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Curious case of the charging capacitor

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ZeroResistance:
I'm going to revisit the classic capacitor charging case, which I just came across

1. Is it true that half the energy fed by the power supply in charging the capacitor is lost, I mean if 100uJ is supplied by the power supply of 10VDC, finally the energy stored in the capacitor @ 10VDC would only be 50uJ.
If this is true can charging a capacitor be considered inefficient?
Following on are there efficient ways charging a capacitor?

2. This case is of charging a capacitor with another capacitor. Eg. If I have a 1uF capacitor charged to 10VDC and another 1uF capacitor uncharged. If I connect both of these in parallel the final derived energy is again half the energy that we started off with. So where did the other half go?
And if this is true then how come charge pumps achieve high efficiencies?

On subsequent searches I also found several posts indicating that this is likely due to energy loss in the resistance
wires.

I would like to hear the views of the learned audience at eevblog on this subject.

TIA

PS: Would the above cases still be true if the capacitor was charged with constant current?

Doc Daneeka:
   The problem is the 'abstract model' of the circuit is not valid - the problem talks about the voltage across the capacitor as if it were directly connected across an idea voltage source, or as if there were a step change in voltage across the ideal capacitor - this is not a mathematically valid set of conditions - its not even that 'the energy is lost in the resistance' there is no resistance in the model. Its just that a step change in voltage gives undefined power into or out of the capacitor.

Yes in a real circuit the energy goes into the resistance. As soon as you add a resistor to the model there is no problem and the model works perfectly

If it is charged with a current source everything works out fine.

Its just the old 1/0 trick in disguise - effectively the problem gives an initial and final condition for the capacitors, but no way for it to go from one to the other.

ZeroResistance:

--- Quote from: Doc Daneeka on February 20, 2019, 06:10:39 am ---   The problem is the 'abstract model' of the circuit is not valid - the problem talks about the voltage across the capacitor as if it were directly connected across an idea voltage source, or as if there were a step change in voltage across the ideal capacitor - this is not a mathematically valid set of conditions - its not even that 'the energy is lost in the resistance' there is no resistance in the model. Its just that a step change in voltage gives undefined power into or out of the capacitor.

Yes in a real circuit the energy goes into the resistance. As soon as you add a resistor to the model there is no problem and the model works perfectly

--- End quote ---

Ok, so lets consider a one ohm resistor into the circuit. What I would like to know is If I put in say "x" amount of Joules in charging the capacitor would the final energy on the capacitor be less that 50% of what was fed in?

Doc Daneeka:
I dont know, do the calculation or try in LT spice! The efficiency does depend on the rate of charging in a slightly surprising way...

Jwillis:
Your greatest loss occurs just at the beginning of charge when the capacitor acts like a dead short.Then when it begins to charge the amount of current absorbing lowers exponentially as the voltage increases exponentially until the capacitors resistance reaches infinity,Then its considered fully charged.
When you disconnect the battery and  attach another fully discharged capacitor parallel to charged capacitor the same thing occurs until both capacitors reach a balanced state then both will discharge because neither have and infinite resistance.

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