Timeline for Led on breadboard burned out why?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 19, 2020 at 8:47 | comment | added | Transistor | @jxramos, see lednique.com/current-voltage-relationships/iv-curves and the accompanying video. Thinking of an LED as "3 V" may get you into trouble. Instead you should think of the current rating and see what the voltage drop will be at that current. | |
| Dec 19, 2020 at 5:58 | comment | added | jxramos | To be sure, a 9V would burn out a 3V LED very quickly right? | |
| Dec 25, 2018 at 20:06 | comment | added | Bob G. | Yep, strange xd. But I know now why my second circuit was wrong. Thanks for the help. | |
| Dec 25, 2018 at 20:00 | comment | added | Transistor | Your mind played a strange trick on you. On the one hand you realised that the power from your battery was being conducted along the red and blue rails because that's what you connected your circuit to. On the other hand you expected it to be broken somewhere between the two ends of the resistor inserted at a random point on the rail! | |
| Dec 25, 2018 at 19:31 | vote | accept | Bob G. | ||
| Dec 25, 2018 at 19:31 | comment | added | Bob G. | So the breadboard goes around then, I did not know that. I thought I wasn't bypassing because the resister was in front of the led essentially. Thanks. | |
| Dec 25, 2018 at 9:25 | history | edited | Transistor | CC BY-SA 4.0 |
added 581 characters in body
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| Dec 25, 2018 at 9:14 | history | answered | Transistor | CC BY-SA 4.0 |