Spotlight dim function with resistor

[quarck]

I'm a beginner in this forum, shoot me if i'm stupid.

i'm going to mount some 10w spot-lights in my car(12v), instead of installing/buying an expensive dimmer, i want to have a 3 way switch which 1. no light. 2. half effect. 3 full effect.

I figured it would work with a simple resistor, can someone help me with which type of resistor i have to use?

excuse my english .

 

[David Bridgen]

You will need a resistor with a value between 10 and 4 or 5 ohms for each light. It is impossible to give an exact figure because the resistance of the light bulb is dependent on the value of current flowing through it

The resistor should also be rated at 5 watts minimum.

It would be useful if you can acquire a wire-wound potentiometer or rheostat with a value up to 10 to 20 ohms and a current rating of at least 1 amp. and experiment. Once you have adjusted it so that the light intensity is what you want you can measure the resistance of the pot and then obtain a fixed resistor of the same or similar value - one for each.

 

[Russlk]

You only have to reduce the voltage on the lamp a little to get half the light output, but I don't know how much that is. At any rate, P=E^2/R, so the resistance of the lamp should be: R = 144/10 = 14.4 ohms. Try about 5 ohms in series and see how that works. A 5 or 10 watt resistor will be OK.

 

[quarck]

Thanks..

..so it have to be on each light, it isn't enough to have it just on the switch?

 

[Russlk]

If you have one switch controlling two lights, then R = 7.2 ohms, so try 2.5 ohms in series. The power rating of the resistor is higher: P=I^2*R = 4*2.5 = 10 watts minimum.