# Capacitance and Electrical Energy, Weird Question

1. Feb 23, 2007

### AznBoi

1. The problem statement, all variables and given/known data
To repair a power supply for a stereo amplifier, an electronics technician needs a 100 micro-F capacitor capable of withstanding a potential difference of 90V between its plates. The only available supply is a box of five 100 micro-F capacitors, each having a maximum voltage capability of 50 V. Can the technician substitue a combination of these capacitors that has the proper electrical characteristics, and if so, what will be the maximum voltage across any of the capacitors used? [Hint: The technician may not have to use all the capacitors in the box}

2. Relevant equations
$$C_{p}=\Sigma C_{i}$$
$$\frac{1}{C_{s}}=\Sigma \frac{1}{C_{i}}$$

3. The attempt at a solution
I just fiddled around with the combinations until I arose to one with a perfect total capacitance of 100 micro-F.

My combination: 2 series capaciators that are parallelly linked by another 2 series capacitors.

2 series capcitors= 50F 2 parallel 50F capacitors= 50F+50F=100F

I also attempted to find the maximum voltage across the capacitors and I came up with 100V for each of the 2 parallel capacitors. Since, in parallel circuits, the velocity is constant, does that mean the maximum voltage across any of the capcitors used is 100V??

If my description of my combination befuddles you, I will be glad to draw a diagram to go along with it. Thanks for your help in advance.

2. Feb 24, 2007

### Tom Mattson

Staff Emeritus
Yes, that works.

How? You're only applying 90V, right? So that's the potential difference across each parallel branch. And since the two capacitors in each branch are identical, it should be straightforward to show that they share that 90V equally.

velocity?

3. Feb 24, 2007

### AznBoi

V is volts. Sorry, I know that the letterings and symbols are soo confusing. Anyways, I don't get the voltage part. It says: capacitor capable of withstanding a potential difference of 90V between its plates. I suppose that means that the capacitators have to have a minimum of 90V?? I found the 100V on each branch by adding the series voltage up so 50V+50V=100V per branch. Since 100V and 100V are parallel, doesn't that make the whole system 100V??

I just don't undestand if they already give you a power supply. Is the battery 90V or what?

4. Feb 24, 2007

### Tom Mattson

Staff Emeritus
No, I don't think you've read the problem right. If the repairman needs a single equivalent capacitance that can handle 90V, then I think you are to assume that he intends to put 90V across it. So you've got something like this:

Code (Text):

-----||-----||-----
|                 |
|                 |
-----||-----||-----
|                 |
|------[90V]------|

5. Feb 24, 2007

### AznBoi

Yep, that's exactly what the combination looks like! However, does that mean you just disregard the part: "each having a maximum voltage capability of 50 V??" Does that piece of information mean anything for part b)? If not, I guess the maximum voltage would be 90V??