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I experimented with charging a capacitor (in my case, 100 microfarad) from by power supply (to 10V) and connecting it parallel to another cap of the same capacitance, with the matching polarities together.

I read this derivation on what should happen:

https://www.edn.com/electronics-blogs/living-analog/4394290/Capacitor-charge-transfer-

I understand that half of the energy in the first, charged capacitor is lost. The voltage across both capacitors should now be half of the original voltage applied across the charged capacitor.

However, this is not what my oscilloscope shows! If I charge the first capacitor to 10V, and connect them, I see around 2 volts across. If I charge the first cap to 8V, I see around 1 volt across the two capacitors! Below is a picture of my setup, can anybody tell what is it that I'm doing wrong or what is happening here? Why am I not seeing half of the original applied voltage across the parallel capacitors?

enter image description here

To clarify, I first took the left capacitor to the left of the breadboard, connected the voltage from my supply across it, and then moved it parallel to the right capacitor. I've checked everything I can think of, all the connections are correct.. And I was careful moving the capacitors, I did not short the leads with my fingers or anything similar.

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    \$\begingroup\$ Were both empty? If yes, swap the capacitors. Is the result still same? You can measure the capacitances measuring the time constant with a resistor, say 220kOhm. Note: the input of the scope must be taken into the account, too. Electrolytics also dig out some charge from slow chemical processes long after they are discharged. \$\endgroup\$ – user287001 Dec 29 '17 at 19:58
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    \$\begingroup\$ User specific part issue solved by experimentation. \$\endgroup\$ – Trevor_G Dec 29 '17 at 20:37
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    \$\begingroup\$ No need to argumentation - it's useless. The energy dissipation in all possible ways are in OP's switching case =50%, no matter occurs there some arcing or not and no matter, what opinions we have. \$\endgroup\$ – user287001 Dec 29 '17 at 21:48
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    \$\begingroup\$ @user287001, half of the energy is lost ... energy does not get lost, it just moves to a different place \$\endgroup\$ – jsotola Dec 29 '17 at 23:34
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    \$\begingroup\$ @jsotola The loss here in my texts = the percentage of energy which get converted to some other form than the electric field in those two capacitors. OP's case = two identical C:s are connected together, one C has zero initial voltage and another has 10 volts. The voltage is assumed to stabilize to a constant DC value. \$\endgroup\$ – user287001 Dec 30 '17 at 13:35
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But here's still no answer. It's corrected now:

The practical tests that I suggested in the comments, mainly swapping the capacitors, showed that they have radically different capacitances. Get new capacitors which are less different, then you get something more like the expected result.

The expected result, when two equal capacitors are connected parallel and one of them had 0V inital voltage and another had 10V: Both have 5V and the total energy stored in the capacitors as static electric field energy is halved, 50% is dissipated as heat and electromagnetic radiation.

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