Question

200 volt dc shunt motor delivers an output of 17 kw with an input of 20 kw.The field winding resistance is 50 ohm and armature resistance is 0.04 ohm.Maximum efficiency will be obtained when the total armature copper losses are equal

Answer 1

total loss = input - output

= 20kW - 17kW = 3kW = 3000 W

current through shunt, Is = V/R = 200/50 = 4A,

input , power = VI

or, 200I = 20kW = 20,000

or, I = 100A

for shunt motor, Ia = I - Is

= 100A - 4A = 96A

copper losses = Ia²Ra

here, Ra = 0.04Ω

so, copper losses = (96)² × 0.04 = 368.64W

so, constant losses = total losses - copper losses

= 3000W - 368.64W

= 2631.36W

hence, maximum efficiency will be obtained when total armature copper losses are equal to 2632 W