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Short shunt compound generator, 100 A

Short answer

Series drop = 10 V, Vsh = 260 V, Ish = 2 A, Ia = 102 A. Adding armature drop and 2 V brush drop yields E = 272.2 V, Ia = 102 A.

A short shunt compound generator supplies a load current of 100 A at 250 V. The generator has the following winding resistances: shunt field 130 Ω, armature 0.1 Ω and the series field 0.1 Ω. Find the emf generated and the armature current, if the brush drop is 1 V per brush.

In a short shunt connection the shunt field is across the armature and the series field is in the load line.

Equivalent circuit

 Ia --[Ra=0.1]--+--[Rse=0.1]--+--> IL=100 A  V=250 V
 (E)             |            |
               Rsh=130 ohm (across armature)

Calculation

  1. Series field drop = 100 × 0.1 = 10 V
  2. Voltage across the shunt branch (armature terminals): V_sh = 250 + 10 = 260 V
  3. Shunt current: I_sh = 260 / 130 = 2 A
  4. Armature current: I_a = 100 + 2 = 102 A
  5. Armature drop = 102 × 0.1 = 10.2 V
  6. Brush drop = 2 × 1 = 2 V
  7. EMF generated: E = 260 + 10.2 + 2 = 272.2 V
StepValue
Series drop10 V
V across armature terminals260 V
I_sh2 A
I_a102 A
Armature drop10.2 V
Brush drop2 V
E272.2 V

Answer: E = 272.2 V and I_a = 102 A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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