An electric motor works from an accumulator battery with an e.m.f. of 12 V With the rotor stalled the current in the circuit is 10 A. What is the motor's power at nominal load, if the respective current is 3 A?

Answer

25W

Explanation

The current in the circuit is `l=(epsi-epsi_("ind"))//R,` where g is the e.m.f. of the accumulator battery, R is the resistance of the circuit (including the internal resistance of the battery) and `e_("ind")` is the e.m.f. induced in the armature in the course of its rotation. When the armature is stopped `e_("ind")=0` and the current is `l_(0)=epsi//R`, from which we find the resistance of the circuit. The power of the motor is `P I epsi-I^2R=I epsi (1-I/I_(0))`.