I understand that when a battery is connected to a capacitor that half the energy is stored in the capacitor while the other half is lost to heat or whatever other losses are involved once the capacitor is charged to the exact same voltage level of the battery.

Only a one time direct connection with a simple switch from open to closed that charges the capacitor up to the voltage level of the battery is what is being referred to. My question is does this occur each and every time?

Once the capacitor is charged to the voltage level of the battery then there should be 7.2 milliJoules of energy in the capacitor.

If the answer is yes that half the energy is transferred into the capacitor each and every time then what would the capacitor voltage have to be in order to equal 100% of the energy if it were to come from the battery?

I understand that in practice this will never happen, which is why the statement equal to 100% of the energy is indicated. (Meaning by some other method of charging means)


simulate this circuit – Schematic created using CircuitLab


closed as unclear what you're asking by brhans, Voltage Spike, winny, Bimpelrekkie, R Drast Nov 2 '17 at 10:37

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Dave Tweed Nov 1 '17 at 20:40
  • \$\begingroup\$ what does half the energy mean? \$\endgroup\$ – jsotola Nov 3 '17 at 3:44

Your statement that "half of the energy being stored in the capacitor and half being lost to dissipation" is not very specific to begin with. It is true that the energy lost to dissipation is equal to the energy stored in the capacitor once it is charged, but only if the capacitor is initially completely discharged. In this case, half of the energy removed from the battery went to the capacitor, and half was dissipated.

As the initial voltage on the capacitor approaches the battery voltage, the ratio between the energy transferred to the capacitor and that removed from the battery approaches 1. This explains why the efficiency of a charge-pump DC-DC converter theoretically approaches 1 as the load current decreases.

  • \$\begingroup\$ How is this not specific? It's nothing but specific, it's a real world circuit presented with a real circuit schematic with that exact result. What am I not getting here? You just said that half the energy is lost, which I assume is happening in this diagram, as well as a real world application. Not talking about anything theoretical in any way, shape, or form. \$\endgroup\$ – Marc Striebeck Nov 1 '17 at 16:29
  • \$\begingroup\$ Man, should I just answer my own question??? \$\endgroup\$ – Marc Striebeck Nov 1 '17 at 16:39
  • \$\begingroup\$ Your original statement is not specific because it's not clear what you mean by "half". I restated the question - as I understood it - to clarify this. The dissipated and transferred energy are equal and they could be described as half the energy removed from the battery, but only in the case where the initial capacitor charge is 0 (which is also not stated in your question). \$\endgroup\$ – user49628 Nov 1 '17 at 16:48
  • \$\begingroup\$ Ok, ok. Fine enough of the nitpicking. I suppose you defined the question even more clear than what it already is. Seriously. So just answer it already, does this happen each and every time, after the cap is discharged of course in order to repeat the process, or even change the capacitance, or even the battery voltage. Then tell me what would the voltage on the capacitor be in order for all the energy to be there, and yes I know the battery can't really do that, and I specified that. \$\endgroup\$ – Marc Striebeck Nov 1 '17 at 16:53
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    \$\begingroup\$ Does anybody here understand anything that I am asking? I need to know if half the energy that is lost and the other half that is stored is an occurrence that happens each and every time. The other part of that question is what voltage would it take to equal all 100% of the energy transferred? I specifically specified that I understood well that this would never happen, as I clearly put in parenthesis (By some other charging method), Man... !!!.... Beginning to get slightly irritated at this. I don't understand the down votes to this question, I haven't asked anything strange or weird here. \$\endgroup\$ – Marc Striebeck Nov 1 '17 at 17:06

This is actually one of the paradoxes in the electrical circuits. Consider the following case:
enter image description here

The figure shows two identical capacitors C1 and C2. Suppose C1 was charged with a potential V and C2 was uncharged and we connected them as shown at t=0. Clearly initial energy of the system is \$\frac{1}{2}CV^2\$. But after charge redistribution we will have equal potential \$V/2\$ on both the capacitors because they are identical. The total energy now becomes \$\frac{1}{2}C(V/2)^2 + \frac{1}{2}C(V/2)^2 = \frac{1}{4}CV^2\$. Where do you think half of the energy went?
Answer is we did not consider all the parasitics and half of the energy is lost in the inductance of the loop. This energy would be lost in the form of electromagnetic field.
Similar is the case with your question work done by the battery is \$CV^2\$ and even if the conditions are ideal, that is there is no resistance, in the circuit you cannot transfer this whole energy into electrical energy some of the energy will be radiated in the form of EM field.
So the answer is you cannot transfer 100% energy into a capacitor.

EDIT Since everyone seems to be questioning that the work done by the battery is not \$CV^2\$. Here is a proof: For the circuit shown below, the voltage across capacitor as a function of time is:

enter image description here $$V_C(t) = V_o(1-e^\frac{-t}{RC})$$ Here \$V_o\$ is the constant potential across the battery. Thus voltage across resistor is \$V_R(t) = V_o e^\frac{-t}{RC}\$. Energy lost in the resistor is: $$E_{diss} = \int_0^ \infty \frac{V_R^2}{R}dt = \int_0^ \infty \frac{V_o^2}{R} e ^\frac{-2t}{RC} dt$$ Integrating and putting the limits, we get, $$E_{diss} = \frac{V_o^2RC}{2R}(1-0) = \frac{1}{2}CV_o^2$$ All this energy is coming from the battery. Thus, work done by battery is $$\frac{1}{2}CV_o^2 + \frac{1}{2}CV_o^2 = CV_o^2$$

  • \$\begingroup\$ I am not asking if I can transfer 100% of the energy into a capacitor in any way shape or form. I understand this is impossible as I have already stated in the question. I am asking if that were to occur, which I have clearly stated in my question, then what would that voltage of the capacitor have to be? Or more like what would the voltage have to be to make it equal to 100%. That's all I was asking, and obviously I know you can't transfer 100% when I have already indicated that half the energy is lost, in which I was asking if that occurs each and every time. \$\endgroup\$ – Marc Striebeck Nov 1 '17 at 17:02
  • \$\begingroup\$ If it is not even theoretically possible how am I suppose to answer that. Imagine whatever you like should happen because I don't know. \$\endgroup\$ – sarthak Nov 1 '17 at 18:06
  • \$\begingroup\$ It's not as difficult as it's made out to be. It's just a battery connected to a capacitor and that's it! Half the energy is lost in transferring the other half into the capacitor, as simple as that. I was just asking if this is a set rule in stone, without any changes made to the circuit. The other question was what would the voltage have to be on the capacitor in order to equal 100%? Or in other words, what would the energy be in the capacitor in order to equal 100%? 14.4mJ? This is all I was asking. I am reformulating the question and making an entire new question, since I feel forced to. \$\endgroup\$ – Marc Striebeck Nov 1 '17 at 18:12
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    \$\begingroup\$ @MarcStriebeck It's a perfect voltage source connected to an ideal capacitor using a switch that closes infinitely fast and wires that have no resistance. At some point "simple" starts to be harder because you have conflicting idealized models. A classic example of this: It's not hard to imagine an unmovable object. That's simple. It's not hard to imagine an unstoppable force. That's simple. What happens when an unmovable object is struck by an unstoppable force? It's just simple pieces put together, right? \$\endgroup\$ – Cort Ammon Nov 1 '17 at 18:55
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    \$\begingroup\$ @Cuadue: A charge pump that fully charges and discharges caps on every cycle would be limited to 50% efficiency. A charge pump which alternates between connecting two caps in parallel to a 10V load and then in series to a 19V load, however, would only charge and discharge the caps by 0.5 volts, thus only wasting energy proportional to that 0.5 volt swing, rather than the entire cap voltage. \$\endgroup\$ – supercat Nov 1 '17 at 22:38

Only a one time direct connection with a simple switch from open to closed that charges the capacitor up to the voltage level of the battery is what is being referred to. My question is does this occur each and every time?

This will occur each and every time you charge a capacitor from a voltage source, with no inductance in the circuit, when the capacitor starts at zero volts.

If the answer is yes that half the energy is transferred into the capacitor each and every time then what would the capacitor voltage have to be in order to equal 100% of the energy if it were to come from the battery?

This is a meaningless question. The capacitor voltage can't equal the energy because one of them is a voltage and the other one is an amount of energy. (also, "the energy" means which energy?)

  • \$\begingroup\$ The answer has already been solved quite easily in Is the 50% loss of energy when charging a cap from a battery a set rule in stone? The question does not require that all 100% of the energy has to come from the battery which is impossible. 12 V is 50% of the energy that came from the battery. What was being asked in reality is that if the 50% energy transferred into the capacitor was doubled then what would the voltage be? It would be 16.97 V was the answer \$\endgroup\$ – Marc Striebeck Nov 2 '17 at 0:32
  • \$\begingroup\$ Sure but if you tried to charge it with a 16.97-volt battery, then 200% of the energy would be extracted from the battery and 100% would be in the capacitor. \$\endgroup\$ – immibis Nov 2 '17 at 0:35
  • \$\begingroup\$ OMG dude. There is no need to complicate things more than they have to be. The answer has already been given and solved. 16.97 V is what the voltage of the capacitor would be if the 12 V or half the energy would be if it were doubled. It is not necessary for the energy to come from the battery, which is impossible. This was clearly indicated in the question. Please reread the question. Thank you and good night. \$\endgroup\$ – Marc Striebeck Nov 2 '17 at 0:41

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