# 2 Capacitor Problems

1. Nov 4, 2009

### lubuntu

1. The problem statement, all variables and given/known data

I think I have a solution to this but I might have made a logical error so I want to check if my reasoning works.

You have two capacitors one charged by some unspecified mean to a paricular voltage. It is connected then connected to another capacitor in parallel with resistor somewhere in the loop. What is the heat dissipated through the resistor?

2. Relevant equations

v = v0 * e^(-t/RC) discharging capacitor
P = V^2/ R heat dissipation through resistor
C = Q/V

3. The attempt at a solution

Charge is conserved in the situation so regardless of the capacitor eventually the potential difference of the 2 capacitors is the same

Q = C1V1

VFinal = Q/(C1+C2)

We can find the time it took for the orginal capacitor to discharge the amount need to reduce the potential difference to Vfinal namely

t = -RC * ln(vf/v0)

then we can look at the infinetessimal amount of heat disappainted at each time

dW = V(t)^2/R dt

then you take the integral from 0 to t as shown and get some number.

I see a potential problem with this though. Since the given potential at time t is that of the capacitor do I instead of to use the current through the resistor at time t and then integrate over dW = I(t)^2/R dt?

Also I feel like I'm discounting whatever changes to the current/ potential to the wire will happen as a result of capacitor 2 charging. I feel that since I know the amount of charge that moved through the resistor I should just easily be able to calculate the energy loss. What principle am I missing?

Last edited: Nov 4, 2009
2. Nov 4, 2009

### Delphi51

I can't understand the text diagram. The actual question hasn't been stated.
Very difficult to work on this . . .

3. Nov 4, 2009

### lubuntu

I'm looking for the heat loss, I have no idea how to fix the diagram.

4. Nov 4, 2009

### Delphi51

If all of the charge dissipates through the resistor, all the energy will be converted to heat. Have to see the circuit to tell if that will happen, though.
Could you draw the diagram in a paint or draw program, save it as a jpg image and upload to a site like photobucket? Then you could put a link to it here.

5. Nov 4, 2009

### lubuntu

its a pretty simple circuit just two capacitors connected in parallel with a resistor between them on one side. What you do mean thought that charge is dissipated, I would a resistor doesn't "lose" charge.

I'm trying to figure out how much energy is lost through the resistor as the capacitors go from one being entirely charged and one having charge zero to them at some time later have the same potential since they are in parallel.

6. Nov 4, 2009

### lubuntu

Does the energy lost even depend at all on the capacitor, could you just calculate the energy of the system before and energy and see that the conversation of energy doesn't hold. So whatever energy you lost, you must have lost in resistance?

7. Nov 4, 2009

### Delphi51

Do you mean like this?

Here there would be a charge of +q on one side of a capacitor and a charge of -q on the other side. Electrons would flow from one side to the other, through the resistor, neutralizing the charge to zero on both sides. All the energy originally in the capacitors would be converted to heat in the resistor.

8. Nov 4, 2009

### lubuntu

No in my problem the resistor is situation on the path between the capacitors not in parallel with them.

9. Nov 4, 2009

### Delphi51

Okay, so no way for the capacitors to lose all their charge. If the two capacitors have the same initial voltage, no current will flow at all. No energy lost to heat.

10. Nov 4, 2009

### lubuntu

Nope, at t=0 all the charge is on one capacitor and begins to flow toward the other until they are in equilibrium and then the they have the same voltage at a later time. Obvious the energy stored in the parallel capacitors doesn't add up to the energy the original one once had. But is this the only way or even a way to calcuate the energy lost. Since this appears to be entirely independent of the resistor being there.

11. Nov 4, 2009

### Delphi51

Okay, all clear now (C1 = C2?). You can use v = v0 * e^(-t/RC) to get the potential on the charged capacitor as a function of time. There will be a similar function for the charge on the other one with the total charge constant as you said before. Knowing the two voltages, you can easily calculate the current through the resistor and then the power loss in it as a function of time. If you know integration, you can integrate that up to the time when the two voltages are equal.

12. Nov 4, 2009

### lubuntu

Thas was my initial though, and it worked through it but then I figured that I might just be able to say that the energy lost through the resistor must be whatever gets lost in changing the state of the capacitor?

13. Nov 4, 2009

### Delphi51

Oh, you have a very clever solution there! Of course it works and the answer pops right out!

14. Nov 4, 2009

### rl.bhat

The total energy of the system before and after sharing the charges does not depends on the resistance. The resistance decides when the capacitors attain the common voltage. If you are interested in th energy loss due to the charge sharing, find the initial energy and the final energy. Then find the difference in the energies.
Ei = 1/2*Q^2/C1
Ef = .....?