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4-pole shunt generator, 3 kW at 120 V

Short answer

Load current 25 A, field 2 A, armature current Ia=27 A. Wave winding (A=2): conductor current=13.5 A. Generated EMF=120+27×0.2+2=127.4 V.

A 4-pole dc shunt generator has wave wound armature. The armature and field winding resistance are 0.2 Ω and 60 Ω respectively. The brush contact drop is 1 V per brush. The generator is delivering a power of 3 kW at 120 V. Calculate: i) Total armature current coming out from the brush ii) Current in each armature conductor iii) Generated EMF (E)

Load current is found from P/V, then field current is added to get the armature current. A wave winding has A = 2 parallel paths whatever the number of poles.

Equivalent circuit

        Ia=27A      Ra=0.2
   +----->---[====]---(E)---+---->--- IL=25A ---+
   |                        |                   |
   |   brush drop 2 x 1 V   |                [LOAD]  V=120 V
   |                        |                   |
   +---> If=2A ---[Rsh=60]--+-------------------+

i) Armature current

  • I_L = 3000 / 120 = 25 A
  • I_sh = 120 / 60 = 2 A
  • I_a = I_L + I_sh = 25 + 2 = 27 A

ii) Current in each armature conductor

Wave winding: A = 2 parallel paths.

I_conductor = I_a / A = 27 / 2 = 13.5 A

iii) Generated EMF

E = V + I_a R_a + brush drop = 120 + (27 × 0.2) + (2 × 1) = 120 + 5.4 + 2 = 127.4 V

QuantityValue
Load current25 A
Field current2 A
Armature current27 A
Armature drop5.4 V
Brush drop (2 brushes)2 V
Generated emf127.4 V

Answer: I_a = 27 A, conductor current = 13.5 A, E = 127.4 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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