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In this link at Activity 0 i read some stuff that make me confuse :

1-At the first step he charges capacitor as we know, no problem enter image description here

2-But when he ties BATTERY's POSITIVE terminal to CAPACITOR's NEGATIVE terminal, he reads 18.48 V! how is that possible? how capacitor acts like that enables this situation? enter image description here

Also same situation is valid here;

1-He charges capacitor as we know in first(1) step,

2-He ties BATTERY's NEGATIVE terminal to CAPACITOR's POSITIVE terminal and we read -9.14V how is this possible also? enter image description here

So to sum up the questions;

Q-1) How these results can occur?

Q-2) How capacitor acts like this to this results to occur? (what provides this situation to occur)

Any help is appreciated, please help me to understand this.

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    \$\begingroup\$ You have just invented a switching boost converter. Or a charge pump. With manual switching. Anyway, when you connect two power supplies in series, their voltages are added. A charged capacitor is a power supply in this sense. \$\endgroup\$ – Eugene Sh. Jul 29 at 20:45
  • \$\begingroup\$ Think of a capacitor as a (very small) rechargable battery. If you connect two batteries in series with the correct polarity, you will get the sum of the battery voltages across the combination. \$\endgroup\$ – Peter Bennett Jul 29 at 20:55
  • \$\begingroup\$ @EugeneSh. and Peter Bennett thank you very much! now all the things are much clear :) \$\endgroup\$ – gm3434 Jul 29 at 21:00
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The key point is, if you disconnect the capacitor, it is still holding a charge, so it still has 9 V across it.

But when he ties BATTERY's POSITIVE terminal to CAPACITOR's NEGATIVE terminal, he reads 18.48 V! how is that possible? how capacitor acts like that enables this situation?

9 V across the battery plus 9 V across the capacitor is 18 V.

This is basic Kirchhoff's Voltage Law.

He ties BATTERY's NEGATIVE terminal to CAPACITOR's POSITIVE terminal and we read -9.14V how is this possible also?

By putting the black lead of the meter on the positive terminal of the capacitor you're choosing to make that terminal your "0 V" reference.

But there's still 9 V across the capacitor. So the negative terminal is 9 V below the positive terminal. Thus it's at -9 V.

The battery is irrelevant to this situation.

How these results can occur?

From the very basic behavior of a capacitor and Kirchhoff's Voltage Law.

How capacitor acts like this to this results to occur? (what provides this situation to occur)

The capacitor's voltage only changes if there's a current through the capacitor.

When you disconnect the capacitor from the rest of the circuit there's no current through it. Therefore the voltage across it remains at 9 V.

(If you do the experiment yourself, you'll see the voltage slowly change as you measure it due to leakage current through the meter)

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  • \$\begingroup\$ thank you very much for your simple and extensive explanations! all the things are clear now! :) \$\endgroup\$ – gm3434 Jul 29 at 21:01
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The volt meter will read voltages in parallel.

Interestingly enough, there is a DC to DC converter that works on this principle. But instead of moving the capacitor by hand, it switches the capacitor in parallel then in series very fast to get more current. In this way you could get 18V from a 9V battery, or -9V if you wanted to invert the voltage.

enter image description here
Source: https://www.analog.com/en/technical-articles/clever-circuit-improves-efficiency-of-charge-pump-plus-linear-regulator-solution.html

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