A bit of help with a capacitor problem

In summary: The total energy stored in each case must be the same (ignoring losses).So in the first case, we have a total capacitance of 16μF, and a charge of 180 μC, giving a PD of 11.25 V.In the second case, we have a total capacitance of 24μF, so the charge on C2 must be 135 μC, giving a PD of 9 V.The change in charge on C2 is therefore -45 μC, or -99 μC (depending on whether you interpret change to mean the difference in charge or the final value).In summary, when the switch is opened in the capacitor circuit shown, the change in charge on
  • #1
bob92
4
0
1. The problem:
In the capacitor circuit shown below C1 = 4␣F, C2 = 12␣F, C3 = 12␣F, and C4 = 3␣F. The switch (S) between A and B is closed and the energy stored in C3 is 1.35x10-3 J. If the switch is now opened calculate the change in the charge on C2. Assume the potential difference between C and D is constant.

3UqNT.png
Here is the work that I have: (sorry the picture is too big to directly link)

http://i.imgur.com/qnTvK.jpg

The question asks what is the change in charge in C2 (the one on the lower left). The correct answer is -99uC but I can't even get my answer close.
 
Last edited:
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  • #2
bob92 said:
1. The problem:
In the capacitor circuit shown below C1 = 4␣F, C2 = 12␣F, C3 = 12␣F, and C4 = 3␣F. The switch (S) between A and B is closed and the energy stored in C3 is 1.35x10-3 J. If the switch is now opened calculate the change in the charge on C2. Assume the potential difference between C and D is constant.

3UqNT.png



Here is the work that I have: (sorry the picture is too big to directly link)

http://i.imgur.com/qnTvK.jpg

The question asks what is the change in charge in C3 (the one on the top right). The correct answer is -99uC but I can't even get my answer close.

You either read the question incorrectly, or typed the question incorrectly - see red above.
 
  • #3
Sorry, I apologize, I mean't C2. C3 is what is given at the start.
 
  • #4
bob92 said:
Sorry, I apologize, I mean't C2. C3 is what is given at the start.

When did you mean C2 - in the question or in the answer?

You have still not removed the ambiguity.
 
  • #5
C2 in the answer. The questions asks for the change in the charge of C2.
 
  • #6
bob92 said:
1. The problem:
In the capacitor circuit shown below C1 = 4␣F, C2 = 12␣F, C3 = 12␣F, and C4 = 3␣F. The switch (S) between A and B is closed and the energy stored in C3 is 1.35x10-3 J. If the switch is now opened calculate the change in the charge on C2. Assume the potential difference between C and D is constant.

3UqNT.png



Here is the work that I have: (sorry the picture is too big to directly link)

http://i.imgur.com/qnTvK.jpg

The question asks what is the change in charge in C2 (the one on the lower left). The correct answer is -99uC but I can't even get my answer close.

Back to square 1 then.

If 1.35 x 10-3 J of energy is stored on C3, which has capacitance 12μF, what is the PD across it, and what charge is stored on it.
 
  • #7
1.35 x 10-3 J would imply a PD of .015 using formula E=(CV^2)/2, so V = .015. Using the formula C = Q/V, gives the charge on c3 to be .18C
 
  • #8
bob92 said:
1.35 x 10-3 J would imply a PD of .015 using formula E=(CV^2)/2, so V = .015. Using the formula C = Q/V, gives the charge on c3 to be .18C

Very small Voltage there. When you stated in the original 12␣F did you mean 12 μF or what?

With 12μF I get 15V [and a charge of 180 μC]

EDIT: DOn't just accept my figures, I could easily be out by a factor of 10
 
  • #9
bob92 said:
1.35 x 10-3 J would imply a PD of .015 using formula E=(CV^2)/2, so V = .015. Using the formula C = Q/V, gives the charge on c3 to be .18C

Just checking that you realize that with the switch closed, we have C1 and C3 connected in parallel, and C2 and C4 connected in parallel, and those two pairs connected in series,

But when the switch is open, we have C1 and C2 in series, and C3 and C4 in series, and each of those pairs connected in parallel.
 

FAQ: A bit of help with a capacitor problem

1. What is a capacitor?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It is made up of two conductive plates separated by an insulating material, known as a dielectric.

2. How does a capacitor work?

A capacitor works by storing electrical energy as charge on its plates. When a voltage is applied to the capacitor, an electric field is created between the plates, causing one plate to become positively charged and the other to become negatively charged. This stored charge can then be released when needed.

3. What types of problems can occur with capacitors?

Some common problems with capacitors include leakage, short-circuiting, and breakdown due to overvoltage. These issues can affect the performance of a circuit and may require replacement of the capacitor.

4. How can I troubleshoot a capacitor problem?

To troubleshoot a capacitor problem, you can use a multimeter to test its capacitance, voltage, and resistance. You can also visually inspect the capacitor for any signs of damage, such as bulging or leaking. If the capacitor is faulty, it will need to be replaced with a new one.

5. Can I repair a faulty capacitor?

In most cases, a faulty capacitor cannot be repaired and will need to be replaced. Attempting to repair a capacitor can be dangerous and should only be done by trained professionals. It is best to replace the capacitor with a new one to ensure the safety and proper functioning of the circuit.

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