CIRCUIT: A 12 volt battery powers a dimmer. The output of the dimmer powers 4 LED light bulbs. They are all in parallel: 6W + 2W + 2W + 2W. The 6W bulb is a spotlight. The 2W light bulbs are under cabinet puck lights. All bulbs are dimmable. This is tested and verified. The dimmer varies its output from 0 - 12 V and all of the bulbs brighten and dim together in unison.

PROBLEM: The 6W spot light, mounted on the ceiling, is much brighter than the other lights. I would like to keep that bulb on the same dimmer but have it be less bright.

IDEA: How about adding a resistor in series with the 6W bulb to reduce the voltage/current, which will cause the bulb to dim?

PLAN: I used my multi-meter to measure the current flowing to the 6W bulb: 0.5 A. Using Ohm's Law, I reckoned that by adding a 5 Ω resistor in series with my bulb I can effect a voltage drop which should reduce the brightness of the bulb. I added a 5 Ω resistor in series to the bulb.

RESULT: The 6W bulb does get slightly dimmer. My ceramic resistor gets "toasty". HOWEVER - my multi-meter now reads 1.2 A for the wire going to the resistor and bulb. This took me completely by surprise. I thought the current would go down to 0.4 A, not up to 1.2 A.

What on earth is going on? Anyone want to take a guess?

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    \$\begingroup\$ TRIAC + battery = does not compute. \$\endgroup\$
    – JRE
    Commented Mar 14 at 18:26
  • \$\begingroup\$ It's already been answered about the current draw increasing do to the LED compensating for the losses introduced by the resistor. But if you want the lights to have the same brightness, why not use lights of equal power? Use all 2W LEDs instead. \$\endgroup\$
    – MOSFET
    Commented Mar 14 at 20:03

1 Answer 1


No, Ohm's law stands. There's also no need to guess what is happening, this is pretty clear:

A lightbulb just simply isn't a simple resistor.

Incandescent light bulbs are approximately constant power sinks, due to the nature of their filament heating up until a thermal equilibrium is reached, which happens at a fixed power.

LED lightbulbs typically have an internal switch-mode power supply and are by design pretty good constant power sinks; their job is to supply the internal LEDs with a constant current at approximately constant voltage, and thus drive them at approximately constant power.

If you draw constant power through a load, adding a series resistor, thus lowering the voltage across the load, the current will go up, as power is the product of current and voltage.

So, it's not Ohm's law that's broken. You just don't realize you're out of line trying to apply it here.

  • \$\begingroup\$ Just for my own edification: my dimmer is of a modern switching (on-off) type, not the old rheostat kind. Can you tell me the proper name for that (apparently not TRIAC). Thanks? So - my trick might have worked for an incandescent light bulb but - not a sneaky "smart" LED bulb that wants more power?! \$\endgroup\$
    – rstross
    Commented Mar 14 at 18:32
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    \$\begingroup\$ no, it would not have worked. I explitly mention that! (I don't know actually what your dimmer does – but I think if you call it "switching", that's a good name; others would probably be "PWM controller") \$\endgroup\$ Commented Mar 14 at 18:34
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    \$\begingroup\$ @MarcusMüller but incandescent bulbs are dimmable, how does that work then? Wouldn't they temporally average the dimming modulation? \$\endgroup\$ Commented Mar 14 at 20:08
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    \$\begingroup\$ oh, keen observation! Due to the switching, the thermal equilibrium that is reached changes. You got me there! \$\endgroup\$ Commented Mar 14 at 20:32
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    \$\begingroup\$ Okay race fans! I had to know, so I ran a little test: dimmer output = 12.4 volts (it's on my camper). Old 1141 side marker bulb rated at 18.4 watts supposedly. I hook it up and notice the bulb is nice and bright. Mr. multi-meter reads 1.3 amps. Next, I put my 5 ohm ceramic resistor in circuit (series of course). Mr. bulb is not happy - he is dimmer but still lit. My multi-meter now reads exactly 1.0 amps. Conclusion: an incandescent bulb will dim in this case but, my smart LED bulb will not because he is too clever for me. Everyone have a great night - thank you for your comments! \$\endgroup\$
    – rstross
    Commented Mar 15 at 0:58

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