Recent content by Edges_sharp

I am only giving the information present in the question

Thanks very much :)

Yes it is copied and should be ohms

Not sure if my method is right but using Ea=VT-IaRa Ea=300-30*0.25=292.5 Then to calculate field current I did If=Ea/Rf=292.6/60=4.875

Yes I am getting it to be 4.875A

An 8.25 kW shunt DC motor is supplied with a terminal DC voltage of 300 V. The armature resistance of the motor Ra is 0.25 , and the field resistance Rf is 40 . The field winding consists of 1500 turns. An adjustable resistance Radj is connected in series to the field winding. Radj may be...

Hi all, I'm a second year Electrical engineering student. Really looking forward to joining this community! Many Thanks