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Lets say I have 2 capacitors, one rated 100 µF, 10 V and the other 100 µF, 300 V.

Now I charge them both with 5 V for 1 min (or until they reach 5 V, fully charged) and connect them to a same load, for example an LED one at a time.

Will they both light that led up for same amount of time?

Please, no tank examples...

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  • \$\begingroup\$ MathJax is your friend. \$\endgroup\$ – skvery Mar 28 at 10:12
  • \$\begingroup\$ Since all the answers are correct and are identical, I randomly chosen the answer. \$\endgroup\$ – asim Mar 28 at 11:36
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Presuming that the capacitors are electrolytic, are in good working condition and their capacitance values are identical (since electrolytic capacitors have a tolerance range of -20% to +80%) they will test identically.

The working voltage of an electrolytic capacitor is the maximum voltage which should never be exceeded. At the same time, using a 300V electrolytic capacitor in a 10V circuit is not a good idea. In an electrolytic capacitor, the dielectric is an insulating oxide layer formed on its aluminium foil anode by electrolytic action when a positive 'forming' voltage is applied to it. The working voltage is decided by the thickness of the oxide layer. During an electrolytic capacitors lifetime, the oxide layer would be maintained only if the working voltage is between 25% and 90% of the forming voltage. If not, the dielectric thickness would reduce leading to capacitor failure.

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  • \$\begingroup\$ Thank you, Asim. \$\endgroup\$ – vu2nan Jun 22 at 10:45
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It depends on the dielectric material. Some dielectric materials have very non-linear characteristics meaning that \$C(V)\$ instead of \$C\$ being constant.

These non-linear characteristics are used by some snubber circuits to reduce losses.

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  • \$\begingroup\$ What about normal most common electrolytic cylindrical capacitors? \$\endgroup\$ – asim Mar 28 at 10:24
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Both capacitors have the same capacitance C, I suppose these capacitors are standard electrolytic ones. The capacity C of each component is nearly constant within the voltage range. The voltage rating printed on the capacitor is the maximum voltage you may charge with. The electric charge Q of each capacitor is (after full charging) Q = C * U that is for each capacitor: Q = 100 µF * 5 V = 0.5 mAs, or 0.5 mC (milli-Coulomb). When connecting a (or both) capacitor(s) to a load, be careful to limit the current e.g. with a resistor. Because when you directly connect the 5 Volts to a diode (LED) a high current will flow for a short time at 5V through the diode, limited only due to the diode characteristic curve and due to internal resistors of the capacitor, leads and transition. This high current could destroy your LED. The voltage (and the current) will both go exponentially towards zero.

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Will they both light that led up for same amount of time?

They might destroy the LED due to the sudden inrush of current into an unprotected LED but, assuming the LED survives this (or is protected against it) then both capacitors will illuminate the LED equally for a short period of time.

The capacitors voltage rating is just telling you not to exceed that voltage; it doesn't tell you anything about how it functions or stores energy i.e. its a stress rating value.

Please, no tank examples...

enter image description here

Sorry, I couldn't help myself. This one is called Deborah.

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  • \$\begingroup\$ Lmao I meant water tanks and pipes examples xD \$\endgroup\$ – asim Aug 27 at 12:12

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