Question

A 100 hp, 250 V, 350 A shunt dc motor with an armature resistance of 0.05 ohms. To limit maximum starting current to twice the rated of its value, what will be the number of stages of starting resistances?

(a) 3

(b) 2

(c) 4

(d) 5

I had been asked this question in exam.

Origin of the question is DC Motor Starting in section DC Motors of Electrical Machines

Answer 1

Right answer is (a) 3

For explanation: n = log(Ra/Rt)/log(Imin/Imax); Rt = Vt/Imax = 250/700 = 0.357 ohms

              = log(0.05/0.357)/log(350/700)

     = 2.84

                          = 3 (approx).