A 100V, 50A and 1200 rpm permanent magnet DC motor has an armature resistance of 0.2 ohms.

a) determine starting resistance= 100V/0.2 = 500 A

b) determine the external resistance that needs to be inserted in the armature to limit the starting current to twice the rated current

Rst = ((VL-EC)/(IR*M))- Ra= ((100-0)/(50*2))-.2= .8ohm (Don't know if I did this right)

c) determine the counter emf when the motor speed reaches 500 rpm

so at 1200 rpm the voltage is 100 V and 50 A. So at 500 rpm the voltage is 41.6 V and 20.8 A. Is this the correct way for finding my values for my formula?

Ec= VL-IA(Ra+Rs) = 41.6V-20.8(.2)= 33.28 V

Any help would be great