A dc shunt motor (50hp, 250V) is connected to a 230V supply and delivers power to a load drawing an armature current of 200A and running at a speed of 1200rpm, given Ra=0.2ohm and the rotational losses = 500W.

Find the value of load torque.

I got 2 different answer with two method. But actually which one is correct?

Method 1
Total mechanical power developed
= Output power + Rotational losses Power
= (50x746) + 500
= 37.8kW

Torque = 9.55x38.8k/ 1200 = 300.83Nm

Method 2
Eb = Vt - IRa = 230 - 200(0.2) = 190V

Power developed by motor = Eb x Ia = 190x200=38kW
Torque = 9.55x38k / 1200 = 302.42Nm

 

The half percent difference in the two estimates is less than real-world variation.