Question

A 50 hp, 250 V, 1200 rpm DC shunt motor with compensanting winding has armature resistance (including brushes and compensating windings and interpoles) of 0.06 ω. Its field circuit has a total resistence Rₐⱼ+R of 50Ω, which produces a no-load speed of 1200 r.p.m. There are 1200 turns per pole on the shunt field winding.

a). Find the speed of this motor when its input current is 100 A

Answer 1

To find the speed of the motor when the input current is 100 A, we can use the principle of electromechanical energy conversion. For a DC motor, the speed is approximately proportional to the back emf, which is given by the formula E = V - I*R, where E is the back emf, V is the applied voltage, I is the current, and R is the resistance.

Step-by-step explanation:

To find the speed of the motor when the input current is 100 A, we can use the principle of electromechanical energy conversion.

For a DC motor, the speed is approximately proportional to the back emf, which is given by the formula E = V - I*R, where E is the back emf, V is the applied voltage, I is the current, and R is the resistance.

In this case, the back emf when the motor is running at no-load speed is 0 since it is not turning. So when the input current is 100 A, the back emf can be calculated as 0.06 * 100 = 6 V. The speed of the motor can be found using the relationship between the back emf and the speed at no-load, which is given by the formula E1 / E2 = N1 / N2, where E1 and N1 are the back emf and speed at no-load, and E2 and N2 are the back emf and speed at the desired current.