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Please, just a quick question, am I right that the reason why LED is not lighting is:

  • LED is connected parallel to the wire

  • The wire has much lower resistance, so nearly all current is flowing through the wire, nothing left for LED

  • There is a short-circuit obviously, so the battery will heat up

Correct or not? :-)

thanks!

enter image description here

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    $\begingroup$ And if you remove the short circuit, you will end up with a blown out LED. Remember: LEDs should be driven with a constant current, not a constant voltage! $\endgroup$ – Massimo Ortolano Feb 4 at 12:57
  • $\begingroup$ Yes, the blown out factor will depend on the voltage, 3V is OK, 9V is direct death :-) $\endgroup$ – Pavel Feb 4 at 13:32
  • $\begingroup$ Pavel, 3 V is NOT OK. $\endgroup$ – Massimo Ortolano Feb 4 at 13:46
  • $\begingroup$ Massimo, you are right, I meant by that the fact that 3V is not a fatal situation for a short period of time. At least what I see when doing experiments. I should definitely use 300Ohms resistor to limit the current to 10mAmp, is that right? $\endgroup$ – Pavel Feb 4 at 14:17
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    $\begingroup$ @Pavel, The LED in the picture appears to be red. The forward voltage of a red LED at it's normal operating point is somewhere in the neighborhood of 1.7 or 1.8V. If you actually forced 3V across the LED, it would be almost instantly destroyed. What can happen though, if you connect the LED to a small, nominally 3V battery, is that the internal resistance of the battery may prevent it from sourcing enough current to damage the LED (Which is another way of saying that the current through the LED is enough to drag the battery voltage down to 1.7 or 1.8V.) $\endgroup$ – Solomon Slow Feb 4 at 17:34
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  • LED is connected parallel to the wire

  • The wire has much lower resistance, so nearly all current is flowing through the wire, nothing left for LED

  • There is a short-circuit obviously, so the battery will heat up

Yes, that is exactly correct.

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  • $\begingroup$ Great! And if we are talking about home voltage (220V) instead of battery, that big current is delivered from the circuit to the fuse and it will disconnect the loop, right? $\endgroup$ – Pavel Feb 4 at 13:36
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    $\begingroup$ @Pavel Do not attempt this with a mains-line voltage ─ it is unsafe and it will seriously damage your breadboard (and, potentially, blow out the LED). $\endgroup$ – Emilio Pisanty Feb 4 at 15:27

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