need to drop a voltage?

Not sure where to start guys, so here goes......

Im putting a backlit number plate on my vehicle. I have bought a 12v interior strip light which has an 8 watt light in it. My original number plate lights have bulbs rated at 5 watt. What i need to do is put some sort of resistor in series on the positive wire to drop the voltage down from 12v to ???? in order to get a 5 watt output from the light. Does this make any sense? I have tried to obtain a 5 watt strip bulb but they are not available :(. Can any of you guys work out for me what resistor i will need to use or give me a formula in order for me to work it out myself?

Many thanks in advance!!

Formula #1

P = watts, V = volt, I = current

8 = 12 * I

I = .667A

Formual #2

R = resistance, V = vot, I = Current

R = 12/.667

R= 18 Ohms

The filament in your 8W bult should be very close to 18 ohms/

5W calculation:

P = V * I

I = .42A

R=V/I

The higher the resistance, the lower the power output.  To get the bulb to only use 5 watts of energy put a 10 ohm resistor in series with it.  You should use at least a 10 watt resistor. 

Why can't you just run the bult at full brightness?  Using a resistor you will still have an 8 watt load, you'll just only have 5 watts worth of light.  The other 3 watts will be dispersed as heat energy.

I don't believe you need to worry about ohms law formulas although KP presented it very nicely. The thing is that "wattage" in this case is only a rating that tells you how much POWER the bulb will use...it does not in itself tell us anything about the voltage rating of the strip light or that a resistor is actually needed. In other words if you replace a 40watt light bulb in your house with a 100watt bulb it does not require a different voltage or a resistor..it simply is brighter and uses more power (heat)..... all else being equal of course.

If the strip is too bright however then knock yourself out..but if brightness is not a concern and the bulb is a 12volt rated I wouldn't be concerned about any problems.