# Finding the charge for Capacitor

Gold Member
1. Homework Statement
Find the charge on C4. (That's the 4.0 uF in the middle). C1=15 uF, C2=10uF, C3=15uF, C4=4uF, battery = 10 v 2. Homework Equations
Q=CV

3. The Attempt at a Solution
The presence of C1 (the one next to the battery) is confusing me. It's in series with C2 and C3. It's also in series with C4, but C4 is parallel to C2,3. I can't find an example on how to do this type of problem.

The answer is 24mC, but I don't know how to get it.
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

Related Introductory Physics Homework Help News on Phys.org
Nabeshin
Deconstruct then reconstruct. Combine all resistors in series and parallel. Start with the inner-most part of the circuit (what looks to be the most complicated) and just start combining capacitors.

Gold Member
I tried that, but with no luck

The 10 and 15 are in series, so they make a combined 6uF capacitor
This new 6 and the 4 are in parallel, so they make a 10
The new 10 and the bottom 15 are in series, so they make a 6.
So the whole circuit simplifies to a single 6uF capacitor.

Q=CV = 6*10 = 60

The 15 on the bottom and the combined 10 are in series, so the charge must be the same. Is this 60? Or does it get divided?

Reconstructing it, the 4 and the combined 15 on the top are in parallel. So assuming that the total is 60, how do I compute the charge on each?

Nabeshin
When computing charge, there are two important things to remember: Q is the same for capacitors in series and V is the same for capacitors in parallel. That should be enough ^^

Gold Member
I still don't get it. I don't see any opportunity to have a voltage other than 10. When I do it, I don't get 24.

Gold Member
Here's my effort showing where I get stuck If you replace two series capacitors by a single one, and then calculate the charge on the replacement capacitor, both of the original capacitors will have the same charge as the replacement capacitor. (because the same current went through them at all times)