Need some help 16v to 12v resistor

Hi everone, I am some what new to electronic.

I am bulding a flashlight.
Here is what i know.
Bulb is 35w 12volts 35w/12v=2.93amps The bulb can handle 12.8 max volt.
battery total voltage would be around 16v +/-
So I want to use a resistor to lower the votage to 12.5v or somewhere around there.

Thanks and have a nice day

Personally I would run the bulb on 12 volts but matters not once you apply a formula. You have your current because 35 watts / 12 volts = 2.93 Amps. So if I want to limit my circuit current to 2.93 Amps I get 16 volts - 12 volts = 4 volts / 3.93 amps = 1.01 Ohms so you place a 1 Ohm resistor in series with the bulb. However, remember that resistor will have a current of about 4 Amps and a voltage across it of 4 volts so 4 * 4 = 16 watts or to be safe and not overheat the resistor you would need about 1 Ohm 30 Watt resistor as the resistor will be dissipating 16 watts. Not very practical but that is how it works out.

Ron

That would be 16v - 12v = 4v, divide that by 2.93A = 1.36 ohms or less, given your bulb's voltage range. The current through the bulb will increase if you increase the voltage, so your 2.93 won't be the same. I would try 1.2 ohms for starters, and since it is passing nearly 3A you should be looking at a resistor rated 12W - that is an awful lot of power to be throwing away, you really should be looking at trying to use a 12V battery instead.

An incandescent light bulb produces mostly heat plus a little light. They are obsolete for a flashlight. Use a powerful LED instead.

12v and 35w sounds like you're wanting to make a flashlight using a quartz-halogen lamp with a built-in reflector --- right? Don't forget that those thing get mighty hot and you'll need a ceramic connector for the thing as well as a heat-resistant mounting. Rather than using a resistor to knock your supply voltage down, why not use PWM circuit that'd be a lot more efficient?

If the battery isnt all that big the lamp will 'pull' the voltage down anyways, so you might not need a power wasting resistor.