Skip to main content

Short shunt compound generator, 30 A

Short answer

Series drop = 0.9 V, Vsh = 220.9 V, Ish = 1.1045 A, Ia = 31.10 A. Total drop including 2 V brush drop gives E = 224.46 V, Ia = 31.10 A.

A short shunt compound generator delivers a load current of 30 A at 220 V and has armature, series and shunt field resistances of 0.05 Ω, 0.03 Ω and 200 Ω respectively. Calculate the induced emf and armature current. Allow 1 V per brush contact drop.

In a short shunt compound generator the shunt field is across the armature only, and the series field carries the load current.

Equivalent circuit

 Ia --[Ra=0.05]--+--[Rse=0.03]--+--> IL=30 A  V=220 V
 (E)              |             |
                 Rsh=200 ohm (across armature)

Calculation

  1. Series field drop = I_L R_se = 30 × 0.03 = 0.9 V
  2. Voltage across shunt (and armature terminals): V_sh = 220 + 0.9 = 220.9 V
  3. Shunt current: I_sh = 220.9 / 200 = 1.1045 A
  4. Armature current: I_a = I_L + I_sh = 30 + 1.1045 = 31.10 A
  5. Armature drop: I_aR_a = 31.1045 × 0.05 = 1.555 V
  6. Brush drop = 2 × 1 V = 2 V (two brushes)
  7. Induced emf: E = V_sh + I_aR_a + brush drop = 220.9 + 1.555 + 2 = 224.46 V
QuantityValue
Series drop0.9 V
Shunt voltage220.9 V
I_sh1.1045 A
I_a31.10 A
E224.46 V

Answer: armature current = 31.10 A and induced emf = 224.46 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Machine chapter 3 questions →