What Determines the Amount of Energy Stored in a Capacitance?

In summary, the energy stored in a capacitor equals to half the energy supplied by the battery. This is counterintuitive because the charging process requires putting out more energy than is stored on the capacitor.
  • #1
Jadaav
175
1
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.
 
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  • #2
Are you familiar with the derivation? I cannot think of a way to "guess" that it is 1/2 other than deriving it.
 
  • #3
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
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
No, the OP is correct. In a RC circuit the resistance dissipates exactly half of the energy.
 
  • #6
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
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
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
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

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.
 
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  • #10
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.
cape14.gif

Transporting differentialcharge dq to the plate of the capacitor requireswork

cape16.gif

cape15.gif


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
No worries, it happens for all of us from time to time.
 
  • #12
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
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
oz93666 said:
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?
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
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)
 
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  • #16
oz93666 said:
DaleSpam! you're right ..you're right and so is OP..I'll just go and crawl back into my hole!

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
oz93666 said:
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?
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
oz93666 said:
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?
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.
oz93666 said:
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)
Yes, that is correct.
 
  • #19
TurtleMeister said:
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.
"
oz93666 said: https://www.physicsforums.com/threads/energy-stored-in-capacitance.779421/goto/post?id=4899993#post-4899993
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)"

Yes, that is correct.

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.
 
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  • #20
I've read through all the debates and I should say that I've learned 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 ?
 
  • #22
Jadaav said:
I've read through all the debates and I should say that I've learned 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 ?

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

1. What is capacitance?

Capacitance is a measure of the ability of a system to store an electrical charge. It is determined by the geometry and material properties of the system, and is measured in units of Farads (F).

2. How is energy stored in capacitance?

When an electric charge is applied to a capacitor, it causes a separation of positive and negative charges on the two plates of the capacitor. This creates an electric field between the plates and stores energy in the form of electric potential energy.

3. What is the formula for calculating energy stored in capacitance?

The formula for calculating energy stored in capacitance is E = 1/2 * C * V^2, where E is the energy in joules, C is the capacitance in Farads, and V is the voltage across the capacitor in volts.

4. How does the capacitance affect the amount of energy stored?

The higher the capacitance, the more energy can be stored in the capacitor. This is because a higher capacitance allows for a larger charge to be stored at a given voltage, resulting in a higher amount of electric potential energy.

5. What factors can affect the energy stored in capacitance?

The energy stored in capacitance can be affected by the size and distance between the two plates, the type of material used for the plates, and the voltage applied. Additionally, the dielectric material between the plates can also impact the energy stored in capacitance.

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