I seem to be expressing myself unclearly or something!Originally posted by drivesafe
1) John, what does the voltage of the battery remaining at 12 volts got to do with the current load being caused by the filaments.
2) Again, using the starter motor as an example. Your vehicle’s battery drops to 8 and in some situations as low as 6 volts while cranking so of cause the battery is not going to remain at 12 volts. As I have already pointed out, the start up current draw is so high that it looks like a dead short at the alternator, when the power is taken directly at the alternator instead of at the battery. A dead short is a low voltage event.
3) The battery still acts as a current supply during the initial start up of the globes and as such cushions the high current draw before it gets to the alternator.
4) No matter what the voltage is or drops to, there is still going to be a huge current draw that is way WAY above what the alternator can produce. This high current draw means there is going to be a voltage drop and this is not what the alternator is designed to work against, it’s designed to try and maintain the voltage and it can’t do that when the current draw is far beyond it’s maximum capability, whereas the battery is specifically designed to caters for high current drains and the very reason everything is connected to the battery and not the alternator.
5) And once again, if this was a better way to wire a vehicle, I’m sure the makers would be doing so.
Cheers.
But looking at your points. (I'll do my best, but you try thinking straight with a five year old and a seven year old demanding explanations of what you are writing - they just got up!)
1) What I was trying to say is that the battery's electrochemical reaction in the battery still produces the same voltage even though the voltage at the terminals is a lot lower due to internal resistance. But its really irrelevant to my point that the startup current is a lot lower per filament with multiple lights because the battery voltage is lower.
2) A dead short is not an exact term, but implies that the resistance of the device is low compared to the wiring resistance. I have just measured the cold resistance of a bulb - the convenient one to measure was a 20w 24v bulb, which has a cold resistance of 3.5 Ohms. This implies a 100w 12v bulb will have a cold resistance of 0.35 Ohms, and an initial current of 35A at 12v. Even six of these would still only be 200A, and in practice less as the available voltage at the filament due to lowering of the battery voltage and resistance of the wiring would be probably eight volts at best, giving perhaps 150amps surge current. This is hardly a dead short, and in fact if your (single 100w lamp considered for clarity) circuit wiring resistance is of the order of 0.35 Ohms the voltage drop at normal current of over eight amps will be close to 3 volts - completely unacceptable - clearly our total wiring resistance must be way under this, so the cold resistance of the filament cannot be considered as a "dead short". If the wiring resistance is low enough for normal operation, then the cold filament resistance must still be high compared to the wiring resistance.
3) Certainly the battery supplies the current surge at startup - but it does this regardless of whether the power feed is taken from the battery or the alternator. The wire between the alternator output is normally the heaviest wire in the system apart from the starter wiring, and the voltage drop along it even at a surge current of several hundred amps will be much less than the drop due to the internal resistance of the battery.
4) As in 3) above - the alternator is not supplying this surge - the battery is, but since the two are connected by a heavy wire with very low resistance the drop across this is marginal and almost the same regardless of which end of the alternator feed the drain is from. The drop in battery voltage is just as large and still appears at the alternator.
5) As I said earlier, the manufacturer's wiring is designed mainly to make manufacture easy, just as long as the results are satisfactory (not best achievable). After all, if the manufacturer's wiring was the best possible, you would not be adding a relay which the manufacturer in their wisdom considered unnecessary.
One point should be noted that although getting the highest practical voltage to the lights will give the brightest light (and even very small increases in voltage make a marked increase in efficiency), operating tungsten filament bulbs at a higher voltage does decrease their life. As against that, note that my 1974 Citroen (which does have relays) still has all its original bulbs, so given good quality bulbs, the life is so long that even operated at a slightly higher voltage, bulb life should not be a problem.



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