Capacitors and energy

  • #1
A friend asked me this question:

There is a capacitor with capacitance C, charge Q on either plate and an electric field E between the plates. The energy stored in the capacitor is then given by (C.V^2)/2, where V is E.(distance between plates).
Now, take another similar capacitor, and connect it across the original capacitor like shown.

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---------| |--------
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Then, due to redistribution of the chrges, the electric field between plates of each capacitor becomes E/2. The equivalent capacitance is 2C and the total energy stored in the 2 capacitors is now (C.V^2)/4.

What happens to the other (C.V^2)/4? I think so much energy is lost during redistribution of charges. But am not sure.. Any help?
 
Last edited:

Answers and Replies

  • #2
1,187
5
Could you provide a more clear visual?
 
  • #3
Sorry, i dint check the preview before posting it. Its just 2 capacitors in parallel.

Code:
       -----||-----
-------|           |-------
       -----||-----

Hopefully its clear now..?
 
  • #4
another person i asked says, that its a fallacy. now i'm really confused.
 
  • #5
another person i asked says, that its a fallacy. now i'm really confused.
 
  • #6
rock.freak667
Homework Helper
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Isn't half of the total energy supplied(QV) by the emf converted to heat? that is why the energy of a capacitor is [itex]\frac{1}{2}QV[/itex]
 
  • #7
i dont think that interpretation is correct. The energy stored in a capacitor is (C.V^2)/2 because it is defined as the integral of the work done in taking charge q on the plates across a potential of dv. When you substitute q=C.V and evaluate it, you get the expression for energy stored. The factor of 1/2 is purely due to the integration.
 
  • #8
Mapes
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What happens to the other (C.V^2)/4? I think so much energy is lost during redistribution of charges. But am not sure.. Any help?

Agreed. It's an irreversible process in which potential energy is converted to thermal energy. I have a further question: how much energy is lost to radiation (if any) as opposed to resistive heating?
 
  • #9
hey, i ve been thinking about this. if there really is a loss of energy in the system by 50% when i charge a capacitor then doesnt it mean that all systems which use a capacitor will necessarily have a low efficiency?
 
  • #10
522
70
Not necessarily.
If you e.g. connect a battery directly to a capacitor, half the energy will be lost in the internal resistance of the battery. When the capacitor is empty all the voltage will drop over the resistance since the voltage of the capacitor is zero. At the end of the charging process all voltage will drop over the capacitor. So in the end you loose half the energy. If however you charge a capacitor with a current source you will loose nearly no energy. An inductor is a current source. It's used in step up converters. The battery is first connected to the inductor for a while. Nearly all of the energy that comes from the battery will then get stored in the inductor. Then the current is turned off and the inductor is connected to a capacitor. It will then pump all it's energy into the capacitor. Such a step up converter can have more then 95% efficiency.
 
  • #11
If you e.g. connect a battery directly to a capacitor, half the energy will be lost in the internal resistance of the battery. When the capacitor is empty all the voltage will drop over the resistance since the voltage of the capacitor is zero. At the end of the charging process all voltage will drop over the capacitor. So in the end you loose half the energy.
Lets not consider the energy lost in the internal resistance of the battery. Plus, since we are talking about charging and discharging a capacitor, lets take the case of a rectifier in which a capacitor is used to smoothen out the ripples(low pass filter).
Here, the capacitor is continously charging and discharging. If there is indeed a loss of 50% energy in one cycle of charging and discharging, then the efficiency of the rectifier will be very low.
However, the theoretical maximum efficiency of a full wave rectifier is 81.2%(without capacitor filter). You put the filter in and the theoretical maximum efficiency drops down to 40%.

Its this which makes me believe that there isnt any loss in energy. i think there is a fallacy somewhere.
 
  • #12
Mapes
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You can't consider just one arrangement (a constant voltage being used to charge an empty capacitor), and then declare that half the energy is lost every time someone uses a capacitor! The rectifier is a different circuit that you need to analyze on its own.
 
  • #13
I suppose when you switch on the battery, current starts flowing through the wires. This leads to deposition of charge on the plates of the capacitor. Now, during derivation of that formula 1/2 C V^2, we have considered a random short interval of time, and then integrated it.
I think that as more and more charge starts getting deposited, a field starts forming in between plates. Moving electrical field then creates a magnetic field. When these two fields are perpendicular to each other, electromagnetic radiation starts getting emitted from between the plates.
That may be responsible for that energy loss while charging
 
  • #14
522
70
Lets not consider the energy lost in the internal resistance of the battery.
But we need to consider that.

Plus, since we are talking about charging and discharging a capacitor, lets take the case of a rectifier in which a capacitor is used to smoothen out the ripples(low pass filter).
Here, the capacitor is continously charging and discharging. If there is indeed a loss of 50% energy in one cycle of charging and discharging, then the efficiency of the rectifier will be very low.

In that case the capacitor is not completely charged and discharged every cycle. It is only charged and discharged a little bit and so only a small amount of energy is lost.
 
  • #15
You can't consider just one arrangement (a constant voltage being used to charge an empty capacitor), and then declare that half the energy is lost every time someone uses a capacitor! The rectifier is a different circuit that you need to analyze on its own.
I just took the rectifier circuit as an example. The question in the OP refers to a fundamental process - the charging and discharging of capacitors which occurs in the rectifier too.

But we need to consider that.
The internal resistance of the source is not needed because the question is about the energy loss during charging and discharging of a capacitor.


In that case the capacitor is not completely charged and discharged every cycle. It is only charged and discharged a little bit and so only a small amount of energy is lost.
I agree, the capacitor is not charged or discharged completely.
But, isnt the resistor the only passive element that loses energy? As far as i know, the capacitor and the the inductor(ideal ones) do not dissipate any energy. Practical ones do that because of some internal resistance in them.
 
  • #16
522
70
If you had no resistance at all there would still be the inductance of the wires. If you connect a battery with no internal resistance to a perfect capacitor with superconducting wires , the charging current would only be limited by the inductance. When the capacitor has reached the same voltage as the battery, half the energy that came from the battery will be in the capacitor. The other half will be in the inductance. The inductance will then force the current to continue to flow and the cicuit will oscillate. The energy will move from the inductance to the capacitor and back again over and over again until the energy has been lost through electromagnetic waves.
Your next question might be - but what if there is also no inductance?
There always is inductance. And if there really wasn't then the acceleration of the electrons would only be limited by their mass. They would get faster and faster until they hit something and loose their energy in form of heat.
 

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