In a long-shunt compound generator, the terminal voltage is 230 V when generator delivers 150 A. Determine (i) induced emf (ii) total power generated (iii) distribution of this power. Given that shunt field, series field, divertor and armature resistance are 92 Ω, 0.015 Ω, 0.03 Ω and 0.032 Ω respectively.

I_sh = 230/92 = 2.5 A
I_a = 150 + 2.5 = 152.5 A

Long Shunt DC compound Generator with divertor resistances

Since series field resistance and divertor resistances are in parallel their combined resistance is = 0.03 × 0.015/0.045 = 0.01 Ω

Total armature circuit resistance is = 0.032 + 0.01 = 0.042 Ω
Voltage drop = 152.5 × 0.042 = 6.4 V

(i) Voltage generated by armature, E_g = 230 + 6.4 = 236.4 V

(ii) Total power generated in armature, E_gI_a = 236.4 × 152.5 = 36,051 W

(iii) Power distribution

Power lost in armature IaRa = 152.52 × 0.032 = 744 W
Power lost in series field and divertor = 152.52 × 0.01 = 232 W
Power dissipated in shunt winding = VI_sh = 230 × 0.01 = 575 W
Power delivered to load = 230 × 150 = 34500 W

Total = 36051 W