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I'm playing around with a very simple setup:

  • 9V battery
  • light diode
  • voltage divider (R1 - three 1 kOhm resistors, R2 - one 1 kOhm resistor), which should give me roughly 3V for the diode

The first time I assembled it, everything was fine, but after a while diode got much dimmer and then stopped emitting light. The battery is ok, as is diode, the problem is in one of the resistors, R2. With power plugged in it shows crazy 10-12 Mega Ohm. When I disconnect the battery it returns to its normal resistance at around 970 Ohm. No wonder diode isn't working given that resistance. Why is the resistance so high? I had another, more complex setup and there was the same problem, I decided to assemble something simpler, but there is this problem with very high resistance again.

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  • \$\begingroup\$ Are you measuring the resistance with your circuit energized? \$\endgroup\$
    – vtolentino
    Commented May 8, 2020 at 11:14
  • \$\begingroup\$ @vtolentino yes \$\endgroup\$
    – Bord81
    Commented May 8, 2020 at 11:14
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    \$\begingroup\$ This is not a problem but an artefact of how DMMs work. TL/DR : don't do that. Make voltage measurements when energised, and infer current from voltage across a known good resistance. \$\endgroup\$
    – user16324
    Commented May 8, 2020 at 11:20
  • \$\begingroup\$ Your voltage divider will not provide around 3V, but around 2.25V (9V / 4) when no LED is connected (do the calculations). The LED itself is a conductor and will be parallel with R2, which will change the resistance of the lower part of your voltage divider (which is then no longer a real voltage divider) to something much lower than 1k Ohm. \$\endgroup\$
    – StarCat
    Commented May 8, 2020 at 11:52
  • \$\begingroup\$ Your LED got dimmer then quit lighting because the battery voltage was dropping as it was used and all the battery current in R1 was going into R2 when the 2.25 V dropped below 2.0V. Without having R2 then all the battery current in R1 goes to the LED. \$\endgroup\$
    – Audioguru
    Commented May 9, 2020 at 1:08

2 Answers 2

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The problem is that you are trying to measure the resistance while the circuit is energized.

First, you have to keep in mind that there is no such thing as measuring resistance directly. The multimeter applies a small current to the resistor being measured and works out the resistance by measuring the voltage drop.

By measuring an energized circuit, you are not only measuring the wrong voltage drop, but also have the risk of damaging the measuring device.

A similar question has been asked before.

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After the "extremely high resistance" of R2 was clarified, I want to draw attention to something typical of LEDs.

You do not need a voltage divider to drive a LED since diodes are current-driven devices. They behave as voltage "sources" (stabilizers) that keep almost constant voltage drop across themselves. The voltage divider behaves also as a kind of voltage "source". As a result, a conflct between two voltage souces in parallel appears in your arrangement.

So remove R2 and drive the LED by the current source consisting of the 9 Vbattery and R1 (probably, only one resistor of 1 k).

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