# Variable Voltage, Two inductor, One capacitor Circuit

1. Sep 18, 2011

### Gasharan

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. Sep 18, 2011

### ehild

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

3. Sep 18, 2011

### Gasharan

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

4. Sep 18, 2011

### ehild

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

5. Sep 18, 2011

### Gasharan

Okay, I got it. Thank you so much for your help.