gauge lighting dimmer

I installed a 25 ohm 3 watt rheostat to adjust 3 gauges seperate from the instrument cluster lighting.  Now the rheostat gets extremely hot. I dont know much about electricity, but Im thinking of putting a 47 ohm  .5 watt resistor before the rheostat.  Will this help dissipate the heat?  Can anyone tell me what is needed to make it cooler?  The power supply is 12 volts and the gauge lighting is LED, but not sure how much they draw. Thanks

Yes this is simply ohms law in effect, adding more resistance will increase your problems, look to a variable DC-DC device.

How did you wire the potentiometer?  Positive voltage on one leg, ground on another and your lights to the wiper?  Or did you wire it as a variable resistor, wired in series with the wire going to the lamps?  Are the lamps incandescent or LED?

power on the left side, ground on the right side, and lights in the middle.  The lights are LED, there is 1 per gauge for a total of 3.  So if I put in a resistor that will increase the heat?  I thought a resistor would help dissipate the heat from the pot by absorbing some heat.  Thanks for all the info.  I know just enough to be dangerous LOL.

You won't reduce the heat given off (Watts) though you may reduce the temperature with heatsinking.

The best solution is a PWM (Pulse Width Modulated) circuit - eg, a 555 timer in 0-100% duty cycle configuration driving a FET or transistor.

Aha - wrong wiring!
Put the pot/rheostat in series with the lamps.
You have connected it across your voltage supply, hence the heat will be V**2/R Watts (ie, Supply Voltage x Voltage divided by Resistance).

With an LED it is going to take WAY more than 25 ohms to make even a slightly noticeable decrease in brightness.  If wired in series.  25 ohms across 13 volts into a 3 watt device is going to get really hot.  Can you get a higher value pot?  100 ohm to 470 ohm at 3 watts will not get hot. 

You really need to know the size, number & wiring of the LEDs.

EG - if the LEDs are 20mA each with a 500R (Ohm) dropping resistor, a 2k resistor will drop the LEDs current four-fold (roughly).
That would be a 1k pot for 2 LEDS, or 500R pot for 4 LEDs etc.

But if they are in series....
EG 4 x 20mA LEDs in series, hence a 200R dropping resistor. Hence 800R = 1k pot for 1/4 current.

Experiment with a 1k or 5k pot. (Heat will depend on LED wiring.)

That's the beauty of a PWM circuit - the topology doesn't matter.
FYI - That's how you dim DC fluoros (eg, 12V) or - with a filament transformer - AC fluoros. (Fluoro tubes etc - not CFLs etc!)

Thanks everyone.  Radio Shack has a 5K-Ohm Linear Taper Potentiometer with Wattage rating of 0.5W.  Would this work?  Also, my current wiring for the current pot is positive on left, ground on right, and output in middle.  If I remove the ground that should make it in series correct?

Yep - just joint the centre arm to either of the ends.
Connect one end to +V & the other to the LEDs.

Whether the 1/2W rating is enough depends on the total current and position of the wiper.

If a 1k pot is cheap enough, get that too.

Can I use the 25 ohm pot and then wire in a few 100 ohm in line resistors to the output side?  Would that work too?  The 5k ohm is the lowest I can get that readily available.  I figure five 100 ohm resistors should work.

Yes - but you'll only vary the current by about 25 in 500 - ie, 1:20 = 5% variation (ie, 25 variable in 5 x 100).