i have led lights (64 in series) around my spotlights, i need an amp reducing device because there power supply is coming of 50amp wire.

Thanks jim

I Don't quite think you understand electricity...

So here we go...

50 amp wire has nothing to do it...

Voltage determines how much current will flow...

I=V/R

So 12 Volts of potential put across 1ohm of load

12/1 = 12

12 amps would flow in the circuit...

People understand water better...

Volts = Pressure
Amps = Volume
Ohms = Cross Section of pipe (these are inverse 1 ohm is big pipe, while 100ohms is small pipe)

So if you stick pressure in one end of the pipe water will flow... how much water (volume) requires a combination of larger pipe or higher pressure... Volts (pressure) therefore make current flow (volume)... People get confused and think current makes voltage happen or that the two aren't linked...

If you run 50 amp wire for that it will not draw 50 amps... It will draw 12... This will mean the wire does not get hot... If you run more current then the wire is mean for you start to heat the copper... Heating the copper is not an issue... Is the fact the insulation will begin to melt... Eventually you'll get a short somewhere in that case...

If you have LEDs which are like 3.5V (check the specs for the individual LED) x 64 in series... That will require 224V DC to illuminate fully plus a current limiting resistor which I'll get to shortly...

You'll find most high quantity led applications may have 3 in series and then join those groups in parallel... This way you are not limiting that much current through the resistor and it can be smaller... if you hook them all in parallel you'll need a huge wattage to dispite the power consumed by it (it gets hot too heat)... Use as many in series as possible (3 would be ideal)... then join you lots of 3 in series in parallel...

Now the current in each lot of three is the same... each LED when run at 3.5V may consume 30mA...

So three in series would require 3.5x3 = 10.5V and they all use the same current so 30mA... As a rough figure allow 2V drop across the resistor So 12.5V total... You can see why this is a common combination...

So for each branch you pick supply voltage (13.8 mean when car is running)... LED voltage (10.5 as above)... And LED current (30mA)...

Then

R= (Vs-Vl)/I

R= (13.8-10.5)/(.03)
= 110ohms so you would use a 120ohm resistor second part is the wattage...

Volts dropped accross resistor is 3.3 (13.8-10.5)... Current is .03A (30mA)

P= VI

P= 3.3x.03
=.099W

So a 1/4 W resistor would satisfy...

Each branch must have its own resistor due to tolerances and the like...

So that'll point you in the general direction...

See here for the same sort of thing worded differently...

http://www.kpsec.freeuk.com/components/led.htm

Cheers,
Tim

thanks for that, dont have time to read it all now but what do i need to put in there so i wont fry the wires?

If you read it I wouldn't have to right it again...

You will only fry the wires when you exceed 50amps...

so exceed 50 amps (assuming your car charges at 13.8Volts)

R=V/I

R=.276

If you connect a load which is any less than .276 ohms you will draw more than 50 amps and begin to fry the wires...

The issue you have my friend is you said you have 64 LED's joined in series... You will never light 64 LED's wired in series... Unless you find a way to draw in excess of 200V from your car battery (yes two hundred)... Again read above where I covered this...

You would have been better off buying those ccfl halos off ebay and fitting them... What you have done so far is a waste of time and money and will need to be all done again...

Even if you have them wired in parallel it will not wire because then each resistor needs a resistor which you would have no done because you are only asking now...

Best to understand and then do is the thing I'm trying to explain...

Cheers,
Tim

well the way they are linked up they do work whn i had them on a battery, so why would it blow wires when its hooked up to my spotlight power wire:S hmm

so they mustn't be linked in series then?

without all the facts and schematics is hard to tell you what you need when you don't know what you have...

And this has nothing to do with the power wire for the third time... You will not blow any wires... Just the diodes if you stick too much juice through them...

ahhhhh hahaha
give up timmy!
bahaha