97modifiedmustangGT

okay so right when i get really comfortable about how to wire things up properly i run into this. I have autometer gauges and i just want a brighter light because with the color cover over the bulbs it gets really dull and even now when i have no cover on the bulb so its just clear it's still dull. I was looking at just replacing the dull bulb with a LED light. I find out that they run on 3.6 volts max so i need a resistor to bring my 12v source down. the trouble is picking the right one. i don't know what they mean, i've saw some that were 1/4 watt and 1/2 watt and i think even 3/4 watt. I've already searched google and all i come up with is the ohms law or whatever it is called. V=IR i think is the formula and i have not a clue what that means so if someone could point me in the right direction of which resistor to pick i would apprciate it.

vanquish

I just did a physics lab with all of this it's not too bad. Voltage equals current times resistance. You need to find the correct resistance (in ohms) of the resistor you need. You can find charts to read the colored bands on resistors.

Since you gave readings in watts which is the unit for power and power= current times voltage you can solve for your current. So now you know current and voltage. So the resistance you need is your voltage divided by your current.

I think it would just be easiest to buy a few resistors and see the voltage drop when you connect them to the power wire with a voltmeter. Move up or down until you get it right.

97modifiedmustangGT

sorry you lost me lol

jwog666

they do make led's that run on 12 volts. just buy those

97modifiedmustangGT

for some reason they don't carry those at radio shack but after some more research i now know what i need to buy. Thanks vanquish, it makes more sense now what you said.

vanquish

Just curious what resistor does it end up being? I'm not a mathematician so i'm just wondering if the math works out.

V=3.6 Volts. If you buy a 1/4 watt resistor and P=IV you have 1/4=I*3.6
Solving for current, I, you get I= 0.0694.
Plugging this I into V=IR gives 3.6=(0.0694)*R
Solving for resistance, R, you get R= 51.87 Ohms so the resistor should be close to that.

97modifiedmustangGT

i came up with a resistor of 370-490 ohms

vanquish

how? What watt (1/2, 1/4, 3/4, etc.) rating resistor are you using?

97modifiedmustangGT

okay i found something that said take the volts the light works on and subtract that from the 12v source i'm using. 12-3.3=8.7 then take the amps the light works on and which is 25amp and somehow that becomes .025 lol. then you take the difference in volts being 8.7 divid that by the .025 which = 348 so i picked up a 470 and a 330 for testing purposes. That probably made little sense but i'm not by the paper that gave me these numbers, so i'll be back later with better detail.

i went into radioshack with printed off instructions and a calculator lol. looked like a complete idiot lol.

vanquish

I was thinking the same thing you're probably right the voltage will be the difference of 12 and the max the LED can handle which is 3.6 (you said 3.6 in the original post). So you need to resist 8.4 volts because that's the actual voltage you want to be running through the circuit I guess. I didn't know that the light's rating was 25 A that makes things easier. I'm getting very confused now too but it looks like you're getting it done.