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Homework Help: Variable Voltage, Two inductor, One capacitor Circuit

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data

    The question asks you to demonstrate what I0(t) and I1(t) are, it gives you the solutions.

    2. Relevant equations


    3. The attempt at a solution

    I've tried to write the differential equations, and run through the algebra to solve them, but I don't know how to cancel my cos(wt), nor do I know i the equations I have are correct.

    I have: L* I0' +L* I1'=Vcos(wt)

    and I0''+w0^2(I1'-I0')=V/L*cos(wt) (where I differentiated once and divided through by L).

    We're allowed to "assume forms of solutions," which would mean I(t)=I0/1*sin(wt)--basically the problem boils down to finding the coefficients that go in front of sin(wt).
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 18, 2011 #2


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    Homework Helper

    You forgot to differentiate the voltage. The right-hand side of your second equation has to be -wVsin(wt).

    Assume the solution in form I0=Asin(wt) and I1=Bsin(wt), get the differentials and plug in. You get two equations for A and B.

  4. Sep 18, 2011 #3
    I tried that-- and I just got A=B= V/wL--which is not the right answer.

    Are my differential equations written wrong, or am I still missing something?

    Thanks for the help
  5. Sep 18, 2011 #4


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    Homework Helper

    Let's start from the beginning. I did not notice that your second equation had some other mistakes.

    The voltage across the capacitor is equal to Vc=Q/C. And

    Vc =Vcos(wt)-L*I0', that is

    Q/C=Vcos(wt)-L *I0'

    The current through the capacitor is Ic=Q', differentiating the previous equation, you get


    According to Kirchhoff's Current Law, I0=Ic+I1, plugging Ic=I0-I1 into the previous equation, you get

    (I0-I1)/C=-Vwsin(wt)-L*I0". Rearranging:

    I0''+(I0-I1)/(LC)=-V(w/L )*sin(wt), but LC=1/w0^2,

    so finally you have the equations for I0 and I1:


    L* I0' +L* I1'=Vcos(wt)

    Try now again, with I0(t)=Asin(wt) and I1(t)=Bsin(wt).

  6. Sep 18, 2011 #5
    Okay, I got it. Thank you so much for your help.
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