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In a shunt DC motor, what are the sequence of events when the field current is lost?

From what I understand the loss of the field current, and thus the field itself, will result in a decrease in back EMF on the armature. This will allow a higher current to flow through the armature, and also cause the motor to speed up.

My question is, if the field current is lost through an open circuit and there is no field will the back EMF drop to zero? And how quickly will the armature current thus increase to the limit imposed only by the DC resistance of the wires, matter of seconds or milliseconds?

Thank you.

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    \$\begingroup\$ If rotor is locked , V/DCR limit is reached in L/DCR=T milliseconds. If not locked. then inertia and load controls rise in RPM and BEMF generated and thus current and torque drops until stable RPM is reached in seconds. \$\endgroup\$ – Tony Stewart EE75 Nov 7 '19 at 15:44
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The iron in the field magnet structure will retain some magnetism even without any current in the winding so the field and hence the back-emf will not drop to zero. The actual remaining amount depends upon the specifics of the material and the magnetic structure. Dynamos rely upon this remanence to bootstrap operation at start-up.

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The back-EMF will drop as rapidly as the field current drops. Then the armature will just appear as a resistance in series with an inductance, and the current will rise according to the applied voltage and the armature's L/R time constant.

For most motors that you can carry, this time constant will be on the order of tens or hundreds of microseconds -- I think a motor with an L/R time constant in the seconds would be bigger than a house.

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