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Car Forum / Dodge / Dodge Trucks / November 2007headlight bulbs
just had a headlight bulb burn out in my 97 ram 4x4 this weekend. i went ahead and replaced both at the same time as usual. as i was logging the service into my service journal i looked back to see when the last time the bulbs were replaced on this truck. at 275,350 miles and 10+ years of rough service this is the first time i have ever put new headlight bulbs in the truck. i was amazed, the factory bulbs lasted a decade. i decided to look at the service records on my other dodge trucks. the 2001 3/4 ton 4x4 diesel at 160k is still on the factory bulbs. the 1991 dakota that has 177k known miles (had a broken odometer for a while) is on its 2nd set of bulbs. the 1993 4x4 dakota with 220k is on its 5th set of bulbs. i am now curious as to why the 93 goes through so many more bulbs than the others. any theories? michael 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 > just had a headlight bulb burn out in my 97 ram 4x4 this weekend. i went > ahead and replaced both at the same time as usual. as i was logging the [quoted text clipped - 10 lines] > than the others. any theories? > michael 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 > 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. [quoted text clipped - 14 lines] > > than the others. any theories? > > michael > Since Power = Volts x Amperes; an increase in the applied voltage will > result in a proportionate increase in current, "Since Volts = Amperes x Resistance" would make more sense. What you wrote has volts and current on the same side of the equation. > 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. 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. >> Since Power = Volts x Amperes; an increase in the applied voltage will >> result in a proportionate increase in current, > > "Since Volts = Amperes x Resistance" would make more sense. What you wrote > has volts and current on the same side of the equation. 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. >> and an increase in the power that is a *product* of the two. An increase >> in consumed power will cause [quoted text clipped - 3 lines] > is brighter than yellow light! Therefore, we can turn down the juice and > just tint the glass blue. That's a discussion for another day. Mike > >> Since Power = Volts x Amperes; an increase in the applied voltage will > >> result in a proportionate increase in current, [quoted text clipped - 20 lines] > > Mike No, come on Mike, lets hear it. > >> Since Power = Volts x Amperes; an increase in the applied voltage will > >> result in a proportionate increase in current, [quoted text clipped - 20 lines] > > Mike 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: http://www.the12volt.com/ohm/ohmslaw.asp Another way of tracking the consumed power (assuming the resistance of the load doesn't change appreciably) would be to use: P = I^2 x R or P = E^2 / R 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: http://tinyurl.com/2opm7l. If concerned, the OP should check the applied voltage with engine running (above idle speed) to verify it is not significantly higher than it should be. Bryan >>>> Since Power = Volts x Amperes; an increase in the applied >>>> voltage will result in a proportionate increase in current, [quoted text clipped - 5 lines] >> You are both correct, kinda sorta. For the purpose of this >> discussion however, Bryan is uisng the "correct" interpretation of Sorry, he had a 50/50 chance, he picked the wrong equation. >> Mr. Ohm's law. Since the resistance of the bulb's filament is a >> "constant" in this case, the question (variable) is the voltage [quoted text clipped - 13 lines] >> >> That's a discussion for another day. "FYI white or blue light is brighter then yellow." http://groups.google.co.in/group/alt.autos.dodge.trucks/msg/a6a2a718fd5687de >> Mike > > Both of Georg Ohm's expressions are true. I was solving for E > (volts) using P (watts) and I (current/amperes). 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. > Ohm's pie and a nifty calculator is here: > http://www.the12volt.com/ohm/ohmslaw.asp [quoted text clipped - 11 lines] > running (above idle speed) to verify it is not significantly higher > than it should be. 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. > Bryan > > Since Power = Volts x Amperes; an increase in the applied voltage will > > result in a proportionate increase in current, [quoted text clipped - 8 lines] > light is brighter than yellow light! Therefore, we can turn down the > juice and just tint the glass blue. Once a cock sucking whore, always a cock sucking whore. But as usual you can do nothing but stir sh.t and prove yourself to be the a.s that you truly are. Additionally, I never once indicated or implied any such thing f.ck nuts. > just had a headlight bulb burn out in my 97 ram 4x4 this weekend. i went > ahead and replaced both at the same time as usual. as i was logging the [quoted text clipped - 10 lines] > than the others. any theories? > michael 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.  SignatureIf at first you don't succeed, you're not cut out for skydiving |
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