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I drew an example of the closed circuit I was thinking of for my question. Below I have the diode stopping the electric charge after it has already passed through the light bulb. My assumption is that the light bulb will flash once. Am I right?

enter image description here

50 points to whoever can find the duplicate question.

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  • \$\begingroup\$ No, because the small die of the diode won't allow the depleted region to expand large enough to allow enough current to flow to heat up the filament to any significant degree. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 18 '16 at 20:20
  • \$\begingroup\$ No. Kirchoffs current law. If your diode is blocking, no current is flowing at all. \$\endgroup\$ – winny Jun 18 '16 at 20:20
  • \$\begingroup\$ For how long do you think the light bulb will flash? \$\endgroup\$ – pipe Jun 18 '16 at 20:25
  • \$\begingroup\$ related answer \$\endgroup\$ – The Photon Jun 19 '16 at 15:56
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My assumption is that the light bulb will flash once. Am I right?

No.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Battery, lamp and diode.

Current flow in a circuit is like a bicycle chain. All the links move together. If one link is stopped then the chain doesn't move.

In your circuit you have correctly understood that the diode blocks current flow as it is reverse biased. Your misunderstanding is that somehow current flows through the lamp and piles up at the diode. It doesn't. No current can flow. The lamp won't flash.

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  • \$\begingroup\$ I disagree, I think he understands that the diode will generally stop the current completely (as opposed to piling up). As the other answer has grasped, I think hes asking about the time in-between the voltage across the diode being 0, and the voltage being the battery voltage. \$\endgroup\$ – BeB00 Jun 19 '16 at 1:01
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In a very technical, nit-picky kind of way, yes… sort of. The depletion layer of a reverse-biased diode has a capacitance. If we assume that capacitor is discharged at t=0, then some current will flow briefly as it charges to the voltage of the battery.

However, that capacitance is very small for most diodes. For instance, it's about 15 pF for a 1N4001 diode. If we assume a 5V source and a 100Ω resistance, this takes about 10 ns to charge; simulate the circuit below to see for yourself.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ And of course that 10ns will not heat the filament to any appreciable temperature. ( however an LED may emit a 10ns burst of light) \$\endgroup\$ – user_1818839 Jun 18 '16 at 21:44

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