Discharging Capacitors into Capacitors

1. Aug 9, 2008

Seph83

Hi Guys,

I hope you can help

A friend and I are having a debate about capcitors and now we are both a bit confused!

So I have a few questions.

Firstly, if you take for example a 10,000uf capcitor charged to 12v, then connect it in parellel with an uncharged capacitor that is also 10,000uf, you end up with 6v in both capcitors but only half the joules you started with...

How is the energy disipitated from the system?

Is charging a capacitor only 50% efficient?

2. Aug 9, 2008

rbj

if there is no resistance in series, i think charge will slosh back and forth from one cap to the other forever.

of course, there is always resistance.

3. Aug 9, 2008

Seph83

hmmm.... so the charge continues "sloshing" between each capacitor unitl the voltages stabalise?

To test that I just tried discharging the cap (as in the example above) through a diode to prevent "sloshing" but the results are the same. Both capacitors rest at 6 volts.

4. Aug 9, 2008

Defennder

This is similar to the loss of KE in an inelastic collision. Where does the lost KE go to? In this case, the charges redistribute themselves to lower the energy of the charge configuration.

Last edited: Aug 9, 2008
5. Aug 9, 2008

Seph83

ok, I think I understand that...

The charge density before the experiment is double the density of the charge after the experiment even though the total charge remains the same, therfore the energy available is half the original... is that right?

6. Aug 9, 2008

Defennder

Ok typo in my previous post. It's supposed to be "inelastic collision", not elastic one. And assuming the two capacitors are identical parallel plate capacitors, the charge density on each plate is halved, as you said.

7. Aug 9, 2008

Seph83

hmm... ok... so bearing that in mind....

Is there a method to take a 10,000uf capacitor charged to 12v (total 0.12 coulombs of charge) and discharge it into another 10,000uf capacitor that's currently on 0v and end up with a charge greater than 0.12 coulombs between the two caps? For example, both capacitors ending up at 7v (0.14 coulombs). Perhaps using a transformer?

8. Aug 9, 2008

Seph83

sorry... was that a stupid question?

9. Aug 9, 2008

Staff: Mentor

This question has come up before, so I decided to do it correctly. I started with the circuit shown below. This circuit is described by the equations
$$\begin{array}{l} 0=\frac{1}{R}(V_1(t)-V_2(t))+C_1 V_1'(t) \\ 0=\frac{1}{R}(V_2(t)-V_1(t))+C_2 V_2'(t) \\ V_1(0)=V_{10} \\ V_2(0)=V_{20} \end{array}$$ (1)

Which have the solution

$$\begin{array}{l} V_1(t)=\frac{C_1 V_{10}+C_2 \left(e^{-\frac{(C_1+C_2) t}{C_1 C_2 R}}(V_{10}-V_{20})+V_{20}\right)} {C_1+C_2} \\ \\ V_2(t)=\frac{C_2 V_{20}+C_1 \left(e^{-\frac{(C_1+C_2) t}{C_1 C_2 R}}(V_{20}-V_{10})+V_{10}\right)} {C_1+C_2} \end{array}$$ (2)

The energy is
$$\begin{array}{rl} \text{Energy in }C_1\text{ at t=0} & \frac{1}{2} C_1 V_{10}^2 \\ \\ \text{Energy in }C_1\text{ at t=}\infty & \frac{C_1(C_1 V_{10}+C_2 V_{20})^2}{2(C_1+C_2)^2} \\ \\ \text{Energy in }C_2\text{ at t=0} & \frac{1}{2} C_2 V_{20}^2 \\ \\ \text{Energy in }C_2\text{ at t=}\infty & \frac{C_2(C_1 V_{10}+C_2 V_{20})^2}{2(C_1+C_2)^2} \\ \\ \text{Energy dissipated by R by t=}\infty & \frac{C_1 C_2 (V_{10}-V_{20})^2}{2(C_1+C_2)} \end{array}$$ (3)

Interestingly, the energy dissipated by the resistor is not dependent on the value of the resistance, so even in the limit as R->0 the same amount of energy is lost to the resistor.

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10. Aug 10, 2008

Seph83

Thank you for your help guys, but just to clarify, is it possible to use a transformer to take a 10,000uf cap charged to 12v to end up with two caps rated at 10,000uf charged to 7 or even 8v each.

Or to put it simply. Is it possible to end up with more coulombs then what you started with?

11. Aug 10, 2008

Defennder

Bear in mind the transformer works on AC voltage, not DC. So the charge on the capacitors will be oscillating all the time. I'm not sure if your question makes sense here. You'll then have to calculate the RMS value of the energy stored in the capacitors.

12. Aug 10, 2008

Seph83

I understand, though all I really want to know is if you can end up with more charge then what you started with. Since half the energy is disipitated as heat when discharging a capacitor into another capacitor you end up with the same amount of coulombs on either side. What my friend and I are argueing about is whether it is theoretically possbile to take the charge from one capacitor and end up with more charge by discharging it into another capacitor by any method.

and I do understand more charge doesn't neccessarily equal more energy we're not trying to perform any magic here. lol

13. Aug 10, 2008

Defennder

The problem with the notion of "more charge" is that the charge isn't a constant. It's time-varying since you're now dealing with AC instead of DC quantities. It's now Q(t) and not just Q. More charge with respect to which point in time?

14. Aug 10, 2008

Seph83

more charge meaning the end results in the capacitors.

Like in the example above. Starting with 0.12 coulombs in one capacitor and 0 in the other, then discharging the charged capacitor into the uncharged capacitor to end up with 0.16 coulombs of combined charge.

15. Aug 10, 2008

Staff: Mentor

Well, you couldn't do it through a resistor, as I showed above, because you lose energy to the resistor. Perhaps you could do it with some fancier circuit using inductors, but I don't know.

Assuming no energy loss you would have each capacitor charged to 1/sqrt(2) of the original voltage. So if you started with each at 10 V you could end with each at a maximum 7.07 V from energy conservation considerations alone.

Note that you are not "creating charge", there is no conservation of matter concern here. All you are doing is separating more charges.

16. Aug 10, 2008

Seph83

Oh well, looks like I lost that bet :rofl: though if the half the energy is lost in charging a capacitor from another capacitor as heat (even in the theoretical scenario of zero resistance) how can an inductor prevent or reduce this loss since an inductor will probably have higher resistance than simply connecting the caps together and will probably emit more energy as EM? I am not asking for a circuit, I am just trying to understand.

17. Aug 10, 2008

Staff: Mentor

The reason that the same amount of energy is dissipated in the resistor regardless of the resistance is that as the resistance goes to zero the current goes to infinity. An inductor prevents that runaway current. Also, an inductor is an energy-storage component, so the energy that goes into the inductor will come back out.

I don't know that it would work, I would have to solve the equations to see, but it is different than just using a resistor.

18. Aug 11, 2008

vanesch

Staff Emeritus
Imagine you have a capacitor C1 and an inductor L. At t = 0, you connect the capacitor to the inductor. The current will rise, and the charge on C1 will lower. At t1, you will have that the charge on C1 is 0, and the current in L is maximal. Later, the charge on C1 will inverse, the current will lower through L and so on.

But let us imagine now that exactly at t1, we connect a bigger capacitor C2 to the inductor, and switch off capacitor C1 "immediately". The current through L will now decay slower, because the voltage is lower (C2 is a bigger capacitor), but nevertheless there will be a moment when C2 will be maximally (inversely) charged, and the current is 0. Call that t2. At that moment, you uncouple L from C2. I didn't do the calculation, but intuitively, I'd say that there is a bigger charge on C2 than there was on C1.
(maybe I'm guessing wrong...)

19. Aug 11, 2008

Staff: Mentor

Hmm, I solved the RLC circuit and got that the two capacitors always end up charged to half of the initial voltage regardless of R and L. I am not sure I trust the result, so I am not going to post the work. That would indicate that some energy is lost to the resistance, but always the same amount. Since there is always some inductance to a circuit it would seem that connecting two identical capacitors (one initially uncharged) will simply split the voltage between the two.

20. Aug 12, 2008

vanesch

Staff Emeritus
That's because you "connect" them and then let them come to rest for t -> infty. But you can do smarter things, like opening and closing switches at the right times.

21. Aug 12, 2008

Staff: Mentor

Yes, that is true. I am not going to try and figure out anything like the perfect timing. Anyway, I looked at my results some more, and I think I believe them now, but I would be glad for anyone to check. I started with a series RLC circuit, with two capacitors, one initially charged (C1) the other initially uncharged (C2). The current is 0 at time 0 so all of the voltage is across C1 and L. I get that current in the circuit is:

$$i(t) = \frac{2 e^{-\frac{R t}{2 L}} V_0 \sinh \left(\frac{\sqrt{R^2-\frac{4 (C_1+C_2) L}{C_1 C_2}} t}{2 L}\right)}{\sqrt{R^2-\frac{4 (C_1+C_2) L}{C_1 C_2}}}$$ (4)

Regardless of the inductance and resistance, as t->infinity the same amount of energy is dissipated in the resistor, and the capacitors each wind up with a voltage equal to V0 C1/(C1+C2) where C1 is the capacitor that is initially charged to V0 and C2 is the initially uncharged capacitor. The fraction of energy lost in this transfer is C2/(C1+C2), so 50% is lost if the capacitors are matched.

Last edited: Aug 12, 2008
22. Aug 12, 2008

uart

You can use a diode to automatically stop the current when the second capacitor reaches it's maximum value (at which point 100% of the energy in the first capacitor is transfered to the second one). If however you want to get some other final voltage on each capacitor then you do need to do some active switching.

With reference to the circuit in the attached figure let me outline its operation (assuming we desire to get half the initial energy on each capacitor).

Initial conditions: Assume v1(0)=V, v2(0)=0 and the inductor current is i(0)=0. I'll also assume that the two capacitors are of equal value.

- Assume the switch is in position one for t<0 and we switch it to position 2 at t=0.
Solving the DE's with the above IC's gives :

$$i = \frac{V}{\omega_1 L} \sin(\omega_1 t)$$

$$v1 = \frac{V}{2} ( 1 + \cos(\omega_1 t) )$$

$$v2 = \frac{V}{2} ( 1 - \cos(\omega_1 t) )$$

where $$\omega_1 = \sqrt{\frac{2}{LC}}$$

- Switch to position 3 at $\cos(w_1 t) = \sqrt{2} -1[/tex] (and therefore $$\sin(w_1 t) = \sqrt{ 2 \sqrt{2} - 2 }$$ ). At this point v1(t) remains constant at [itex]V/ \sqrt{2}$, so it retains half of it's initial energy.

(For convenience I'll reset the time variable to zero for the next phase of the analysis).

The initial conditions for the next transient are therefore $$i(0)=\frac{V}{w_1 L} \sqrt{2 \sqrt{2} - 2}$$
and $$v2(0)=V*( 1 - 1 / \sqrt{2} )$$

Now $$V2 = A cos(w_2 t) + B sin(w_2 t)$$

where $w_2 = 1/ \sqrt{LC}$, $A = v2(0)$ and $B = i(0)/(w_2 C)$
{ B = V* sqrt(2 sqrt2 - 2)/(w1 w2 LC) = v* sqrt(sqrt2 - 1) }

So $$v2 = V \, (1-1/ \sqrt{2}) \, \cos(w_2 t) + V \, \sqrt{\sqrt{2} - 1} \, \sin(w_2 t)$$

The maximum voltage reached on capacitor C2 is $v_2 \max = \sqrt{A^2 + B^2}$

So $$v_2 \max = V \sqrt{\sqrt{2} - 1 + 1 + 1/2 -\sqrt{2}} = V/ \sqrt{2}$$.

At this point i=0 and the other half of the inital energy is stored in c2. If the switch is returned to position 1 at this time then c1 and c2 will each retain a voltage of V/sqrt(2).

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