A car starter circuit pulls around 500A to start a car.
Does this mean the resistance of this circuit is around 12 / 500 = 0.024 ohms? (Assuming the battery is ideal)
\$\begingroup\$
\$\endgroup\$
Add a comment
|
2 Answers
\$\begingroup\$
\$\endgroup\$
All electrical motors have a high starting current momentarily which is several times higher than the normal running current. This is due to the fact that when a motor is switched on the total applied voltage drives heavy current through the winding depending on its impedance which is quite low. As the motor picks up speed it generates a counter EMF (voltage) which is in opposition to the applied voltage and the value of current (I)drops quickly (I= Applied voltage- counter Voltage/Impedance). Therefore it is seen that if the motor does not pickup required speed quickly the insulation of its winding melts away due to high heat generated by high current and a short circuit occurs which is generally referred to as motor burning in common parlance. Your calculation is perfectly correct. Starting motors (Starters) of cars are designed for heavy current as the load they have to drive is quite high but only for a very short duration. This heavy current is drawn from car battery only for a short duration. If the engine does not start in a short duration we stop the starter for giving rest to the battery. A week battery fails to operate the starter due to its inability to supply requisite current.
\$\begingroup\$
\$\endgroup\$
I know that starters draw a lot of current, don't know if it is indeed 500 A, but yes, that does mean that the circuit impedance would be 0.024 ohm.
Keep in mind that this is not the resistance of the wire or anything - a lot of this resistance comes from the counter-EMF created in the motor winding.
