Question

An automatic starting circuit was designed for a parallel-current DC motor with the following rated operating modes:
power 15hp, voltage 240V and current 60A. The motor armature resistance is 0.15Ω and the field resistance is 40Ω. The motor must start at no more than 250% of rated current and as soon as the current falls to its rated value a portion of the starting resistor must be disconnected. Find out how many segments we need to divide the starting resistor and what the value of each segment will be.

Answer 1

The first thing to understand here is that the starting resistance comprises of segments Armature resistance and a starting external resistance. So here to limit the current going into the motor, we shall have to find starting resistance and then,

Starting resistance = External added resistance + Amature resistance.

So we have that the Input current 60A . (At full load)

Field current will be

Which Means Full load Armature current will be 60A-6A = 54A

The 250% of this above current is

Now, starting resistance that needs to be in circuit is,

This starting resistance is equal to Amature resistance plus the externally added resistance

SO the two segments of the Starting resistance would be,