headlight bulbs

Most likely, higher system voltage. This will be a product of the voltage regulator on the alternator. You are probably over 14 volts when running. Steve

That could be caused by many things. The two primary ones are a loose headlight assembly and / or a voltage problem. If the headlight assembly or the bulb socket is loose, that will cause the bulb to vibrate excessivly which can cause a significant reduction in bulb life. If the regulator is not controlling the voltage properly and either allowing an excessivly high voltage or voltage spikes, that to can cause a significant reduction in bulb life.

Since Power = Volts x Amperes; an increase in the applied voltage will result in a proportionate increase in current, and an increase in the power that is a *product* of the two. An increase in consumed power will cause the filament(s) to burn hotter, and reduce the life of the lamp.

Another factor is the number of on/off cycles. A lamp filament undergoes an extreme change in temperature from off to on, causing a mechanical stress in the filament. These mechanical stresses in the filament decrease the life of the filament.

Bryan

Steve Lusardi wrote:

"Since Volts = Amperes x Resistance" would make more sense. What you wrote has volts and current on the same side of the equation.

consumed power will cause

Punkin, a.a.d.t's resident electrical engineer, taught us that blue light is brighter than yellow light! Therefore, we can turn down the juice and just tint the glass blue.

You are both correct, kinda sorta. For the purpose of this discussion however, Bryan is uisng the "correct" interpretation of Mr. Ohm's law. Since the resistance of the bulb's filament is a "constant" in this case, the question (variable) is the voltage which ultimately determines the current and thus the filament temperature which if too high will result in an early failure.

That's a discussion for another day.

Mike

No, come on Mike, lets hear it.

Both of Georg Ohm's expressions are true. I was solving for E (volts) using P (watts) and I (current/amperes). Ohm's pie and a nifty calculator is here:

The only trouble is, AFIK, no lamp manufacturer specifies lifetime vs power consumption. However, suffice to say it will be less if the specified voltage (and current) is exceeded, and longer if the voltage (and current) is less than specified. There is some mention of lifetime vs applied voltage here:

Sorry, he had a 50/50 chance, he picked the wrong equation.

"FYI white or blue light is brighter then yellow."

That's NOT what you wrote. You were (weren't) showing that an increase in applied voltage will result in a proportionate increase in current.

Some here don't know that the voltage should be checked across the bulb, while the bulb is in the socket and turned on. They'll take out the bulb, measure 12V at the empty socket, and think it's all bitchen.