Why trying to store energy in a capacitor can be less efficient than you think

[MrAl]

Hi again and thanks for the reply,

When you say "circuit" do you mean the circuit with RC or the circuit with RLC or the circuit with zero resistance with or without inductance?  We've been talking about some four or so different circuits.

Also, when you say "change" do you mean increase or decrease the 50 percent, or both increase and decrease?

[IanB]

What still isn't clear to me is whether the radiation losses are in addition to the 50% or included in the 50%, nor the % in which energy is lost as heat in resistances vs the losses due to EM radiation, nor if a 50% loss is inevitable. Some have suggested in this thread that slow charging trough an inductor can reduce the losses below that 50%, is it true?

OK, let's take this step by step with a two capacitor example.

Problem statement: We have two identical capacitors, one charged to voltage V₁, one discharged. We connect them together and wait for the system to stop changing. What is the final result?

To solve such a problem we need to state our assumptions.

Assumption #1: Charge is conserved in the system (no charge is added or removed from outside)
Assumption #2: The system eventually settles out and the voltages stop changing (the system has at least some source of dissipative loss so that it doesn't oscillate forever)
Assumption #3: The capacitance of each capacitor is constant and does not change
Assumption #4: The capacitors are ideal and have no charge leakage between their plates

Given these assumptions we can proceed.

Firstly, determine the initial charge in the system:

    Q = CV₁

This is merely the charge stored in the first capacitor. The second capacitor has no charge.

Secondly, determine the final voltage V₂ when the two capacitors have equalized. By assumption #1 we must have:

    Q = 2CV₂

Therefore:

    V₂ = ½V₁

Thirdly, determine change in energy between the initial and final states. We have:

    E₁ = ½CV₁²
    E₂ = 2(½CV₂²) = CV₂²

But since V₂ = ½V₁, we find that:

    E₂ = CV₂² = ¼CV₁² = ½E₁

As you can see from the working, the change in stored energy is exactly 50%. Neither 49.999% nor 50.001%, but 50% exactly. The only way to change this result is to change the assumptions.

Does that help?

[dannyf]

E2 = CV2² = ¼CV1² = ½E1

So every time you split your capacitor into halves, you would double your energy.

As the capacitors get smaller and smaller, your total energy gets doubled, and redoubled and so on and so forth.

Don't you love math when you don't understand it?

[T3sl4co1l]

Hi again and thanks for the reply,

When you say "circuit" do you mean the circuit with RC or the circuit with RLC or the circuit with zero resistance with or without inductance?  We've been talking about some four or so different circuits.

Also, when you say "change" do you mean increase or decrease the 50 percent, or both increase and decrease?

Anything topologically equivalent; so, a series circuit between a discharged cap, a charged cap, and some (non capacitive) impedance, or a discharged cap, a voltage source, and some (non capacitive) impedance.

Obvious "no good" cases include anything where the capacitor is shunted by a resistor, thus its charge bleeds away at t --> infty.

Speaking of, note that switching converter circuits break this rule, because they exhibit switch leakage to ground or whatever.  The leakage is small enough to be of great value in the short term (microseconds to years), or under continuous operation (power transfer rather than energy transfer).  The latter of course is beyond the scope of thermodynamics, which deals with the balance of energy, not the flow of energy (power).

Tim

[CatalinaWOW]

E2 = CV2² = ¼CV1² = ½E1

So every time you split your capacitor into halves, you would double your energy.

As the capacitors get smaller and smaller, your total energy gets doubled, and redoubled and so on and so forth.

Don't you love math when you don't understand it?

I love it when a comment actually answers itself

[Galenbo]

...
What's indeed striking, is the independence of this phenomenon from the resistance .. that's a very strong indication, that there is a deeper physical effect buried in the grains, something like a "universal" physical effect.

We handled this in university with a mechanical example: Making a shaft turn, by coupling a clutch, has an efficiency of 50%.
Independent of the time that is taken to do it, independent of rpm, Nm and so on.