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Short shunt compound generator, 80 A

Short answer

For a short shunt generator, the series field drop (80×0.05 = 4 V) makes shunt field voltage 224 V. Shunt current is 2.24 A, armature current 82.24 A. Armature drop 82.24×0.1 = 8.224 V; generated emf E = 224 + 8.224 = 232.2 V.

A short shunt compound generator delivers a current of 80 A to the load at 220 V. The shunt field, series and armature winding resistances are 100 Ω, 0.05 Ω and 0.1 Ω respectively. Calculate the emf generated by the armature.

For a short shunt compound generator, the shunt field is connected across the armature and the series field is in series with the load. No brush drop is given, so it is neglected.

Equivalent circuit

 Ia --[Ra=0.1]--+--[Rse=0.05]--+--> IL=80 A  V=220 V
 (E)             |             |
               Rsh=100 ohm (across armature)

Calculation

  1. Drop in series field: I_L R_se = 80 × 0.05 = 4 V
  2. Voltage across shunt field: V_sh = 220 + 4 = 224 V
  3. Shunt field current: I_sh = 224 / 100 = 2.24 A
  4. Armature current: I_a = I_L + I_sh = 80 + 2.24 = 82.24 A
  5. Armature drop: I_aR_a = 82.24 × 0.1 = 8.224 V
  6. Generated emf: E = V_sh + I_aR_a = 224 + 8.224 = 232.22 V

Answer: emf generated by the armature = 232.2 V (with I_a = 82.24 A). If a brush drop of 1 V per brush were included, E would be 234.2 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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