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electric resistance!!


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Sounds like a homework question, so I can't just tell you the answer. Are you familiar with basic electrical circuits? With Ohm's law? If so, try drawing out the equivalent circuit, both with the starter motor on; and with it off. Don't forget the internal resistance of the battery. What, with respect to the headlights, is the difference between these two?

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To hopefully clarify, what I think npts2020 means is to look at this intuitively and consider that the battery has a limited amount of power it can produce. With and without the starter, how is this power distributed? If you think about it you should be able to come up with the answer easily.

 

If you still have trouble, or need to explain this in a report, you can solve for this mathematically using ohms law per my suggestion; P = VI and V = IR and come to the same conclusion. Let V = 12V for the car battery. There is a resistance in the light; in the battery; and two resistances for the starter (on and off). I'll let you decide appropriate values for these (a google search might be helpful). So calculate the current and voltage (and therefore power) in the light when the starter is on, and when it is off.

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quite simply because the starter uses more juice than the battery and loom can supply at ~13.8V, it`s right at the very limits, that`s why it`s never a good idea to keep trying to turn the engine over on fault.

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  • 3 weeks later...

The power source (car battery) has internal impedance.

What this means to you is that when any load is placed on the battery its voltage drops.

The bigger the load the more it drops. A starter is a very significant load and thus the voltage drop is significant.

Anyway if you are running the starter the engine is not running and the alternator is not trying to hold the battery voltage up.

Even touching the brake lights will dim the headlights when the engine is not running just not as much as the starter.

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  • 2 weeks later...

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