electric/resistor guru's ... a question

**krazE** · 11-16-2004, 01:12 PM

what type of resistor would i use to reduce a 90 watt headlight bulb to say 45 watts?

i know it's kind of offtopic, but i've been searching day and night and haven't been able to find anything

. The guys at radioshack didn't even know what a resistor did.
thanks much

**Marsupial** · 11-16-2004, 01:18 PM

you want to reduce the wattage of the lightbulb?

**ColdPhreze** · 11-16-2004, 03:52 PM

You will use a power resistor, and calculate the value using some standard electronics formulas:

This has been edited and does confirm Nic's data (thanks Nic!):

To determine equilivent resistance of lightbulb:
Power = (voltage * voltage) / resistance
therefore
resistance = (voltage * voltage) / Power
which gives me
resistance = (12 * 12) / 90 = 1.6 ohms

Now, to drop the power dissapated by the light bulb and the current consumption, put a resistor in series with the light bulb. To calculate, use the following:
current = sqrt( power wanted / resistance of bulb) =
current = sqrt( 45watts / 1.6ohms) = 5.3amps
Now, total resistance = total voltage / current =
which gives me:
resistor + 1.6 = 12v / 5.3amps -> resistor = (12 / 5.3) - 1.6 = .664ohms

Power of resistor:
Power = current * current * resistor =
power = 5.3 * 5.3 * .664 = 18.65 watts

NOW, a 20 watt resistor is still kinda big, so it would be better to use multipe 10 watt resistors. 6 4ohm 10watt resistors in parallel will give you 0.667 ohm and 60 watts of power handeling, which should work. (To calculate other parallel resistances, use the formula:
resistance of paralled resistors = 1 / [(1 / resistor) * number of resistors in parallel]
power of paralled resistors = power of resistor * number of resistors in parallel

However, I don't know that this will all work the way you are thinking. The color of the light will likely change, it will be much dimmer, etc.

What are you trying to do? Dim the bulb? Why?

The above has been edited and corrected, as my previous calculations were flawed as Nic graciously pointed out.

ColdPhreze

**Das** · 11-16-2004, 04:01 PM

Would it be feasible to make your own power supply? Seeing as hower the power supplied to a car is already DC? Whats not to say you make your own. This is a new concept! to me... Has anyone ever attempted anything/seen anything like this?
Hrm my truck runs @ 12 volts with a 105A alternator..

**ColdPhreze** · 11-16-2004, 04:22 PM

Das, what exactly do you mean by make your own power supply? How does this have refrence to this thread? Slightly confused by your question. If you are referrring to making a DC-DC power supply for a carpc, then goto www.mastero.tk. It's a DIY carpc power supply for pcs.

ColdPhreze

**krazE** · 11-16-2004, 04:49 PM

Coldphreze, thank you so much!

To answer your questions, yes I want to dim the light bulb. Why? Because I have a motorcycle that runs 1 headlight during the day and it looks like the other one is out, when in actuality it's for high beam. To offset this, I am going to use a yellow light bulb on the high beam but at a lower wattage (so I don't blind people, like you all

)

**krazE** · 11-16-2004, 05:07 PM

So just to clarify, I go to RadioShack and purchase (9) 15ohm 10watt resistors and solder those to the power line of the bulb (it has 1 ground and 1 positive)? Thanks!

**Superduck** · 11-17-2004, 01:14 PM

Is it a CBR by chance? The manager at my work has an F4i and was complaining about the same thing. I think he installed a couple of white LED's inside the the housing, so he didn't have to mess with the high beam.

Anyway, I guess you're going to run an extra wire to the high beam, so you'll still be able to use it as a high beam? I'm thinking you'll be glad you did if you're on a dark road at night.

Doesn't anyone make a replacement housing that's got a high/low on that side, so you can switch it?

Cheers,

Kris

**ColdPhreze** · 11-17-2004, 01:54 PM

That is correct. Also, those resistors can go on either side of the bulb:

power from bike-----/\/\/\/Resistors-----Light(+) Light(-)-----ground

or

power from bike-----Light(+) Light(-)----/\/\/\/Resistors-----ground

Just for all you perfectionists, please realize I know halogen and most other lights don't have a positive and negative side, I am simply referring to how it is wired in the bike, which side the positive line comes in and which side of the light is connected to ground.

ColdPhreze

**Nic** · 11-18-2004, 10:50 PM

Originally Posted by ColdPhreze

You will use a power resistor, and calculate the value using some standard electronics formulas:

To determine equilivent resistance of lightbulb:
Power = (voltage * voltage) / resistance
therefore
resistance = (voltage * voltage) / Power
which gives me
resistance = (12 * 12) / 90 = 1.6 ohms

Now, to drop the power dissapated by the light bulb and the current consumption, put a resistor in series with the light bulb. To calculate, use the following:
resistor = [(volts * volts) / power] - above calculated resistance of light bulb
so you get:
resistor = (12 * 12 / power you want the light to dissapate) - 1.6 = 1.6ohms

NOW, a 45 watt resistor is HUGH, so it would be better to use multipe 10 watt resistors. 9 15ohm 10watt resistors in parallel will give you 1.667 ohm and 90 watts of power handeling, which should work. (To calculate other parallel resistances, use the formula:
resistance of paralled resistors = 1 / [(1 / resistor) * number of resistors in parallel]
power of paralled resistors = power of resistor * number of resistors in parallel

However, I don't know that this will all work the way you are thinking. The color of the light will likely change, it will be much dimmer, etc.

What are you trying to do? Dim the bulb? Why?

ColdPhreze

if you add a 1.6 ohm resistor, the 2 resistors put together (one being the light globe) will use 45watts, both resistors being equal means that they each are using 22.5watts each, so you are efectivley reducing the power output of the light globe to 22.5 watts and dissapating 22.5 watts in the resistor