# Ratio between 2 capacitors from known energy ratio

1. Sep 30, 2012

### david13579

1. The problem statement, all variables and given/known data
Two capacitors connected in series store energy U. Now the two capacitors are connected in parallel to the same voltage source and have energy 5U. What is the ratio of the larger to small capacitor?

2. Relevant equations

For series capacitors Ceq=C1C2/(C1+C2)

For parallel capacitors Ceq=C1+C2
U=Q2/2C=V2C/2
3. The attempt at a solution
I've tried everything that has come to mind:
I've tried using the U ratio as the answer but that is not it
I've tried solving for C1 and C2 but i could not find the info without knowing at least one equivalent capacitance.
I've tried playing around with the symbols, setting things equal etc but i end up going in circles.

2. Sep 30, 2012

### david13579

Oh I found the answer on my own.
I had already come to the equation (C1+C2)2 / C1C2=5 but it was impossible for me to solve. Wolfram Alpha now gave me an approximate answer and it is the ratio indeed.

Is there another way of going about this? I doubt a professor would want students to solve such an odd equation.

3. Sep 30, 2012

### Staff: Mentor

You're looking for a ratio. So set C2 = r*C1. Write the energy expression for both cases. Set one equal to the appropriate multiple of the other. Cancel appropriately and solve for r.

4. Sep 30, 2012

### SammyS

Staff Emeritus
It's not that difficult to solve. It results in a quadratic equation in (C1/C2).

$\displaystyle \frac{(C_1+C_2)^2}{C_1C_2}=5$

Multiply both sides by C1C2 and square the numerator.

$\displaystyle C_1^2+2C_1C_2+C_2^2=5C_1C_2$

Subtract 5C1C2 from both sides then divide all by (C2)2 .

5. Sep 30, 2012

### david13579

Hmmm dividing all by C22... that is a bit that never occurred to me, making the actual ratio the variable. I'd end up with 2 solutions, reciprocal to one another. So I guess that is what Wolfram did then, since it does give the answer in terms of the 2 variables and I just solved for the ratio (simple division).

Thanks.

Still doesn't seem like a nice thing, for the professor to ask for that.

Last edited: Oct 1, 2012
6. Oct 1, 2012

### david13579

OMG, that's it, that's it. Thanks. This is something very simple and obvious, I don't know how I could have missed it. It is also easy enough that I can picture a professor asking for it. Now I feel embarrassed by not realizing myself.

Interesting enough, the ratio found using the method I originally used is slightly different but close enough. I wonder why.

7. Oct 1, 2012

### SammyS

Staff Emeritus
... and ...

What was that ratio ?

8. Oct 1, 2012

### david13579

2.62 for large to small 0.382 for small to large