1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Variable Voltage, Two inductor, One capacitor Circuit

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data
    http://imageshack.us/photo/my-images/101/circuitk.jpg/

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


    2. Relevant equations
    V(t)=Vcos(wt)

    I0(t)=(V/wL)((w0^2-w^2)/(2wo^2-w^2))
    I1(t)=(V/wL)((w0^2)/(2wo^2-w^2))



    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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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.

    ehild
     
  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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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

    Ic/C=-Vwsin(wt)-L*I0''.

    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:

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

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


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

    ehild
     
  6. Sep 18, 2011 #5
    Okay, I got it. Thank you so much for your help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook