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schematic

simulate this circuit – Schematic created using CircuitLab

How does the capacitor behave here?

I tried and the led flashed. Does the capacitor work like a diode after charging?

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    \$\begingroup\$ Does the capacitor work like a diode after charging? - it works like an open-circuit after charging. \$\endgroup\$ – Eugene Sh. Oct 15 at 20:10
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    \$\begingroup\$ In fact, it behaves like a capacitor. Really, the capacitor is component that's way more fundamental to understanding the temporal behaviour of electrical circuits than a diode, so I can only encourage you, Jundullah, to read an introduction to linear electrical circuits. There's a reason you learn about what a capacitor is, mathematically, in the first three weeks of any Electrical Engineering program: It's just so fundamental for the rest of electronics. \$\endgroup\$ – Marcus Müller Oct 15 at 20:25
  • \$\begingroup\$ AFter cap charges up , no more current for LED until it self discharges or another LED in parallel reversed to LED 1 is added. Then it flashes alternately. A big Cap is like 1milliFarads. While a small Li-Ion cell battery is like a huge cap of 10kFarads. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 15 at 20:25
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    \$\begingroup\$ Added HOMEWORK tag to discourage folks from giving outright answers to the question. \$\endgroup\$ – Michael Karas Oct 15 at 20:27
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    \$\begingroup\$ Carefully look at the capacitor schematic symbol. Note how there's two parallel lines that do not connect. This means that direct current cannot flow across this gap. Changing current can, because the charge on the capacitor is building or depleting. As Marcus said, this is fundamental. Research a bit more, then feel free to post a most specific question if you have gaps in your understanding. (Gap. Like the capacitor. See what I did there? I'll see myself out...) \$\endgroup\$ – JYelton Oct 15 at 20:35
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The capacitor acts like a capacitor. It charges up like it normally does, but when the voltage potential measured at the anode reaches the LED's turn on voltage, the diode's junction internal resistance goes down, and the current flows through the diode, discharges the capacitor until the voltage potential has lowered past the junction off voltage, then the capacitor charges up at the rate of the dark current (normal junction leakage current when not turned on)

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