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Capacitor charging loss (Not the two capacitor issue.)

  1. Feb 16, 2009 #1
    I got some guy who says when a capacitor is fully charged, that half of the input energy from the power source gets lost as heat. Now I do understand the two cap problem (links below), but I think he is wrong because one can use dc switching and inductors to improve the 50% loss of energy. (but he says 1/2 loss is still happening.)

    I would find it hard to believe that for DL ultracaps/supercapacitors used on EV's/hybrid's for quick storage would be loosing 1/2 of the power to charge them as heat.

    He's asking me to prove him wrong (he has same links I've listed below), is there any other references that anyone has that I use to back up my argument?

    http://puhep1.princeton.edu/~mcdonald/examples/EM/powell_ajp_47_460_79.pdf [Broken]
    http://puhep1.princeton.edu/~mcdonald/examples/EM/mita_ajp_67_737_99.pdf [Broken]

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 16, 2009 #2


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    Hi spaceball3000! :smile:

    Up to 50% is lost even with one capacitor , unless it's done through an inductor …

    see this from the PF Library (unfortunately, no details for the inductor part) …

    Energy loss:

    Energy lost (to heat in the resistor):

    [tex]\int\,I^2(t)\,R\,dt\ =\ \frac{1}{2}\,C (V_1\,-\,V_0)^2[/itex]

    Efficiency (energy lost per total energy):

    [tex]\frac{V_1^2\,-\,V_0^2}{V_1^2\,-\,V_0^2\,+\,(V_1\,-\,V_0)^2}\ =\ \frac{1}{2}\,\left(1\,+\,\frac{V_0}{V_1}\right)[/tex]

    Accordingly, charging a capacitor through a resistor is very inefficient unless the applied voltage stays close to the voltage across the capacitor.

    But there is no energy loss on charging a capacitor through an inductor, basically because the applied voltage then appears across the inductor instead of across the capacitor.​
  4. Feb 16, 2009 #3
    Thanks Tiny-tim,

    I hope that will be enough for him to see the light.
  5. Feb 19, 2009 #4
    Well that didn't convince the guy at all.

    I even talked about LC Tank circuits and how every LC Oscillation periods C is being charge and discharged. And that there doesn't seem to be a rule where there has to be an minimum of 50% energy loss with every LC oscillation period in a LC Circuit. Where the charge has is moved once from the inductor to capacitor (or the opposite), the total energy in the capacitor by his logic it would have to be reduced by a minimum of 50% each time period (when the capacitor charged.)

    Can't someone provide some proof so that he doesn't confuse more people that how capacitors work.
    Last edited: Feb 19, 2009
  6. Feb 19, 2009 #5

    The Electrician

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    The third reference you gave:
    http://puhep1.princeton.edu/~mcdonald/examples/EM/mita_ajp_67_737_99.pdf [Broken]

    explains things. On the second page, near equation (11), the author discusses the situation where an inductor in introduced into the current path. He mentions in footnote 7 that he considers the case where only a single cycle of the oscillation that occurs in this situation is allowed to happen.

    In the case where the circuit includes a small parasitic resistance as well as the inductance, and where a voltage source is suddenly connected to a capacitor (plus L and parasitic R in series), the current will oscillate until the oscillations finally die out. In that case, I think the parasitic resistance will in fact dissipate half the energy delivered by the power supply.

    But, if the oscillations are stopped after one cycle by the simple expedient of placing a diode in the current path, then the capacitor will contain essentially all of the energy delivered by the power supply.

    Furthermore, if a variable power supply is used as the source, starting with the supply set to zero volts out and connecting a capacitor to the supply, if the voltage is turned up gradually then the capacitor will contain essentially all of the energy delivered by the supply.
    Last edited by a moderator: May 4, 2017
  7. Feb 19, 2009 #6
    Hi 'The Electrician',

    I 100% agree on your thoughts on this subject.

    He basically ignored that third reference saying it's "contradicts itself", and saying

    Then later on he points me to this link --> http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html#c1 saying

    "the formula from the University of Georgia is a physical principle, it is not limited to how the energy is fed, you assumed a constant voltage, but could also be a constant current, or via any other element etc. This is physics: it is like that because it is like that!"

    He does seems smart, but I think he is missing something, the equation in the link he gave, seems to only apply to scenarios when a capacitor is being charging through a resistor. Thus the equation he gave (I think) would need to be changed to take into account when the inductor is added.

    Now adding the inductor, doesn't prevent the oscillations (which would generate RF/EM, i.e. losses) and you are correct a diode would prevent that.

    So in a nutshell, can someone find proof that his equation (that he linked) Does Not apply to the scenario where an capacitor is being charged via inductor?
  8. Feb 19, 2009 #7

    The Electrician

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    As near as I can tell, the link http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html#c1

    merely gives the classical formula for the energy stored in a capacitor, c*v^2/2, or q*v/2 which is the same thing. This formula is correct no matter how the capacitor gets charged.

    However, it isn't correct to say that whenever a capacitor is charged from some kind of energy source, without exception, only half the energy provided by the source ends up in the capacitor.

    Sometimes it is true, as in the case where a source of constant voltage is suddenly connected to a series combination of a resistor and capacitor.

    If the circuit consists of an inductor in series with a capacitor, with negligible parasitic resistance, then suddenly connecting this circuit to a constant voltage source will transfer almost all of the energy provided by the source to the capacitor, IF the voltage source is disconnected at the right time, namely when the current goes to zero at the end of one half cycle of oscillation. A simple diode can provide this disconnect.

    Another way to ensure that all the energy provided by the source ends up on the capacitor is to connect the capacitor to a variable voltage source initially set to zero, and then gradually turn up the voltage to some final value.

    Even if there is a resistor of non-negligible size in series, if the voltage is turned up slowly so that the charging current remains small at all times, most of the energy ends up in the capacitor, with very little dissipated in the resistor.

    This is what is described in your original third reference, in the vicinity of equations 13 and 14. By making the voltage steps smaller and more numerous, the effect is the same as turning up a variable voltage source gradually.
  9. Feb 19, 2009 #8
    Ah, that makes sense to me.

    Me just repeating this to him will likely not convince him at all, I'm trying to find a logical error in his thinking.

    So just to confirm, the classical formula for the energy stored in a capacitor never changes, but to calculate the energy lost he likely using this equation from --> Energy lost (to heat in the resistor): (look under "Energy lost (to heat in the resistor):"

    But since there is no energy loss on charging a capacitor through an inductor, so does this same energy loss (resistor) equation still apply (newly added inductor)?
  10. Feb 20, 2009 #9

    The Electrician

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    If there's no resistance (even the smallest parasitic) in the circuit, with an inductor and capacitor, the oscillations will go on forever, and in that case the resistor-capacitor equation doesn't apply. But then how can we say that the capacitor reaches a final state of charge?

    I've attached an analysis of the problem for the R-C and the R-L-C cases.

    But, the fact that the energy that ends up in the capacitor in these two cases is only half the energy provided by the voltage source doesn't mean that every method of charging the capacitor loses half the energy from the source. I described two methods that don't suffer from that loss in a previous post.

    Attached Files:

  11. Feb 20, 2009 #10
    Wonderful! Think this will help a lot, Thanks!
    Last edited: Feb 20, 2009
  12. Feb 22, 2009 #11
    You only need one contrary example to disprove an assertion. But it's not very convincing to say that no energy is lost while the current is still sloshing back and forth between capacitor and inductor. Will adding a switch change this?
  13. Feb 22, 2009 #12
    Ok dropped the word sloshing.

    In an LC Tank circuit when the energy is being transferred from one to another, disconnecting L when it's empty (i.e. switch) will prevent any more oscillations. At that point most all of the energy that was in L should be transferred to C.
    Last edited: Feb 22, 2009
  14. Feb 22, 2009 #13


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    Yes if you are allowed to use an inductor and a switch it's easy to do.

    Just get a voltage_source, an inductor, a switch and the capacitor all in series. Let the initial voltage on the capacitor be zero and the value of the voltage source be one half of the final voltage that you wish to charge the capacitor to.

    Close the switch and then re-open it on the next zero crossing of the curent waveform (at [itex]t = \pi \sqrt{LC}[/itex]). The capacitor will then be charged to twice the voltage source value, ideally without loss. This is the basis of how a resonant mode DC-DC converter operates.
    Last edited: Feb 22, 2009
  15. Feb 22, 2009 #14
    It's a tank, it's sposta slosh. :smile:

    As uart was saying, the voltage across the capacitor will be double the supply voltatge. Connect a switch, an inductor and a capacitor across 10V.

    Initially, the current in the inductor is zero. The voltage across the capacitor is zero. Close the switch at time t=0.

    The voltage across the capacitor will be
    [tex] V_C = 10 - 10 cos ( 2 \pi f t ) [/tex]

    The current through the inductor will be
    [tex] I_L = I_{peak} sin ( 2 \pi f t ) [/tex]

    The peak current can be determined from conservation of energy.
    [itex] L \cdot I_{peak}{}^2 + C \cdot 10^2 = C \cdot 20^2 [/itex] Joules.

    Open the switch after half a cycle, as uart was saying, at time t = 1/(2f).

    The energy stored in the capacitor will be
    [tex](1/2)C \cdot 20^2[/tex]

    The energy delivered by the 10 volt source will be
    [tex]\int_{0}^{1/(2F)}10I_L dt[/tex]
    Last edited: Feb 22, 2009
  16. Feb 23, 2009 #15
    Again, you guys have been great!

    Honestly this has helped me more than the guy who I was trying to convince (I learned a few new things), he will not admit he is wrong in lieu of the facts. Only reason I can think hes ignoring the facts, is that he trying to convince people that all capacitors have 50% losses when charging (all methods including an inductor.) So that gives his new battery/capacitor invention an edge, i.e. he says his capacitor doesn't have this 50% problem.

    If your curious more on this discussion, just Google with these keywords "quantum battery rolf" , and look at the various forums.
    Last edited: Feb 23, 2009
  17. Feb 23, 2009 #16
    Low loss translation of current and voltage using inductors and capacitors and switches (transistors and diodes), done in a dozen different ways, is the basis of switched mode power supplies.
  18. Mar 3, 2009 #17
    he says I'm wrong again, anyone interested to point out his math errors?
    See attached file.

    Attached Files:

  19. Mar 3, 2009 #18


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    Yes he's totally wrong on the 8th line :

    Code (Text):
    Work Performed = Uc Qc
    This should read :

    Power supplied by Battery = Ub Qc = 2 Q Ub, exactly the same as the energy delived to the capacitor.
  20. Mar 4, 2009 #19
    Well I mentioned what you said, and also saying 4 Q Ub is wrong and that 2 Q Ub is the Work performed. He replied (below) and also added and additional equation.

    I'm going to assume he messed up like in his first example. Taking a look I see the equation where Ipeak has 4Ub, I'm pretty sure it should be 2Ub, am I correct?

    Attached Files:

  21. Mar 4, 2009 #20


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    Yeah there are quite a few errors here as well. He seems to like just throwing in factors of 2 and 4 at will without any explaination to make the results fit his theory.

    The first error is that Ipeak is only one times Ub w c, not 4 times as claimed (line 4). Interestingly however he then makes a mistake in the integration giving a factor of 1/2 (integral over half period of w sin(wt) dt ) whereas the correct value for this integral is 2 (line 7). So it turns out that these two mistakes cancel out and that the expression he obtains, Wc = 2 Q Ub is actually correct (where Q = C Ub).

    Note however that Wb = 2 Q Ub (not 4 Q Ub as claimed by this guy) so once again we have Wb = Wc.

    So (apart from some new mistakes that fortunately cancelled out) the common mistake from both posts is his belief that Wb = 4 C Ub^2 whereas the correct value is only 2 C Ub^2. This is pretty trivial, the charge drawn by the capacitor (up until t=T/2) is C Uc(T/2) = 2C Ub. So the energy drawn from the DC supply is just Ub times the charge drawn which gives the value of 2 C Ub^2. I have no idea where he gets the 4 from here, and he gives no explaination, so it's all very dubious.
    Last edited: Mar 4, 2009
  22. Mar 5, 2009 #21

    The Electrician

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    I've worked out the full-bore mathematical analysis of an underdamped RLC circuit, with a suddenly applied voltage step.

    I've plotted the energy loss in the resistor vs. time. It can be plainly seen that if the oscillations are allowed to continue until they damp out, the loss in the resistor reaches a final value of c*v^2/2. But, it can also be seen that in the first half cycle, the loss is nowhere near this much, so that if a switch is opened to stop the oscillations (such as a diode) after one half cycle, the efficiency of charging the capacitor can be much higher than 50%.

    Attached Files:

  23. Mar 5, 2009 #22

    The Electrician

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    I tried to post an image that is too big, and the forum shrunk it.

    I'll try posting in two parts.

    Attached Files:

  24. Mar 5, 2009 #23
    Thanks for the RLC info I'll pass that on.

    He says he fixed his mistake (see attached file), and says that your guys math is wrong (see his response below.) I'm sure he trying to hide/confuse the math even more now, I'm having hard time following his new cryptic changes though.


    Attached Files:

    Last edited: Mar 5, 2009
  25. Mar 5, 2009 #24

    The Electrician

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    I added a plot of the energy in the capacitor vs. time to the plot.

    You can see that after many oscillations, the energy lost in the resistor equals the energy stored in the capacitor, but if the oscillations are stopped after one half cycle, the energy lost in the resistor is a small fraction of the energy in the cap.

    Attached Files:

    Last edited: Mar 5, 2009
  26. Mar 5, 2009 #25


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    Hi Spaceball. Once again he gets the current wrong. Once again he arbitrarily includes a factor of two without explaination (and where it doesn't belong).

    Specifically he claims that :

    [tex] I_{peak} = 2 w C U_B[/tex]

    btw previously he claimed it was 4 w C Ub and that was wrong too.

    The correct value is :

    [tex] I_{peak} = w C U_B[/tex]

    BTW, this is a 100% certainly, no if's no but's. There is absolutely no controversy about this result, it could be found in any text book and could be derived by just about anyone who's completed even half of an EE degree.

    Specifically the DE's are (where "i" is the inductor current and "v" is the capacitor voltage)

    [tex]di/dt = (U_B - v)/L[/tex]

    [tex]dv/dt = i/C[/tex]

    with initial conditions : [itex]i(0) = 0[/itex] and [itex]v(0)=0[/itex], hence [itex]di/dt (0) = U_B/L[/itex]

    Combining the two DE's gives :

    [tex] d^2 i / d t^2 = - i /(LC)[/tex]

    Which has the solution :

    [tex]i = A \cos(w t) + B \sin(wt) [/tex]

    where "A" and "B" are constants and [itex] w = 1 / \sqrt{LC}[/itex]

    Applying the initial condition on the current we get [itex]A=0[/itex] and from the initial condition on the current derivative we get [itex]B w = U_B/L[/itex].

    Rearranging the last equation gives : [itex]B = U_B\,\sqrt{C/L} = w C \,U_B[/itex].

    That is :

    [tex]i = w C U_B \sin(wt)[/tex].
    Last edited: Mar 5, 2009
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