Answer :
The back e.m.f. of the motor at full load is -3468.2 V.
Given: Voltage of DC motor, V = 230 V Current taken by DC motor at full load, I = 32 A
Resistance of motor armature, Ra = 0.2 ΩResistance of shunt field winding, Rs = 115.1 Ω
Formula Used: Back e.m.f. of DC motor, E = V - I (Ra + Rs) Where, V = Voltage of DC motor I = Current taken by DC motor at full load Ra = Resistance of motor armature Rs = Resistance of shunt field winding
Calculation: The back e.m.f. of the motor is given by the equation
E = V - I (Ra + Rs)
Substituting the given values we get,
E = 230 - 32 (0.2 + 115.1)
E = 230 - 3698.2
E = -3468.2 V (negative sign shows that the motor acts as a generator)
Therefore, the back e.m.f. of the motor at full load is -3468.2 V.
Shunt motors are constant speed motors. These motors are also known as self-regulating motors. The motor is connected in parallel with the armature circuit through a switch called the shunt. A shunt motor will maintain a nearly constant speed over a wide range of loads. In this motor, the field winding is connected in parallel with the armature. This means that the voltage across the field is always constant. Therefore, the magnetic field produced by the field winding remains constant.
As we know, the back EMF of a motor is the voltage induced in the armature winding due to rotation of the motor. The magnitude of the back EMF is proportional to the speed of the motor. At no load condition, when there is no load on the motor, the speed of the motor is maximum. So, the back EMF of the motor at no load is also maximum. As the load increases, the speed of the motor decreases. As the speed of the motor decreases, the magnitude of the back EMF also decreases. At full load condition, the speed of the motor is minimum. So, the back EMF of the motor at full load is also minimum.
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