Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

[hanlu]

Hi to all. What seemingly was originally thought as an extremely easy question was not well understood. So it's written very specifically..
It is meant to be understood that a capacitor starts with 0J and 0V in it. Next the switch is closed in order to connect the battery to the capacitor in order to charge it up to the exact same voltage level of that of the battery.
There is now 7.2 milliJoules of energy stored in the capacitor.

I am only referring to the schematic presented, which works exactly like what the schematic portrays. This is a real world circuit done on the work bench. It has losses due to resistance in the wires, capacitor, battery, and switch.

It is not lossless and so it is not perfect, there are losses involved as mentioned immediately above.

Half the energy is lost in doing work to transfer the other half into the capacitor. In other words the amount of work it takes to transfer half the energy into the capacitor is exactly equal to the amount of energy stored in the capacitor. This is a well known fact of life and electronics.

In other words, if 2 Joules of energy came out of a battery, then only 1 of those Joules would make it into the capacitor.

What I am asking is: When charging a capacitor from a battery, is this 50% loss of energy and 50% stored energy in the capacitor a set rule in stone?

I am referring to this exact circuit only or any other similar circuit with different capacitance's or different battery voltages with the exact same process of simply closing the switch just one time only.

I am not referring to ramp charging, step charging, or anything other type of charging. No other components are involved or added into the circuit. Just a simple closing of a switch to connect the battery to the capacitor only.

I know that this is another question but it's directly related to the first question, I don't want to ask it in another question and risk having a duplicate question, which it shouldn't anyways.

Other question: If the 50% loss of energy in order to transfer the other 50% energy is a set rule in stone then what voltage would the capacitor have to be in order to double the energy in the capacitor to make it equal to 100%?

In other words, what voltage should the capacitor be when it has 14.4mJ of energy in it, which should be double or twice the amount of the 50%, if I am getting that correct at all.

It should not be inferred that I am implying that the capacitor is capable of receiving 100% of the energy transferred from the battery as this is completely impossible and preposterous with the given circuit, but only asking what voltage would the capacitor have to be in order to double the 50% of the energy transferred into it?
Have you any ideas?
Thanks all!

[eccerr0r]

Re: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

Unfortunately it requires calculus to prove that half is lost during charge. Also there's no voltage that will result in less than 50% loss, it will always be lost.

The boiled down version is that if the charge on a capacitor is Q, the work done (energy burned) by the battery is W=V*Q, and since Q=CV, work done is W=C*V^2.

However we know that the energy on a capacitor is 0.5*C*V^2. We lost half regardless of the resistance.

The key thing is that traditional math V=IR doesn't work here. The voltage across a capacitor cannot change instantaneously. You get a divide by zero situation here with R=0 and algebraic math no longer works, thus requiring the calculus.

[budm]

Re: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

What ever the resistance that is in series with the cap will determine how fast the cap will charge up after 5 TC. So if the wires resistance, switch resistance are known then you can calculate the TC.
You are not losing energy once the cap is fully charged. No wires or switch has ZERO resistance.
Look up R/C time constant.
You can put 10 Ohms, 100 Ohms resistor in series with the cap, the cap will still charge up close to the supply Voltage but the amount of time will be due to the resistance.
https://www.eecs.tufts.edu/~dsculley...al/rc/rc3.html
https://www.digikey.com/en/resources...-time-constant
https://micro.magnet.fsu.edu/electro...ant/index.html

[stj]

Re: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

interesting thread, kind of puts a stop to ideas of using caps in electric cars!!

[eccerr0r]

Re: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

I wouldn't call this phenomenon "interesting," it's the zero resistance paradox that shows up again and again.

We know we can charge capacitors/batteries(exact same problem!) without loss. Just not with the circuit presented.

(Oh, I should mention, sometimes we just don't care and eat the loss, it costs too much to save that energy...)

I should make a correction about the statement "no voltage that's less than 50% loss" ... If the capacitor/battery is already partially charged, charging the rest of the way will lose less energy, though you will still lose energy up until the point the voltage of the capacitor equals that of the source, and at that point there's no more loss. And no more charging.

[rievax_60]

Re: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

There will be no zero resistance paradox because of the unavoidable circuit inductance. The circuit will oscillate indefinitely if every part has no resistance and no other losses.

[budm]

Re: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

Conductor has resistance, we are not talking about super conductor here, juts wires and switch contact resistance even in milli-Ohms range it is still resistance.

[rievax_60]

Re: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

The charging of the capacitor could be made more efficient by purposely placing a large inductance in series. I would expect the capacitor to be charged to close to double the supply voltage at which point the switch has to be turned off before the capacitor begins to discharge.

[rievax_60]

Re: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

"In other words, what voltage should the capacitor be when it has 14.4mJ of energy in it, which should be double or twice the amount of the 50%, if I am getting that correct at all."
The 100uf capacitor would have to be charged to 16.97v for 14.4mj of stored energy.
V=√(E/0.5*C)

[eccerr0r]

Re: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

The OP was theorizing ideal wires - so no inductance, no resistance, so how do you explain things?

I guess there's also the alternate thought scenario. Replace the capacitor with an inductor and the switch with a normally closed switch. Let the inductor charge, then open the switch at time t. Again assume all ideal components including an ideal switch that goes from 0 to infinite resistance (meaning, impossible to arc over) at time t with no bounce and no contact resistance.

What happened to the 0.5*L*I^2? A dangling wire should have NO current flowing through and should be 0 energy.

[rievax_60]

Re: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

It is ok to have no resistance, superconductors do exist. Inductance can never be eliminated and can't always be ignored.

[eccerr0r]

Re: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

You can't make superconducting switches and all power sources have ESR and that capacitor has ESR too... so no, even if we have superconductors, we still have resistance losses. But the premise has nothing to do with resistance losses, there's loss here that can't be described by normal math. Same problem with the inductor example, which is even weirder because inductors are weird as it is...

[redwire]

Re: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

It appears to be set in stone.

"...So the energy supplied by the battery is E = CVb^22, but only half that is on the capacitor - the other half has been lost to heat, or in the extremely low charging resistance case, to heat and electromagnetic energy."
http://hyperphysics.phy-astr.gsu.edu...apeng2.html#c4

If you do get to zero or very low resistance, you end up generating electromagnetic energy.

The case of two capacitors is a similar problem with the same result, 1/2 the energy is lost.

[rievax_60]

Re: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?

I don't think inductors are weird. They just do everything in an opposite way to capacitors.
Capacitors store energy electrostatically and cause current flow while attempting to oppose voltage change.
Inductors store energy magnetically and produce voltage while attempting to oppose current change.
These opposite characteristics are what cause LC resonance to exist.