# why does headlights dim when a starter motor is turned on

okay first of all, what exactly does the starter motor do? does it increase the current flow and decrease voltage? and does starter motor have its own internal resistance? i understand how there is a voltage drop in the battery terminals, so does the starter motor lower the potential difference to become less than the voltage drop?

so for example if the voltage is 12V and the voltage drop is 2V. The potential difference is now 10V. what is the voltage when the starter motor is used? does it become less than 10V? but then after the engine starts the starter motor turns off and the lights return to normal. So is normal 10V?

Basically, a "12 V" battery is not a perfect voltage source and the starter is a large load.

Due to the very large current (100 A or more) a starter motor can draw, it causes the battery voltage to sag a bit. Dropping 2 V or so during the brief period the starter motor is on would not be out of line. You can think of the battery as being a perfect voltage source with some resistance in series with it. If the battery output voltage drops 2 V when 100 A is being drawn, then this internal resistance is 20 mΩ.

If you have ordinary incandescent headlights that are just directly tied to the battery voltage, then they will dim because they are getting less power at 10 V than at 12 V. Look carefully and you may actually see them get even brighter than before starting once the engine is running. That is because now the alternator is making a higher voltage to charge the battery. The normal "12 V" bus level when the engine is running is usually around 13.6 V.

• I have edited my answer in the light of the calculation you did: you are very welcome to take my last paragraph and put it in your answer and I shall delete it from mine: it more belongs in your answer because I was too lazy to do it when I should have. I was also a little surprised: my gut feeling was that the voltage drop should be much more. BTW: the magic number 13.6V - I kenned it as soon as I saw it having long forgotten it: is this the electrochemical potential difference? I'm having trouble tallying it as a whole number of cells described on the lead acid battery Wiki page. Aug 28 '13 at 1:15
• There are typically 6cells in a "12V" battery. Open circuit voltage at typical temperatures is about 2.1V charging voltage is around 2.25-2.28V hence 6*cells = 13.6V Aug 28 '13 at 1:32

A DC motor uses current to produce torque using essentially the moment of the Lorentz force $q \mathbf{v}\wedge \mathbf{B}$ on the conductors. When it is stationary, there is almost no resistance: if driven by a source with low source resistance, there is theoretically infinite current (in practice very big: IIRC a car's starter motor draws in the neighbourhood of 500A) and huge torques as a result. The current only decreases as the motor's rotational speed increases: the conductors moving through magnetic fields produce back voltages in accordance with Faraday's law: thus the incoming current does work against the opposing voltage as the motor's output torque $\tau$ does work at a rate of $\tau \omega$, where $\omega$ is the rotation speed (the works of course are equal). Indeed if you work out the relevant equations grounded on Faraday's law and plot torque of a series-connected DC motor (stator and rotor windings in series) you get $\tau \propto I \propto \omega^{-1}$, where $I$ is the current flowing. This torque inversely proportional to speed is ideal for "getting things going", which is why DC motors are so useful for lugging trains with. Or for starting cars with: the compression in an internal combustion engine means that you need big torques for starting, thus the need for huge currents.

The upshot of all this is that when you start the car, you're dropping an almost dead short circuit across the battery - at least for a fleeting moment before the rotor begins to spin. This is what you want: it yields the immense current you need to start the car with. But it also means that, for a short moment, the potential difference across the motor is almost nought. This means that everything else switched on is almost shorted out as the near-zero resistance motor draws almost all the current.

The potential difference across the battery output equals that through the whole circuit it supplies. Any potential drops below the battery's open circuit potential difference are owing to source resistance. Look up Thévenin equivalent circuits and source impedance. Car batteries are designed to have very low source impedance - a tiny fraction of an ohm - because this is what limits the all important current to the starter motor.

Edit to be read together with Olin's answer: The following expands on Olin's answer below to show in pictures what he has done - I am only doing this now because he has put me to shame by showing what I should have done for you in the first place: Below is a circuit diagram (taken from the Wiki page for source impedance) of your battery with the total "load" $Z_L$ (it is made up of everything in the car switched on wired in parallel with the starter motor) and the battery, which here is comprises its Thévenin equivalent of an "ideal" electrochemical cell $V_S$ (an infinitely big one so that electrolysis products negligibly hinder current flow) together with an in-series source resistance $Z_S$. If you do the calculation Olin suggests, the voltage you measure at the battery terminals is $V_L$, i.e. the 12V less the current times the source resistance. I didn't quite understand at first your 10V - my gut feel was that it should be lower - but Olin's calculation makes it quite realistic, although I thought that many starter motors draw more than 100A.

• I have changed my answer in the light of Olin's - he shows that the 10V is quite realistic so I now "understand it". Aug 28 '13 at 1:12