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Homework Help: ODE RC Circuit Problem

  1. Sep 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose an RC series circuit has a variable resistor. If the resistance at time t is given by by R = a + bt, where a and b are known positive constants then the charge q(t) on the capacitor satisfies

    (a+bt) q' + (1/C)q = V

    where V is some constant. Also q(0) = q_0
    Find q(t) as an explicit function of t.

    2. Relevant equations

    Now I have obtained the answer, however my main question is: am I allowed to treat C (capacitance) as a constant in this equation. It doesn't specify in the question, but to my knowledge (unless I am wrong of course) capacitance is a constant value and only depends on material and physical parameters of the capacitor itself (how it is build).

    If I can't treat C as a constant then I believe there is no way to evaluate the integral in integrating factor and I'd have to leave it as it is.

    Process is simple from there, I rewrite the equation in standart form, find the integrating factor and obtain a formula for q(t), evaluate an integration constant with q(0) = q_0 and obtain the overall solution q(t).

    The answer looks quite frightening btw

    3. My solution:
    attached pdf file

    Attached Files:

  2. jcsd
  3. Sep 15, 2015 #2
    The capacitance must be constant during the all process, which just depends on its material and geometric shape, as what you say above,
  4. Sep 16, 2015 #3


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    Is there some reason you're not simplifying ##\frac{(a+bt)^k}{a+bt}## to ##(a+bt)^{k-1}## and integrating the righthand side? Also, the ##t## in the integral is a dummy variable, so it's not correct to set it to 0.
  5. Sep 17, 2015 #4
    That's what my prof does. By setting it from 0 to t, we ensure evaluated integral is equal to the true value with t as a variable, how would setting it to something arbitary as t_o (I'm assuming that's what you mean)be any better? My understanding of it is that we need to choose some "convenient" interval, what's wrong with 0 to t?

    And thank you for pointing out I can actually simplify it further, I completely missed that.
  6. Sep 17, 2015 #5


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    What I'm saying is you can't do something like this:
    $$\int (a+bt)^2\,dt = \int a\,dt$$ by claiming you're setting ##t=0##. The ##t## inside the integral isn't the same ##t## that appears elsewhere.
  7. Sep 17, 2015 #6
    I see what you mean, my mistake, thank you for pointing it out. I see holes in my knowledge about understanding of basic integration now, will have to fill them in :)
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