The 2 capacitor paradox

[ratstar]

I just tried it out, i used 28.9 volts, and ended up with 11 volts on each cap. I tried it with a resistor and it was the same just took more time.

It was enough for a strong blip from both capacitors.

I got 25% loss each time. (add 11 and 11, you get 22.) so its 25% loss on starting with 28 or so.

So... where does this 25% loss come from then? and how come the resistor didnt cause more energy loss?

because its rubbish what u read.

 

[AnalogKid]

Does it ever occur to you that no one has any idea what you are talking about?

ak

 

[Nanren888]

What did you do?

 

[Nanren888]

So you put two capacitors in series across a voltage source.
Then removed the supply?
Then measured the voltage on each, after some time, hence internal leakage, with a voltmeter of finite resistance, hence drawing current. and measured less than the voltage divided by two? Is that it?
.
Would need those two parameters to answer the question. Rate of decay in the capacitor & impedance of the voltmeter.
Your measurements device, voltmeter, will draw current and normal RC decay will apply, even if maybe somewhat nonlinear if the effective load changes with voltage if it's a smart meter.
.
Are my guesses close?

 

[ratstar]

Yep, and i even added a diode.

 

[Nanren888]

where does this 25% loss come from then?

So, capacitors leak & meters drain current.
Measure it with the power connected. The sum of voltages all the way round should be zero, so the battery drops in stages round the circuit. If it's just two capacitors and no current to speak of, then the capacitors charge to sum to the battery voltage.
.
If you disconnect and immediately put a meter across a capacitor, then it starts at that voltage and drops as the meter takes current, drains charge from the capacitor and the capacitor volage drops.
If it is a normal meter, it reads some sort of average or sample of this dropping voltage. Got access to a scope? If so, you will see the discharge, voltage dropping away.

If not:
Connect the meter while the battery is on, & leave it connected when you disconnect the battery. Depending strongly on what sort of meter, you;ll see the voltage drop away.
Maybe google RC discharge.
.
Hope this helps. Enjoy.

I guess, better yet, if you have access to a square wave generator & a scope you can see it repeating and effectively steady on the screen.

 

[Ratch]

You can ask the same question when energizing a cap and then shorting it out with a wire. Where did the energy go? It went away in a electromagnetic burst of static, arcs, and heat. Whether you de-energize the full cap through a straight wire, another cap, or resistor, the same thing happens to more or less the same degree. Energy is lost in the in the path taken by the current transfer. So the energy received by the second cap will never add up to the energy lost by the full cap. Can't happen.

Ratcj