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When there is a short in the car-starter, the resistance of the winding decreases so the starter consumes more current and the starter cranks slowly, while there is a risk of eventually burning the windings.

When the brushes are worn or there are bad connections along the starter line, the resistance along the line will increase and the starter will also crank slowly. Does the starter consume more current, or or the starter just get a lower voltage (which subtracted by the voltage drops along the bad connections/worn brushes etc.) and createsa lower current that cranks the starter slowly but with no risk of burning the windings (since the current doesn't increase?)

What is correct? What exactly happens in that case?

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When the brushes are worn they create a high resistance (assuming that they are still in contact). This, in addition to more arcing, will produce a lot of heat. Heat is wasted energy. So the windings do not get the amount of current needed to turn the motor as intended, and the brushes continue to heat. The heat along with arcing will create carbon deposits, which further degrade the brush's contact resistance (which further reduces the current to the windings). Eventually, the contact resistance will be high enough that no appreciable current will flow, and the motor will cease to turn.

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  • \$\begingroup\$ So opposed to short in the winding that caused the voltage to drop very low in the battery due to excessive current draw of the starter,shouldn't the voltage of the battery drop just a a little when cranking,since the current draw is lower when the brushes are worn or when there is a bad connection? \$\endgroup\$
    – Gary
    Commented Oct 13, 2021 at 15:30
  • \$\begingroup\$ @Gary Yes, if motor doesn't move due to high resistance of brushes, then the battery voltage will not move much. When there is a shorted winding, then that will both prevent the motor from turning and will cause the battery voltage to sag as a lot of current is being pulled. \$\endgroup\$
    – Aaron
    Commented Oct 13, 2021 at 19:54
  • \$\begingroup\$ You referred to the case that the motor doesn't move at all due to the brush resistance,but if the motor will move,but very slowly due to the brush resistance,will the battery voltage also won't move much? \$\endgroup\$
    – Gary
    Commented Oct 14, 2021 at 7:58
  • \$\begingroup\$ @Gary yes, if the motor moves slowly due to high brush resistance, then the battery voltage will not change much. \$\endgroup\$
    – Aaron
    Commented Oct 14, 2021 at 13:28

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