# Capacitance in series

1. Jan 30, 2006

### jgonzon

I have a problem where there is a capacitor (10 mF) charged to 15V, that is then placed in series with another capacitor (5 mF) and then an emf of 50V is added. I need to find the the potential across both capacitors.

My question is can I combine the capacitors and make one since they share the charge? But then what happens to the charge when we add the 50V emf? Can I add those together?

Any suggestions would be greatly appreciated.

Jose
:surprised

2. Jan 30, 2006

### chroot

Staff Emeritus
Capacitances in series add as follows:

$\frac{1}{C_\textrm{total}} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n}$

- Warren

3. Jan 30, 2006

### Staff: Mentor

It's hard to picture what you are asking. In the end, do you have two capacitors in series and the 50V power supply in series as well? Or is the 50V power supply connected across the two series capacitors? Can you maybe sketch the problem?

4. Jan 31, 2006

### jgonzon

Pic of circuit

Here is what I am talking about. The 10 mF is charged first before being added to the circuit. I can combine the two capacitors to get the one, but what happens to the charge all ready on the 10 mF? Does it get absorbed by the emf? Since the two capacitors share the same charge do I add it to the emf?

#### Attached Files:

• ###### circuit.png
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5. Jan 31, 2006

### Staff: Mentor

The attachment is still pending approval at the moment (plus I don't know what format a *.png file is....), but to answer the capacitor part of your question, just use the equation Q=CV. If you have a capacitor C1 charged up to V1, that means that you have forced a charge Q1 onto the plates of the capacitor. If you then hooked an uncharged cap C2 in parallel withi C1, then the charge will redistribute across the two capacitors. Since the caps are in parallel, they will have the same voltage across them, but the charge will redistribute between the two caps based on the capacitance of each (using the equation above).

That's part of the confusion I have about your problem description -- hooking the caps up in series makes no sense, unless the power supply somehow completes the circuit. In any case, try to use the Q=CV equation to help out.

6. Feb 2, 2006

### Staff: Mentor

Ah, the attachment came through. Think about what happens when you connect up two capacitors like that to the power supply, but neither cap is charged initially. You get a voltage divider across the caps, so you have 50V on one side, 0V on the other side, and some other voltage in the middle. Think in terms of charge and capacitace, and use Q=CV to figure out what the middle voltage is. Keep in mind that some charge is getting pushed off of the - plate of the 5mF cap and onto the + plate of the 10mF cap next to it. That charge re-distribution is what gives the middle node its voltage between 50V and ground.

Now if the 10mF cap is initially charged to 15V before being connected in this circuit, it will have some initial charge separation across it, again using Q=CV to understand how much charge. Depending on which way the polarity is on the 10mF cap when you connect it, you will get two different answers. Hopefully in the homework question they showed the polarity of the 15V on the cap initially. For example, if the cap is connected so that the + is to the left, then the voltage across the cap will go up because of the charge that is pulled off the - pin of the 5mF cap by the overall EMF of the circuit.

7. Feb 3, 2006

### jgonzon

Thanks

Thanks to everyone that posted to my question. I will use that and wait until I get to class to see how it all works out. It is hard to see it sometimes.

Thanks, again

8. Jun 21, 2006

### meldave00

Capacitors in series

I have a question in regards to capacitors in series. We are taught that the charge must be equal on all plates of the capacitors in the case of a series connection. However, we also know that Q = CV. If we increase V, then Q must increase. How is this possible for the node between the two capacitors. Does it not have a finite amount of charge? Also, does not all the charge at the middle node just immediately rush to the perspective plates no matter what voltage is applied? Or does just some charge rush to the plates?

Hope this isn't too confusing. I will clarify more if people respond.

regards,

David

9. Jun 21, 2006

### Staff: Mentor

Welcome to PF, Dave. One quick tip as you get started -- it's generally better to start a new thread instead of reviving an old one if you have a question. When people see that there have been several replies to a thread, they are less likely to open it up to see if they should respond. Once a thread gets several responses, the OP's (original poster's) question is likely answered, so often we will go right past those threads without opening them.

But, to answer your questions.... First of all, I don't know who said that the charge has to be the same on all caps in series -- that's simply not true. Maybe you misunderstood their statement. Capacitors in series do not interact with each other, other than to have their voltages add in series, and their total capacitance reduced by the one-over one-over addition equation. Think of each capacitor as a separation of charge between just its plates, and that's what generates the voltage according to the equation you listed, Q=CV.

10. Jun 21, 2006

### meldave00

Berkeman,
I've started a new thread on this topic so if you don't mind try to respond on that one if you can. But back to the subject at hand:

There are dozens of reading and resources that state that the charge on both caps are equal in a series situation. I'm assuming that this is the case for the positive charge and negative charges on both plates of the capacitor.

Have you ever tried to calculate the voltage at the conductor (node) between two caps. In order to get the correct result, you have to first assume that the charge Q is equal on both caps. The you use the Q=CV calculation of both to figure out the individual voltage drops across each. This can only be achieved if you first assume that the charge is equal on both caps.