Timeline for How can Ohms law be used to calculate the resistor value for an LED when multiple voltage/current pairings give the same resistance?
Current License: CC BY-SA 4.0
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| May 2, 2019 at 17:51 | comment | added | jonk | @G36 Although that choice was mentioned earlier in the answer, I just updated the bottom of the answer to re-address your question so that the computation I used should be more immediately clear than it was before. Thanks. | |
| May 2, 2019 at 17:50 | history | edited | jonk | CC BY-SA 4.0 |
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| May 2, 2019 at 17:49 | comment | added | G36 | Thank for the info. | |
| May 2, 2019 at 17:45 | comment | added | jonk | @G36 The OP said that \$V_\text{CC}=9\:\text{V}\$ and that \$V_\text{LED}=3\:\text{V}\$. But I substituted in a datasheet to make it real and the LED in that datasheet says \$3.0\:\text{V} \le V_\text{LED}\le 3.6\:\text{V}\$, so I used the midpoint for the calculation, or \$V_\text{LED}=3.3\:\text{V}\$. | |
| May 2, 2019 at 17:39 | comment | added | G36 | I have a stupid question. What values did you use to compute \$\frac{\%\,I_\text{LED}}{\%\,V_\text{CC}}=1.58\$ ? \$3V\$ and \$9V\$ ? | |
| May 1, 2019 at 5:16 | history | edited | jonk | CC BY-SA 4.0 |
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| May 1, 2019 at 3:27 | history | edited | jonk | CC BY-SA 4.0 |
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| Apr 30, 2019 at 23:23 | history | edited | jonk | CC BY-SA 4.0 |
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| Apr 30, 2019 at 21:59 | history | edited | jonk | CC BY-SA 4.0 |
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| Apr 30, 2019 at 21:28 | history | answered | jonk | CC BY-SA 4.0 |