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I have placed a simple picture of the circuit I am trying to determine the behavior of. It is just a simple voltage source like a battery in series with a capacitor in series with an LED.

To my understanding, the capacitor would charge up, then release a large charge quickly which would make the LED very bright, but then begin to fade away quickly, and the process would repeat. However, I’m not sure this would be true. So what exactly would happen in this circuit?

enter image description here

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    \$\begingroup\$ Current will flow until capacitor is fully charged. Then current won't flow any more, and capacitor stays charged. \$\endgroup\$ – Justme Jun 29 at 20:41
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The process will not repeat. You will get a brief flash of light (if the capacitor is large enough, but not so large that the LED is destroyed). Once the voltage across the capacitor reaches its maximum value, which is roughly the same as the supply/battery voltage, then no more current will flow and nothing happens.

Of course, the battery voltage must be more than 1V for normal LEDs. I'm assuming that was just a placeholder value.

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Probably nothing, since the drop across an LED is usually more than 1V. If you increased the voltage (and added a limiting resistor in series) the LED would light up for a short time until the capacitor was charged to Vin-Vled. It would then remain off until someone intervened and discharged the capacitor again.

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