# Where did it go?

1. Nov 5, 2005

### Martin

Two ideal capacitors (i.e., purely capacitance—no resistance or inductance), each with a capacitance of C, are connected together through ideal wires (zero resistance), and an ideal switch (i.e., when open, the switch offers infinite resistance; when closed, it offers zero resistance), which initially (say, for t<0) is open.

A charge Q is placed on one of the capacitors; the other has zero charge. Since the energy stored in an ideal capacitor is given by

Energy = (1/2) (1/capacitance) X (charge)^2

we see that the energy stored in the two capacitors is

E(1st Capacitor, t<0) = (1/2) (1/C) X Q^2

and

E(2nd Capacitor, t<0) = 0.

Hence the total energy stored in the circuit while the switch is open is

E(total, t<0) = (1/2) (1/C) X Q^2.

Suppose (say, at t=0) the switch is then closed. The charge Q will flow from the 1st (charged) capacitor onto the 2nd (uncharged) capacitor until it’s evenly distributed between the two capacitors (because each has the same capacity to store charge). Hence, each will end up storing a charge of Q/2.

The energy stored in the two capacitors after the switch is closed is, therefore,

E(1st Capacitor, t>0) = (1/8) (1/C) X Q^2

and

E(2nd Capacitor, t>0) = (1/8) (1/C) X Q^2.

Hence the total energy stored in the circuit after the switch is closed is

E(total, t>0) = (1/4) (1/C) X Q^2 .

Clearly, half of the initial (before the switch is closed) energy is lost after the switch is closed.

Where did it go?

(The answer is not at all obvious.)

Last edited: Nov 5, 2005
2. Nov 5, 2005

### Reality_Patrol

This is actually quite a practical problem, this loss places limits on the peformance of charge-pumps (DC-DC converters). The answer is entropy, but as far as I know the details of the energy transfer mechanism are unknown.
Don't get me wrong, it clearly produces heat in charge-pump circuits, but whether that heat comes from "transient" conduction losses, or "IR" type radiation between the plates is uncertain. Do you know?

3. Nov 5, 2005

### Martin

We are dealing with an “ideal” circuit (i.e., lumped circuit theory), which means that you must analyze the circuit’s behavior in terms of idealized circuit elements—that is, electrical properties (resistance, capacitance, and inductance) as opposed to actual (physical) electrical devices (resistor, capacitor, and inductor). In this instance, the circuit specifies that the only circuit elements present are capacitances. The property “capacitance” accounts only for electrical energy storage. Nevertheless, if you carefully apply Kirchoff’s Laws, you can arrive at the answer.

Last edited: Nov 5, 2005
4. Nov 6, 2005

### SGT

Your first capacitor has a voltage $$V=\frac{Q}{C}$$, while the second one has a voltage of zero. Connecting both of them in parallel violates Kirchoff's Voltage Law, so you are not allowed to do it.
Of course, with real components, there will be a resistence associated, the charge of the second capacitor will not be instantaneous and there will be energy dissipated as heat in the resistive part of the circuit, so no paradox exists.

5. Nov 6, 2005

### Martin

The “violation” is only apparent, and no “paradox” exists.

6. Nov 6, 2005

### SGT

Is this an opinion, or do you have arguments to support your affirmation?

7. Nov 6, 2005

### Martin

Lumped circuit theory is self consistent: By insisting that Kirchoff’s Laws hold, there will be no violations, and the circuit will be forced to respond in such a manner that avoids any apparent paradox.

8. Nov 6, 2005

### SGT

According to KVL, two elements in parallel must have the same voltage. You cannot link together two capacitors with different voltages, because it violates that law. You can find this in any circuits theory book. In Basic circuits Theory by Charles Desoer and Ernest Kuh it is specifically written. My edition is in Portuguese, so I don't know if the page reference would be valid in the American edition, but it is in item 5.2, page 89.

9. Nov 6, 2005

### Martin

The capacitors are simultaneously in parallel as well as in series.

By insisting that Kirchoff’s Voltage Law holds, the circuit will respond by forcing the charge to redistribute itself such that the capacitors attain the same voltage.

I am not familiar with that particular textbook. I can assure you, however, that there is nothing inherent in lumped circuit theory that prohibits “link[ing] together two capacitors with different voltages.”

In lumped circuit theory, there are no “disallowed” circuit configurations: As long as you properly apply the voltage/current relationships for each circuit element and enforce Kirchoff’s Laws throughout the circuit, you will arrive at self-consistent results.

10. Nov 6, 2005

### SGT

can you explain how lumped circuit theory allows two different branches of a parallel circuit to have different voltages?

11. Nov 6, 2005

### Martin

Lumped circuit theory does not allow parallel branches to have different voltages. Quite the contrary, the theory insists—via Kirchoff’s Voltage Law—that they have the same voltage. This is the key to analyzing the above circuit and answering the question that I posed.

12. Nov 6, 2005

### Reality_Patrol

I'm not exactly sure where you're going with all of this. I'd say the problem formulation given in your initial description contained the result of applying circuit laws: there is a loss of stored energy.
So, you've got me, why not just share your answer? I think you might have found something interesting; something mathematically correct but physically inaccurate since heat production is the experimentally found answer. Models are just that, the real thing is what it is. Still, I'd like to see it. Come on, don't be scared.

13. Nov 6, 2005

### SGT

Are capacitors allowed to change voltage instantaneously in that theory?

14. Nov 6, 2005

### Martin

Why not at least attempt to find a solution analytically, the way you would go about determining the solution to any circuit problem? By doing so, you might get some insight into what makes this such an interesting problem.

15. Nov 6, 2005

### Martin

Why shouldn’t a capacitor’s voltage be able to change instantaneously?

16. Nov 6, 2005

### Corneo

I believe you would need infinite energy. One of the conditions used in transient analysis is $$v_c({t_0}^{-}) = v_c(t_0^+)$$

17. Nov 6, 2005

### SGT

Because the current in a capacitor is $$i=C\frac{dV}{dt}$$. If $$dt$$ is zero, $$i$$ is infinite.

18. Nov 6, 2005

### Martin

How so? We begin with a finite amount of energy ((1/2) (1/C) X Q^2), and we end up with a finite amount of energy ((1/4) (1/C) X Q^2).

Where does this come from? Are there any situations where this is not true?

19. Nov 6, 2005

### SGT

Yes, if you have an impulse of current. Infinite amplitude and zero time, so the energy is finite.

20. Nov 6, 2005

### Martin

What is wrong with infinite current?