Conceptual problem on capacitors

In summary: Note that the total charge must be the same as before and that the voltages across the C1 and the C2+C3 pair must be the same.In summary, the problem is to determine the final charges and potential difference of three charged capacitors connected in a circuit. One approach is to replace each capacitor with an equivalent circuit consisting of an uncharged capacitor in series with a voltage source equal to the initial voltage. Then, sum the voltages to find the net EMF and find the net capacitance. The ΔQ for that net capacitance should then be applied to the original capacitors with their initial charges. This method takes into account the initial conditions of the capacitors and the effect of the switch being closed to complete
  • #1
tsw99
35
0
It is neither a homework nor coursework problem, but I really want to know the solution.

Homework Statement


3 charged capacitors are connected as shown in the diagram. What will be their final charges, potential difference when equilibrium is reached? Units are arbitrary.
http://imageshack.us/photo/my-images/19/cap1zu.jpg

Homework Equations


Q=CV obviously


The Attempt at a Solution


(1) Should I use the charge conservation?
E.g. The total charge on left plates of C1 and C2 is 1+4=5 units, on right plate of C2 and left plate of C3 is -4+9=5 units, on right plates of C1 and C3 is -1-9=-10 units
(2) p.d. across C1 = p.d. across C2 and C3?
Equivalent capacitance across C2 and C3 = 1.2 unit

Then I cannot continue...

How about the case if the polarity of one of the capacitors is reversed? I feel really, really confused.
 

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  • #2
The way the circuit is drawn and the polarities shown mean, I would say, that C2 and C3 are in series and C1 is in parallel with that combination.
Do you know how to calculate combined capacitance?
 
  • #3
One way to approach such a problem, where capacitors have some initial charge, is to replace each capacitor by an equivalent circuit consisting of an uncharged capacitor of the same value in series with a voltage source equal to the initial voltage. Take care to maintain the correct polarity for the combination. Thus in your case:

attachment.php?attachmentid=43052&stc=1&d=1327355624.gif


Then to see how much charge is going to move in the circuit (and since they are all connected in series the same amount of current should flow through each), sum the voltages to find the net EMF and find the net capacitance. The ΔQ for that net capacitance should then be applied to the original capacitors with their initial charges. Pay attention to the current direction and thus whether charge will be added to or subtracted from individual capacitors.
 

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  • #4
I would say that you are correct to recognise that V1 =V2 +V3. This means that V1 is in parallel with (V2+V3) in series. This also means that Q2 = Q3 and the total charge = Q1 + Q2. (or Q1+Q3)
Therefore the combined capacitance = (Q1+Q2)/V1.
I would say that C2 and C3 are in series (as drawn) so the charge on C2 = charge on C3 = charge in the series arm of the circuit.
I would say the combined capacitance is C = (C1C2 + C1C3 +C2C3)/(C2 + C3)
 
  • #5
technician said:
I would say that you are correct to recognise that V1 =V2 +V3. This means that V1 is in parallel with (V2+V3) in series. This also means that Q2 = Q3 and the total charge = Q1 + Q2. (or Q1+Q3)
Therefore the combined capacitance = (Q1+Q2)/V1.
I would say that C2 and C3 are in series (as drawn) so the charge on C2 = charge on C3 = charge in the series arm of the circuit.
I would say the combined capacitance is C = (C1C2 + C1C3 +C2C3)/(C2 + C3)

Are you sure C2 and C3 are in series? Because they don't have same charges.
 
  • #6
gneill said:
One way to approach such a problem, where capacitors have some initial charge, is to replace each capacitor by an equivalent circuit consisting of an uncharged capacitor of the same value in series with a voltage source equal to the initial voltage. Take care to maintain the correct polarity for the combination. Thus in your case:

attachment.php?attachmentid=43052&stc=1&d=1327355624.gif


Then to see how much charge is going to move in the circuit (and since they are all connected in series the same amount of current should flow through each), sum the voltages to find the net EMF and find the net capacitance. The ΔQ for that net capacitance should then be applied to the original capacitors with their initial charges. Pay attention to the current direction and thus whether charge will be added to or subtracted from individual capacitors.
Thanks for your suggested solution, but I completely do not understand how it works. Could you please elaborate on it?
Using your method, I figure out the net EMF is 4V, but I do not understand "how much charge is going to move in the circuit", there should be none right? Because in fact it is an open circuit.
 
  • #7
tsw99 said:
Thanks for your suggested solution, but I completely do not understand how it works. Could you please elaborate on it?
Using your method, I figure out the net EMF is 4V, but I do not understand "how much charge is going to move in the circuit", there should be none right? Because in fact it is an open circuit.

I put the open switch in there to make the point that there are initial conditions on the capacitors --- they're given some charge prior to be connected together as a completed circuit. No charges will move or current flow until the switch is closed, after which the circuit is identical to the one you described.

So, if there's an EMF of 4V and all the capacitors are uncharged to begin with, a while after the switch closes what will the be the individual charges on each capacitor? One way to figure this out is to first determine the equivalent capacitance of the three capacitors combined (They are all three in series). Then find the charge that the voltage supply will put on that equivalent capacitor. Add (or subtract, accordingly) that amount of charge to the original charges on the three capacitors.
 

1. What is a capacitor and how does it work?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied to the capacitor, one plate becomes positively charged and the other becomes negatively charged, creating an electric field between them. This stored energy can then be released when needed.

2. What factors affect the capacitance of a capacitor?

The capacitance of a capacitor is affected by three main factors: the surface area of the plates, the distance between the plates, and the type of dielectric material used. A larger surface area and smaller distance between the plates result in a higher capacitance, while using a material with a higher dielectric constant also increases capacitance.

3. How does a capacitor behave in a DC circuit?

In a DC circuit, a capacitor acts as an open circuit, meaning that it does not allow current to flow through it. This is because the electric field between the plates prevents the flow of current. However, when a voltage is first applied to the capacitor, it behaves like a short circuit and allows current to flow until it is fully charged.

4. What is the difference between a series and parallel capacitor circuit?

In a series capacitor circuit, multiple capacitors are connected end to end, with the positive plate of one capacitor connected to the negative plate of the next. This results in a total capacitance that is less than any of the individual capacitors. In a parallel capacitor circuit, the capacitors are connected side by side, with the positive plates connected together and the negative plates connected together. This results in a total capacitance that is equal to the sum of all the individual capacitors.

5. How can capacitors be used in electronic circuits?

Capacitors have many uses in electronic circuits, including filtering out noise signals, storing energy for later use, and smoothing out voltage fluctuations. They can also be used in timing circuits, oscillators, and power supply circuits. In addition, capacitors are often used in combination with other components, such as resistors and diodes, to create more complex circuits and functions.

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