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My components are

  • 2x 1.5 V batteries
  • 25V 2200uF capacitor
  • 3v Led
  • copper cables

My circuit looks like the following.

schematic

simulate this circuit – Schematic created using CircuitLab

My understanding here is that as the two batteries are in linked in series, we now have 3v for our circuit.

When completing the circuit the LED starts bright then slowly fades and fades until there is no light at all. Also interestingly if I dissconnect the batteries, wait a minute, and reconnect them completing the circuit, the LED stays off - It does not start bright and fade again.

What is actually happening here? Is the capacitor consuming all the power? If I left this circuit complete for a while, would the capacitor gradually consume all the power until it reaches its 25V and then suddenly release 25V instantly? (blowing the LED of course)

I actually ran this test because I wanted to try to fill the capacitor up, and then dissconect the batteries hopeing to see the LED stay alight for a while running from the capacitors consumed power.(if this is even possible, perhaps my understanding is not correct)

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  • \$\begingroup\$ [1] Draw a schematic of your circuit (wiring diagram which you drew doesn't count as a schematic, although it's a start). EE.SE has got a built-in schematic editor. Maybe the answer will jump at you. [2] You've got a polarized capacitor. Be sure not to reverse the polarity of this capacitor, otherwise it will blow up. \$\endgroup\$ – Nick Alexeev Aug 12 '15 at 3:22
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    \$\begingroup\$ You absolutely have to connect the LED through a 100-200 Ohm resistor. It is not negotiable - if you have a LED in your circuit have a resistor, period. \$\endgroup\$ – ilkhd Aug 12 '15 at 3:57
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As cowbydan stated, placing the capacitor in series to the LED will block DC currents. To achieve the desired effect of a glowing LED after removing the batteries you need a circuit similar to this :

schematic

simulate this circuit – Schematic created using CircuitLab

The capacitor needs to be in parallel to the LED and the polarity of the capacitor needs to match the polarity of the connected battery. The resistor R1 is not mandatory in your particular application but it reduces the current which the LED consumes. A larger resistor value will increase the time the LED glows after you remove the battery but will also decrease its brightness. Just test it with some different values. Choosing a LED with a lower voltage rating would require a resistor in series to it to reduce the voltage.

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Capacitor works like a toilet tank: it keeps the water flowing, until it fills. Then you can discharge it (flush). After you flushed water flows faster, but slows down. Capacitors do similar thing with electricity.

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Your capacitor is connected in series with the LED instead of in parallel with the anode. The "fading" is the AC coupled impulse from connecting the battery to the capacitor. Capacitors block DC current, so you are starving the LED for current.

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  • Nominal 25V on the capacitor means it can tolerant up to 25V input I think. What happen to the capacitor is that it store the electricity, not consume, until it is charged up to 3V in the opposite direction to the batteries, meaning that the longer tail of the capacitor becomes 3V. So then the voltage drop between the LED gradually(the speed depends on the capacitance of the capacitor) becomes 0V and the LED.
  • When you remove the batteries and INSTANTLY connect the LED with the LED on the other side, the capacitor might source current for a little while, but the electricity stored in the capacitor might be too small to source enough current to LED you
    can notice on eyes. Remember, the voltage itself does not tell how much
    current the device can source and light up the LED.
  • Please connect resistors BTW.
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A capacitor acts kind of like a break in the circuit, so no current could flow through your circuit. How is it then, that your led lights up for a short time?

A capacitor has other characteristics, when you apply DC to it, it starts to charge. Charging means, that is slowly attempts to reach the supply voltage. (3V in your case) the 25V on that capacitor is the maximum voltage you can apply to it before it blows up.

So because your capacitor stores electrical energy(charge), when you charge it, it lets current flow, this current causes the led to light up. But as the voltage on the capacitor increases, the less current flows, and the led will fade.

Why doesn't this happen after disconnecting and reconnecting? Because the capacitor kept its charge. Capacitors don't magically discharge, when they are charged they act like small fast-depleting batteries. You could discharge your capacitor by shorting it with a small value resistor(not with a wire, as that could cause it to be damaged). Then the led would start fading again.

Why doesn't your LED light up again when you disconnect the battery? Because you leave the other end of the led disconnected. If you would connect it to the capacitor with correct polarity, it would light up. (it would need reverse polarity, because currently the positive side of the capacitor is connected to the negative side of the LED.) BUT DON'T DO THIS WITHOUT A RESISTOR, limit the current flowing thorough your LED.

Also see @Grebu 's answer for how to make the circuit behave as you first expected it to.

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So i was also questioning this and did some research and i came up with something that i'm not sure is correct

Electron flow is the reason the LED lights up then fades but not from electrons that originated from the battery, but rather from the metal plates located inside the capacitor

keep note the series circuit is not actually closed, the capacitor has dielectric material keeping the current from flowing through

Before anything, both plates inside the capacitor are neutral, but once you connect the battery, the electrons from the plates in the capacitor get attracted to the positively charged side of the battery. and that its self is current(flow of electrons)

when the capacitors plates change in polarity, and electrons leave one of the neutral plates(p1) to go to the positively charged side of the battery and repelled from the negatively charged side of the battery, to the other neutral plate(p2), causing it to become negatively charged, those electrons create current, lighting up the LED.

but the plate doesn't have a lot of electrons, so eventually the neutral plate loses most or all of its electrons and the other plate gains electrons. when one of the plate runs out of electrons, current flow stops , explaining why the LED eventually stops lighting

and then after you have charged your capacitor, one plate is has a lot of electrons(negatively charged) and the other has a lack of electrons so more protons(positively charged) and they are both attracted to each other inside the capacitor, they are eager to "connect" or "touch"(for the lack of a better word) but cant because of the dielectric material

so the only way for them to touch or reach each other is if the terminals are connected by some conductive material so when you connect the LED to the circuit, in series, the electrons now have a way to reach the protons(positive plate) so they flow inside the conductor, creating current so when you put a load, such as an LED, current flows though, making the LED light up, but for a short while

then the 2 plates both reach a neutral state, ones the positively charged plate has gained electrons and the negatively charged plate has lost electrons. then the current stops, when the flow of electrons stop

i watched ElectroBOOM's video on capacitors and a few google searches it helps with the understanding

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  • \$\begingroup\$ Plates do not run out of electrons per se. When enough excess electrons accumulate on one side, the voltage difference between the plates cancels the voltage difference generated by the battery and causes current to stop. Also the "before and after" circuit connections you mention don't seem to match what is described by OP. \$\endgroup\$ – AJN Aug 9 at 6:03

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