Compound generator with 80 lamps

Load current

Load power = 80 × 60 = 4800 W at 220 V (terminal voltage V = 220 V).

I_L = 4800 / 220 = 21.82 A

Brush drop is not given, so it is neglected.

i) Long shunt

 Ia --[Ra=0.3]--[Rse=0.2]--+------> IL=21.82 A
 (E)                       |
                    Rsh=60 (across armature+series)

The shunt field is connected across the load terminals (series field in the armature circuit).

• I_sh = V / R_sh = 220 / 60 = 3.667 A
• I_a = I_se = I_L + I_sh = 21.82 + 3.667 = 25.49 A
• E = V + I_a(R_a + R_se) = 220 + 25.49 × (0.3 + 0.2) = 220 + 12.74 = 232.74 V

ii) Short shunt

 Ia --[Ra=0.3]--+--[Rse=0.2]--+--> IL
 (E)             |            |
                Rsh=60 (across armature only)

The shunt is connected across the armature, and the series field is in the line.

• Voltage drop in series field = I_L R_se = 21.82 × 0.2 = 4.364 V
• Shunt voltage = V_sh = 220 + 4.364 = 224.36 V
• I_sh = 224.36 / 60 = 3.739 A
• I_a = I_L + I_sh = 21.82 + 3.739 = 25.56 A
• E = V_sh + I_aR_a = 224.36 + 25.56 × 0.3 = 224.36 + 7.67 = 232.03 V

Answer: Long shunt: I_a = 25.49 A, E = 232.74 V. Short shunt: I_a = 25.56 A, E = 232.03 V.