Energy transferred from 1 to 2 capacitors

Hi,
I had a simple problem (probably well known) that was asked to me in an interview:

Say we have two capacitors with equal value C.
One is charged up to V, so the stored charge is Q = CV,
and the other cap contains no charge so Q = 0.
We connect the two together, so the charge from the 1st capacitor redistributes
among itself and the second capacitor.
Due to charge conservation and the fact that the final voltage on both must be the same (since they're shorted together), Q/2 goes to each, meaning V_final = V/2 on both capacitors.
So the initial energy stored on both caps is E_initial = 1/2*C*V^2 + 0
But the final energy stored is E_final = 1/2*C*(V/2)^2 + 1/2*C*(V/2)^2 = 1/4*C*V^2
We find that E_final is less than E_initial by a factor of two.

Where did the energy go?

Is it radiation?

patricky said:

Where did the energy go?

Click to expand...

Heat in the switch. If you take 5V and a couple of 1nF capacitors and simulate it you will find the voltage drops from 5V to 2.5V as expected. The energy was 12.5nJ and is now 6.25nJ. If you look at the power dissipation in the switch you will find the lost 6.25nJ. The switch resistance doesn't matter. If you lower the resistance, the instantaneous power goes up and the time goes down but the lost energy is the same.

Keith.

thanks. this sounds good.

it looks like the energy is being dissipated through the switch resistance?
so if the switch were ideal (R_sw = 0 when on; R_sw = inf when off), there would be an infinite amount of current passing through the switch in a very infinitesimal time?

Yes, it is lost in the switch, but you cannot work with infinity and zero and make any sense of it. I am sure you could generate the equation for the power in the switch resistor versus time and integrate it to get the lost energy.

I prefer to deal with simple values which make practical sense. So, with the example of two capacitors, one charged to 5V, the other to zero, with 1k coupling the peak current will be 5mA and so the power 25mW. This will drop as the voltage on the capacitors equalise. Reducing the resistor simply makes the current increase and the duration decrease so the total energy lost is the same.

Keith

keith1200rs said:

Yes, it is lost in the switch, but you cannot work with infinity and zero and make any sense of it. I am sure you could generate the equation for the power in the switch resistor versus time and integrate it to get the lost energy.

I prefer to deal with simple values which make practical sense. So, with the example of two capacitors, one charged to 5V, the other to zero, with 1k coupling the peak current will be 5mA and so the power 25mW. This will drop as the voltage on the capacitors equalise. Reducing the resistor simply makes the current increase and the duration decrease so the total energy lost is the same.

Keith

Click to expand...

Keith - this is a very interesting question, and not that trivial. From practical viewpoint, you are absolutely right.

However, let's cool the whole system to a very low temperature - so low that the conductor connecting the capacitors (and capacitor plates themselves) become superconductors, i.e. their resistance is not just small, but is exactly zero. This is a very real situation, there are many materials that are known to be superconductors at low temperature.

So, now we do not have Joule heat AT ALL.

Where does (half) the energy go when charge redistributed between the capacitors?

I will let someone else deal with such hypothetical questions! I am an engineer. If you connected to two capacitors with a lossless energy transfer system (a high efficiency switching regulator, which does exist - not 100%, but close) then energy must be preserved so the end voltage would be higher (3.53V, I think)

Keith

patricky said:

so if the switch were ideal (R_sw = 0 when on; R_sw = inf when off), there would be an infinite amount of current passing through the switch in a very infinitesimal time?

Click to expand...

timof said:

So, now we do not have Joule heat AT ALL.
Where does (half) the energy go when charge redistributed between the capacitors

Click to expand...

Physicists - who are used to deal with zero and infinite values ;-) - would explain it as a current pulse of infinitely high value during an infinitely low time span - a so-called Dirac pulse. Using Fourier transform, this corresponds to a radiation spectrum of zero to infinitely high frequencies with infinitely low - but not zero! - energy per frequency. Integration over all frequencies results in a totally radiated power of exactly the missing value.

With a totally surrounding radiation meter - a bolometer - it should be possible to measure this power pulse as totally radiated power.

erikl said:

patricky said:

so if the switch were ideal (R_sw = 0 when on; R_sw = inf when off), there would be an infinite amount of current passing through the switch in a very infinitesimal time?

Click to expand...

timof said:

So, now we do not have Joule heat AT ALL.
Where does (half) the energy go when charge redistributed between the capacitors

Click to expand...

Physicists - who are used to deal with zero and infinite values ;-) - would explain it as a current pulse of infinitely high value during an infinitely low time span - a so-called Dirac pulse. Using Fourier transform, this corresponds to a radiation spectrum of zero to infinitely high frequencies with infinitely low - but not zero! - energy per frequency. Integration over all frequencies results in a totally radiated power of exactly the missing value.

With a totally surrounding radiation meter - a bolometer - it should be possible to measure this power pulse as totally radiated power.

Click to expand...

Erik -

actually, "zero" and "infinity" as a concepts reflecting very small or very large values are normally used by mathematicians.

"Very low" but non-zero, and exactly zero may have quite different consequences - for example superconductivity or superfluidity.

You are right regarding the power dissipation mechanism - if the conductors connecting the capacitors have exactly zero resistance AND zero inductance, then the power will be dissipated by emission of electromagnetic radiation. If the conductors have some inductance, then the energy will be oscillating between the two forms - magnetic energy in the inductor, and electrostatic energy in capacitors (there will be sinusoidal current oscillations - damping with time due to Joule heat and electromagnetic radiation).

Hi everyone,

I assumed that your capacitors are in a lossless system, i.e. the wires have no resistance, and came up with the conclusion that no energy was lost in the process, because all energy should be conserved.

At first we have two capacitors, C1 and C2 of capacitance C.
C1 has a charge Q1, resulting in a voltage V1 as Q1 = C * V1
C2 has a charge of 0, resulting in a voltage V2 as 0 = C * V2 => V2 = 0
If I was to calculate the total energy in this initial system, I would get:
E = 1/2 * C * V^2 (general equation)
E1 = 1/2 * C * (V1)^2 = 1/2 * C * (Q/C)^2 = 1/2 * Q^2 / C (energy in C1)
E2 = 1/2 * C * (V2)^2 = 0 (energy in C2)
E_total = E1 + E2 = 1/2 * Q^2 / C

Now let us attach the two capacitors. On doing so, the effective capacitance is doubled as they are now in parallel.
The C of C1 and C2 combined = 2 * C
As the charge on each capacitor is now Q/2, the voltage should be halved on either capacitor, as everyone agrees on that.
Now, if I were to calculate the total energy in the final system, I would get:
E = 1/2 * (2 * C) * (V1/2)^2 -> This is 2 * C because the effective capacitance is doubled
V1 = Q/C -> From the initial system's capacitor C1
=> E = 1/2 * (2 * C) * (Q/ (C * 2))^2
=> E = 1/2 * Q^2 / C

Therefore, no energy was lost, or gained, in the process.
Q = C * V
Since Q needs to be conserved, and C is doubled, V was halved as a result.

Please let me know if I made any errors, thanks.

yes you made an arithmetic error: 2/4=1/2 not 1.

You are so right.. I fail at life.
So I guess half of the initial energy is lost, even though I tried to account for a completely ideal situation.

Edit:
This is weird.. If you add the capacitors in parallel, the energy is halved. But if you add the capacitors in series, the energy is doubled. Can someone shine some light on this oddity?

lordgraviton said:

This is weird.. If you add the capacitors in parallel, the energy is halved. But if you add the capacitors in series, the energy is doubled. Can someone shine some light on this oddity?

Click to expand...

If you connect the two capacitors in series nothing happens. The first will stay charged at its voltage, the second will stay with zero charge.

I can only imagine a real system with resistance so there will be losses and that is where the losses go as explained earlier. I will leave the hypothetical lossless theory to others!

Keith.

Keith, even if there is a resistance and loss of energy by heat radiated, more than half of the energy is dissipated with the original half unaccounted for. In my search for a solution, I came across this:
https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=01214769&authDecision=-203

Apparently it is a famous paradox with many more papers on the topic. And the way I was solving it showed that by assuming conservation of charge, does not result in conservation of energy. This paper proposes a different method for solving the problem, albeit, thoroughly convoluted.

I cannot access that paper.

I did a quick simulation of the principle and there was no paradox. Charge is conserved, the lost power is in the switch resistance. The numbers add up and agree with my hand calculations. That is for a practical system with some resistance and hence a finite time for the voltages to equalize. I will let other people slug out the hypothetical "zero resistance" theory!

Keith.

Conservation of energy is only threatened, if you forget to model the system correctly. Apart from the fact, that zero resistance
would be an illusion. You may want to assume it for a while. You still have a capacitor inner and connection inductance. With
zero resistance, the respective oscillation will never end, which is simply another way to store the energy.

You may want to design the system in a way, that part of the energy is actually radiated (as electromagnetic wave). Whatever
you do, it doesn't vanish.