# Capacitors as energy storage devices

My colleague and I have a disagreement about the storage of electrical energy using capacitors in DC circuitry. It all has to do with the equation for the capacitive energy storage, namely the energy stored in a capacitor with capacitance C charged from a power supply at some constant potential V is given by:

E=0.5*CV2

This represents half of the energy drawn from the power supply during charging, with the other half being "used up" as the charging energy. In other words, E=CV2 is drawn from the power supply, but only half of it ends up stored in the capacitor. I have never been able to find a detailed description of where the energy used in charging the capacitor goes. One website I found mentioned something about "resistive losses in the wires", but that doesn't seem correct to me.

1) what is the mechanism by which the charging energy is "used up"?

2) When you try to recover the stored energy from the capacitor by discharging it, are those same loss mechanisms in place? In other words, if I wanted to use the stored energy to power a device, would I be able to extract the entire 0.5*CV2 stored in the charged capacitor, or would I again lose half upon discharging? Does the answer depend on how the capacitor is coupled to the circuit when it is discharged?

marcusl
Gold Member
In other words, E=CV2 is drawn from the power supply, but only half of it ends up stored in the capacitor.

In other words, E=CV2 is drawn from the power supply, but only half of it ends up stored in the capacitor.

I can't remember where I read it initially, but if you look at the hyperphysics page http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c1", it seems to agree:

"Note that the total energy stored QV/2 is exactly half of the energy QV which is supplied by the battery, independent of R!"

That also may be where I got the idea that the energy that is used up is dissipated due to resistance, since it also says http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html" [Broken]:

"For a finite resistance, one can show that half of the energy supplied by the battery for the charging of the capacitor is dissipated as heat in the resistor, regardless of the size of the resistor. "

Last edited by a moderator:
For a finite resistance, one can show that half of the energy supplied by the battery for the charging of the capacitor is dissipated as heat in the resistor, regardless of the size of the resistor.
Yes, that is true. Energy can also be lost to radiation. You may find this web page interesting: http://www.smpstech.com/charge.htm" [Broken]

Last edited by a moderator:
1) what is the mechanism by which the charging energy is "used up"?

2) When you try to recover the stored energy from the capacitor by discharging it, are those same loss mechanisms in place? In other words, if I wanted to use the stored energy to power a device, would I be able to extract the entire 0.5*CV2 stored in the charged capacitor, or would I again lose half upon discharging? Does the answer depend on how the capacitor is coupled to the circuit when it is discharged?

1) When the capacitive power equals that of the charge, no further charge is allowed.

2) You will lose discharge potential ONLY related to the resistance in the discharge circuit(minus charge losses of course)

Ok, thanks for the responses. Just to make sure that I understand, is it then correct to say that, if you were to discharge a charged capacitor into a ideal circuit, then you would be able to re-use all of the stored energy (0.5*CV2)?

Not exactly.
There are losses associated with any electron movement(barring superconductors)
As such, even the event of discharge by a capacitor involves conductive elements, so there will be a loss prior to the "ideal circuit"

.... would I again lose half upon discharging?

From the second reference cited, the formula shows charging a capacitor of EQUAL SIZE results in half the power being lost. But the proportion of energy lost depends on the size of the second capacitor relative to the first capacitor. This must be due to different quantities of charge being transferred.

Does the answer depend on how the capacitor is coupled to the circuit when it is discharged?

As noted above, it seems to depend on the amount of charge removed from the source moving through connecting wires: The more charge transferred the more the I2R loss...

I checked my undergraduate physics text Physics for Students of Science and Engineering, Halliday and Resnick (1961), and sure enough they discuss the loss as heat and radiation!!!! (as cited above) I even highlighted that sentence with a question mark....still seems a bit odd.

This discussion was surprising for me, I think because I have forgotten so much about this issue over the years. However, I think remember doing a proof in college that showed the maximum electrical power transfer from source to load takes place when the impedance of the load matches the impedance of the source.....

Anyone have an online reference with a discussion of that?? I'm wondering if that relates properly to the current discussion because from the formula for capacitance power transfer above, the maximum power resulting power (source plus load) occurs when the second capacitor is very small compared with the first. That does make sense because the less current moved the smaller the losses to resistance.

Found it myself:
http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

and I did not state it quite right above:
The theorem can be extended to AC circuits that include reactance, and states that maximum power transfer occurs when the load impedance is equal to the complex conjugate of the source impedance.

Have to go, will read later...

marcusl
Gold Member
I can't remember where I read it initially, but if you look at the hyperphysics page http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c1", it seems to agree:

"Note that the total energy stored QV/2 is exactly half of the energy QV which is supplied by the battery, independent of R!"

That also may be where I got the idea that the energy that is used up is dissipated due to resistance, since it also says http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html" [Broken]:

"For a finite resistance, one can show that half of the energy supplied by the battery for the charging of the capacitor is dissipated as heat in the resistor, regardless of the size of the resistor. "
This is correct, but is a property of the battery, not the capacitor. Every physical battery has an internal impedance Z. The real part of Z is a resistance R that dissipates heat whenever current is supplied, regardless of the type of load. Naty1 points out that maximum power is transferred from source to load when the load impedance is the complex conjugate Z* of the source. This means that the load resistance should always be R, and it's reactance be the negative of the source reactance (if any). Z for a battery is usually purely real (R) to a good approximation.

To summarize again, internal heat dissipation is a property of the source, not of the capacitor or other load.

Last edited by a moderator:
Wikipedia notes the distinction between power transfer and efficiency:

The theorem results in maximum power transfer, and not maximum efficiency. If the resistance of the load is made larger than the resistance of the source, then efficiency is higher, since a higher percentage of the source power is transferred to the load, but the magnitude of the load power is lower since the total circuit resistance goes up

... a property of the battery, not the capacitor.... internal heat dissipation is a property of the source, not of the capacitor or other load.

This seems a bit too simplistic....or maybe I should say too subject to misinterpretaton. Of course a battery DOES have internal resistance and therefore losses when delivering power.

But the source from above, Energy Loss in Charging a Capacitor, and my own old texbook clearly show similar losses when one capacitor charges another. When one capacitor charges another, losses have nothing to do specifically with the source as distinct from the load. Both references explain such losses a bit more generally: losses due to heating of the connecting wire or via electromagnetic radiation. If you want to consider such connecting wires part of the source instead of part of the load, ok.

marcusl
Gold Member
The load was purely reactive (capacitive) in the hyperphysics example, so lead resistance is included as part of the source. A more realistic model would include some loss in the load, as you suggest.

I was so surprised that nobody actually worked the problem that I just had to register and explain

in a capacitor:
I = c*dv/dt

anywhere in an electric circuit
power = I * V

therefore

power into capacitor = I*V = V*C*dV/dt

Energy Stored = power integrated with respect to time

Pdt = V*C*dV

integrate both sides

P*t = Energy = 1/2*CV^2

as with kinetic energy, and a myriad of other physical equations, the 1/2 is a consequence of integration, not of efficiency or theoretical limits

sophiecentaur
Gold Member
2020 Award
There's a standard 'think about it' question which asks students to account for energy loss when sharing charge by connecting one charged capacitor to an identical uncharged capacitor. It's guaranteed to cause them a bit of confusion because it's an unreal situation. The voltage will not change instantly and resistance or radiation is always involved. The circuit will 'ring' and the volts will decay to a steady half volts, eventually.
When you charge a capacitor through a resistor (whatever value), an equal amount of energy will be dissipated by the resistor and stored in the capacitor.
However, this only applies when there is no inductance in the circuit, so it's an artificial situation and the initial statement really only applies for a very slow charging rate. Charging a 1uF capacitor through a 1M resistor 'really will' divide the energy half and half.

If you want to charge a capacitor much more efficiently, it is quite possible to avoid the majority of these resistive losses by using an inductor and switching the components at the appropriate time so you can say that 'no' energy need be lost, ideally.

The third diagram from the above reference (Post #3) shows the voltage change as a function of multiple time constants:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c1

I was asked a similar question during a job interview many years ago; When you disconnect a charged capacitor from a battery, say via a switch, what happens to the charge/energy.....it seems to instantly disappear????.

Took me a few minutes to answer "spark at the switch".....

How much energy from the source gets stored in the capacitor depends on two factors:
• Relative values of source-voltage rise-time (tr) and circuit time constant (RC)
• Whether the capacitor is linear or non-linear (in most cases it is linear though)

A brief article on this issue: http://chronicengineer.blogspot.com/...ng-things.html [Broken]

A more detailed analysis on the same issue:

Last edited by a moderator:
sophiecentaur
Gold Member
2020 Award
The energy stored in a capacitor is just CVsquared/2; independent of source resistance. But, of course, the RC time constant will affect how long it takes to charge.

A non-linear capacitor would be a bit like a battery, possibly? I've not come across a non linear capacitor when used for storage but reverse biased diodes have a variable capacity, depending on the bias voltage. But they only have very low capacities - in the order of tens of pF; useful as tuning devices etc. but not for storage.

You have to consider the physics to come up with your answer intuitively. What is happening? Mass is increasing on one plate, causing an electrostatic charge differential between the plates. As the mass builds, it repels further flow because like charges on the that plate repel further current flow unless the pressure (voltage) is increased. Reactance is built in. Does that help? If the voltage source is removed, with no sparking, and no loss due to leakage in the system, and no leakage externally, the charge could theoretically remain. At the BBC a guy who thought these big capacitors were dangerous because he thought they were charged up, or could charge up. He put wires across them...cost a fortune in batteries...that is what they were.

sophiecentaur
Gold Member
2020 Award
You have to consider the physics to come up with your answer intuitively. What is happening? Mass is increasing on one plate, causing an electrostatic charge differential between the plates. As the mass builds, it repels further flow because like charges on the that plate repel further current flow unless the pressure (voltage) is increased. Reactance is built in. .

You seem to be putting the cart before the horse here. Charge builds up because of the applied PD. This is because of minuscule net movement of electrons, which will constitute some 0.001% of the mass of the surface layer of the electrodes. (one electron compared with the atomic masses involved). Could you actually measure this displacement? I doubt it.

You seem to be putting the cart before the horse here. Charge builds up because of the applied PD. This is because of minuscule net movement of electrons, which will constitute some 0.001% of the mass of the surface layer of the electrodes. (one electron compared with the atomic masses involved). Could you actually measure this displacement? I doubt it.
I stand corrected, I should have more correctly stated "mass charge". I am old, I still visualize this stuff the way I learned it long ago...when I worked in a small office reading about people's ideas for perpetual motion, or the alchemy concept of making gold from lead...thanks for the correction.

sophiecentaur
Gold Member
2020 Award
In the context of most of electronics (operational) what is the point of discussing anything other than charge? What's "mass charge" when it's at home? Electrons just get in the way and are an additional layer which we can do without. Q=CV works and doesn't discuss numbers of electrons or quantum energy levels.

In the context of most of electronics (operational) what is the point of discussing anything other than charge? What's "mass charge" when it's at home? Electrons just get in the way and are an additional layer which we can do without. Q=CV works and doesn't discuss numbers of electrons or quantum energy levels.
Again I stand corrected. I was using the term "mass" in my correction to indicate a scalar quantity, rather than as a physical descriptor. I haven't done this stuff for years, it's not my field since being involved in the electronics necessary to locate Russian subs using magnetic anomaly detection back in the 70's. Perhaps I'll stick to the "Ask a stupid question..." forum. Thanks for the correction.

RonL
Gold Member
Again I stand corrected. I was using the term "mass" in my correction to indicate a scalar quantity, rather than as a physical descriptor. I haven't done this stuff for years, it's not my field since being involved in the electronics necessary to locate Russian subs using magnetic anomaly detection back in the 70's. Perhaps I'll stick to the "Ask a stupid question..." forum. Thanks for the correction.

Hope you stick around for a while, I have survived here for several years now and still tend to make comments that would indicate I have learned nothing, however that is a false condition.
The first thing in my mind when I read the OP, was an old electronics book, that had a very brief description of a "Tank Circuit" in which the statement was made, "the closest thing to perpetual motion, an electronic version of a Flywheel".
Naty1 gave a link that to me seemed close to this. As I recall, the circuit is charged and during this charge an electromagnetic field is building as a resistance (the payment made) but when the field collapses, "almost" the exact same energy is returned (the return investment). As I understand this happens at "Resonance Frequency" and is a tuning filter.
So as you have indicated, our clocks might be old fashion and only correct based on our frame of reference
I get the impression you might have worked as a patent attorney or in the patent office ?
Anyhow hope you continue here in PF.

Ron

Hope you stick around for a while, I have survived here for several years now and still tend to make comments that would indicate I have learned nothing, however that is a false condition.
The first thing in my mind when I read the OP, was an old electronics book, that had a very brief description of a "Tank Circuit" in which the statement was made, "the closest thing to perpetual motion, an electronic version of a Flywheel".
Naty1 gave a link that to me seemed close to this. As I recall, the circuit is charged and during this charge an electromagnetic field is building as a resistance (the payment made) but when the field collapses, "almost" the exact same energy is returned (the return investment). As I understand this happens at "Resonance Frequency" and is a tuning filter.
So as you have indicated, our clocks might be old fashion and only correct based on our frame of reference
I get the impression you might have worked as a patent attorney or in the patent office ?
Anyhow hope you continue here in PF.

Ron
Thanks Ron... Yeah, you caught me. I was told to wiggle out the best way you can when you get caught... Ha! This stuff has changed since I delved in it. I need to read more. I was taught that electrons moved through wires, never questioning it, or Newton's Gravity. Einstein's Theory is still a mystery to me. I was reading about photons now I am totally confused. What is one photon, is that a mystery too? If you can see one photon then it must have a duration because my eye detects a specific range of frequencies, not just an energy burst, but a range of frequencies, Right! So what is the length of a photon? Anyway, we'll see... but thanks for the reply.