RC / RL Differential Equations - beginner

Discussion in 'Homework Help' started by kvnsmith.at.uw, May 16, 2012.

  1. kvnsmith.at.uw

    Thread Starter New Member

    May 16, 2012
    4
    0
    First, I hope this topic doesn't already exist. I've searched and searched and Google'd and compiled everything that seemed to make sense.

    Just getting started on these, I don't know differential equations but I don't think first-order are that hard so I'm not too worried. What I do want to know is that the work and figuring that I have done is not in error.

    My goal was to get a GENERAL equation. I don't know why it was so hard. My solution ended up as this:

    y(t) = (y(0-) - y(∞))e^-at + y(∞)

    Where y can be either current or voltage and a can be either the time constant for an RC or for an RL. From what I've found, this should work BUT I really don't know what I'm doing and maybe it's pure falsehood.

    The only real problem I have "solving" these after I get this form set up is what to do with y(∞). There's a table in the textbook where if that value is a constant you just replace it with a different constant and a couple other if-else situations. The ones we did in class were super confusing with 800 more letters in them, so I'm not confident I should say (for a small circuit with a switch, a Vs, an R(4) and a C(.25) where Vc(0-) is 2) that the solution is:

    Vc(t) = (2 - K)e^-t + K

    I know that's probably hard to follow, but the general equation is what I'm really after for now.
     
  2. panic mode

    Senior Member

    Oct 10, 2011
    1,320
    304
    if you are only using first order circuit you can use

    y(t)=N+M*e^-t/τ

    where
    τ (tau) is time constant

    if you know initial condition (t=0) then you know N+M because

    e^0=1 so

    y(0)=N+M

    if you know t=∞, then e^-∞=0 so y(∞)=N

    in between you get exponential function. N and M are constants (which you find using evaluation of t=0 and t=∞)

    for example you have series RC circuit that will be connected to 5VDC, and you are measuring voltage across capacitor as output variable. initially circuit is not powered so C is fully discharged. therefore at t=0 we get y(0)=0 and at infinity we get capacitor fully charged so y(∞)=5V

    y(0)=0=N+M*e^0
    0=N+M
    N=-M

    y(∞)=5 so

    5=N+M*e^-∞=N+M*0
    therefore N=5
    and so it must be that M=-5

    now we just need RC constant

    τ=RC so our result is equation

    y(t)=5(1-e^(-t/RC))
     
    kvnsmith.at.uw likes this.
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