I've built a circuit on a breadboard using a 10K potentiometer, a 220 ohm resistor and a LED. They are connected in series. Using Ohm's law, you find that:

Total resistance \$R = 10000 + 220 = 10220\Omega\$

Battery: 9V.

\$I = \frac{V}{R} = \frac{9}{10220} \approx 0.88mA\$

Isn't that current too tiny for a LED? But even tough, it lights up at this resistance. Can anyone explain?

  • \$\begingroup\$ The blue led in youtube.com/watch?v=yxoKb4d3gFA is powered at 0.3mA, 300 microamps. It's clearly visible in full daylight, when the white leds are on, and in the dark. I have other leds that are reasonably still on at 100 microamps and below. \$\endgroup\$ – Passerby Mar 25 '16 at 3:52
  • \$\begingroup\$ I messed with an LED once that you could see glow by attaching one side to ground and holding the other side. Basically it was acting as a diode detector for all the line voltage noise that my body was picking up from the mains. \$\endgroup\$ – Daniel Mar 25 '16 at 3:54

Since the pot is in series you should be able to vary the current- assuming the LED drops 2V you would have from (9-2)/220 = 32mA to (9-2)/10.2K = 0.7mA.

Note that there is no need to have more than a couple significant figures for the current calculations- the battery voltage , LED drop and resistances are not that accurately defined.

Modern LEDs can be fairly bright with 0.7mA DC so if you have a good LED it could well appear reasonably well illuminated. On the other hand, 32mA may be too much current for a small LED.

|improve this answer|||||

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.