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Registered Member #2906
Joined: Sun Jun 06 2010, 02:20AM
Location: Dresden, Germany
Posts: 727
Charging a capacitor via a resistor is always just 50% efficiency. This resistor is the internal ESR, actually the sum of both ESRs of the both capacitors. Also charging a capacitor from any other source is only 50% efficient too, since there is always the ESR in some way. The equation changes btw, if the other capacitor is not completely empty. This is why charge-pumps can have efficiency close to 100% despite using resistive capacitor charging. The initial cycles suck, but if you charge a capacitor from 4.98V to 5V from another 5V capacitor that is basically 100% efficient.
The problem you describe is generally known as capacitor paradox. Give Wikipedia some love:
Registered Member #11591
Joined: Wed Mar 20 2013, 08:20PM
Location: UK
Posts: 556
Indeed, another way of thinking about it is you are putting two differing voltage sources in parallel, which would cause (V1 - V2)/0 A of current to flow, which is impossible, and therefore you have made some unuseful simplifications, and need to add some more realistic mathematical models in there, like adding resistance and inductance (if you are interested in what would happen over time).
Registered Member #2099
Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1714
The youtube video in OP is not wrong. The charge Q initially on one capacitor becomes shared equally between the two capacitors -- no charge is lost. Half of the initial CV^2/2 energy is lost, of course. I bet that's mostly from ESR, in ordinary cases. The relevant law says that charge, not electrical energy, is conserved. Could also say that initial energy is Q^2/2C, and state 2 has same total charge spread over twice as much capacitance.
Analogous to connecting a tank of compressed air to an empty tank of the same volume. No air is lost when you open the valve, but half of the initial vp^2/2 energy gets turned into heat.
You can get practically all of the charge from C1 moved to C2 without loss of energy. Connect the two capacitors through an inductor, and break the connection at the instant when current stops after 1/2 cycle. Or just include an ideal diode in series with the inductor.
Registered Member #54278
Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367
Brilliant! When my shop is back up and running, I think I will try the "series diode" technique..."learn something new every day" as I was recently told!
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2S2P series/parallel capacitor + extra connection
