to be the same solution, as en is not equal to -e-n.


    Can somebody please help me figure out what I've missed?

    Oh, and hello everybody. I'm a junior engineer but work in a field that barely ever touches on a lot of what I learned in school...I've decided to start studying again to make sure I retain the fundamentals, especially if I do switch into a hardware design position.

    Thank you.
     
  2. jcsd
  3. Jun 3, 2009 #2
    I think I spotted the mistake,

    (d/dx)ln(x - KT) = 1/(KT - x)

    but

    (d/dx)ln(KT - x) = -1/(KT - x)

    so

    dx/(KT - x) ==> (-1)ln(KT - x)

    hope that helps :)

    ---
     
    Last edited by a moderator: Aug 6, 2009
  4. Jun 3, 2009 #3
    You're right! I was thinking about the whole denominator when I should have been careful and integrated with respect to x.
     
    Last edited: Jun 3, 2009
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