# Energy stored in capacitance

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1. Nov 1, 2014

My question is why is the energy stored in a capacitance equals to half the energy supplied by the battery.

I already know that there is heat loss in wires. But what I really want to know is that why is it only half of it ?
It could have been any other fraction.

2. Nov 1, 2014

### Staff: Mentor

Are you familiar with the derivation? I cannot think of a way to "guess" that it is 1/2 other than deriving it.

3. Nov 1, 2014

### TurtleMeister

If you connect an inductor of sufficient value in series with the capacitor then the loss will be less than 1/2. See this document for details.

4. Nov 1, 2014

### oz93666

It isn't... (virtually) all the energy supplied by the battery is stored in the capacitor.

There maybe confusion here because the formula for energy in a cap. is E=1/2 xCxVxV ... this is also the energy supplied by the battery(except for , normally, very small loss due to resistance in connecting wire, and internal resistance of cap.)

5. Nov 1, 2014

### Staff: Mentor

No, the OP is correct. In a RC circuit the resistance dissipates exactly half of the energy.

6. Nov 1, 2014

### oz93666

That was not the query....In an RC circuit the resistance dissipates ALL the energy , not half ...his question was ..." why the energy stored in the capacitor is half supplied" ALL energy supplied by the battery is stored, except perhaps a few % (in normal circumstances) lost in connecting wires and int. resistance.

7. Nov 1, 2014

### Staff: Mentor

No, that is incorrect. In a RC circuit half of the energy supplied by the battery is dissipated by the resistor and the other half of the energy supplied by the battery is stored in the capacitor.

8. Nov 1, 2014

### oz93666

My friend ... this is not true. first an RC circuit is an RC circuit not an RC battery circuit!

An RC circuit is just that , a capacitor charged up . then a resistance is put across the cap, the current oscillates around, and dies to nothing, ALL the energy dissipated in the resistance.

9. Nov 1, 2014

### Staff: Mentor

A circuit with a resistor, a capacitor, and a voltage source is also called an RC circuit. This is clearly the type of RC circuit described by the OP.
http://faculty.wwu.edu/vawter/PhysicsNet/Topics/DC-Current/RCSeries.html [Broken]

Here is the proof that half of the energy goes to the capacitor and half to the resistor.
https://www.physicsforums.com/threa...gy-stored-in-a-capacitor.692537/#post-4388239

Also, there is no oscillation of the current in a RC circuit. It is an exponential decay.

Last edited by a moderator: May 7, 2017
10. Nov 1, 2014

### oz93666

DaleSpam!! you're right ..you're right and so is OP..I'll just go and crawl back into my hole! How can this be possible ?( well now I know), it just doesn't seem fair or reasonable, super caps are used for regenerative braking in some buses ,half is wasted!! why did god make the world so unfair?

Where did half of the capacitor charging energy go?
The problem of the "energy stored on a capacitor" is a classic one because it has some counterintuitive elements. To be sure, the battery puts out energy QVb in the process of charging the capacitor to equilibrium at battery voltage Vb. But half of that energy is dissipated in heat in the resistance of the charging pathway, and only QVb/2 is finally stored on the capacitor at equilibrium. The counter-intuitive part starts when you say "That's too much loss to tolerate. I'm just going to lower the resistance of the charging pathway so I will get more energy on the capacitor." This doesn't work, because the energy loss rate in the resistance I2R increases dramatically, even though you do charge the capacitor more rapidly. It's not at all intuitive in this exponential charging process that you will still lose half the energy into heat, so this classic problem becomes an excellent example of the value of calculus and the integral as an engineering tool.

Part of the intuitive part that goes into setting up the integral is that getting the first element of charge dq onto the capacitor plates takes much less work because most of the battery voltage is dropping across the resistance R and only a tiny energy dU = dqV is stored on the capacitor. Proceeding with the integral, which takes a quadratic form in q, gives a summed energy on the capacitor Q2/2C = CVb2/2 = QVb/2 where the Vbhere is the battery voltage. So the bottom line is that you have to put out 2 joules from the battery to put 1 joule on the capacitor, the other joule having been irretrievably lost to heat - the 2nd Law of Thermodynamics bites you again, regardless of your charging rate. The non-intuitive nature of this problem is the reason that the integral approach is valuable.

Transporting differentialcharge dq to the plate of the capacitor requireswork

Though it will not be shown here, if you proceed further with this problem by making the charging resistance so small that the initial charging current is extremely high, a sizable fraction of the charging energy is actually radiated away as electromagnetic energy. This crosses the threshold into antenna theory because not all the loss in charging was thermodynamic - but still the loss in the process was half the energy supplied by the battery in charging the capacitor.

So the energy supplied by the battery is E = CVb2, 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.

11. Nov 1, 2014

### Staff: Mentor

No worries, it happens for all of us from time to time.

12. Nov 1, 2014

### TurtleMeister

oz93666, don't forget that if the inductance is low and the resistor resistance is low then the rest of the energy will be dissipated by the internal resistance of the battery itself. One method of solving this energy loss is by using an inductor in series with the capacitor. You should read the link I provided in my first post.

13. Nov 1, 2014

### oz93666

Have just read your link TM... I can't see how putting an inductor in series, when charging a capacitor will help get around that 50% loss (although it seems it does as the link looks reliable).. can you explain in simple terms?

14. Nov 2, 2014

### TurtleMeister

When resistor value is low the losses occur as a result of trying to instantaneously charge the capacitor. By using an inductor the initial voltage appears across the inductor instead of the capacitor. The energy is temporarily stored in the magnetic field of the inductor where it is more gradually released to the capacitor when the magnetic field collapses.

15. Nov 2, 2014

### oz93666

Ah...got it (I think), it's because loss is I SQUARED R, where as charging the cap is about just I (not squared).. keeping that I low does the job ... thanks

I wonder in practice how well this works out? Because now you have the additional resistance of the inductor. I guess they must use inductors in series with the caps. in regenerative braking for electric buses. I wonder what % of the electricity being fed to the cap. actually reaches it in this arrangement?

So then it should follow if you charged with a variable voltage supply starting low and increasing, you would charge more efficiently? (neglecting losses in supply side)

Last edited: Nov 2, 2014
16. Nov 2, 2014

### phinds

Don't feel bad ... I had EXACTLY the same situation a few weeks back and had to eat my hat and I used to be an EE (about 100 years ago).

17. Nov 2, 2014

### sophiecentaur

You can transfer 100% of the energy in one capacitor into another capacitor (at least in principle). Introducing an inductor can reduce the energy loss to (nominally) zero. If you discharge a C into an L (an LC resonant circuit doesn't lose energy inherently) and switch the circuit to a different C, just as the volts across the C are zero., All the energy will be transferred. (Neglecting any radiated energy, of course) [Edited to make it read sense]
I can't think of any application for this, except to prove a point.

18. Nov 2, 2014

### TurtleMeister

Yes, the resistance of the inductor and also the internal resistance of the battery or source will limit the efficiency, but not as much as you might think. Wikipedia states a conversion efficiency of 75% to 98% for switch-mode conversion. However, this applies to switch-mode power supplies, not regenerative braking.
Yes, that is correct.

19. Nov 2, 2014

### sophiecentaur

Except you would really need to include what goes on in the power supply, to work out the overall efficiency. Anything other calculation would hardly mean anything, imo.

Last edited by a moderator: May 7, 2017
20. Nov 3, 2014

I've read through all the debates and I should say that I've learnt quite a few things from you guys.

How does an inductor stores energy in its magnetic field ? Also how does the inductor decreases the energy losses ?

21. Nov 3, 2014

### Staff: Mentor

22. Nov 3, 2014

### sophiecentaur

You seem happy enough to accept that a Capacitor stores energy by virtue of the E field so why question the idea that an Inductor stores energy by virtue of the H field.? Work, in the form of Electrical Energy has to be done in order to establish the fields and, as it is not dissipated, it has to be there, in the fields.

23. Nov 3, 2014