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Shunt motor speed at 30 A armature current

Short answer

For a shunt motor, flux is constant, so speed is proportional to back emf. With armature current 30 A, Eb = 224 V, giving speed = 600 × 224/218.4 ≈ 615 rpm.

A dc shunt motor runs at 600 RPM taking 60 A from a 230 V supply. Armature resistance is 0.2 Ω and field resistance is 115 Ω. Find the speed when the current through the armature is 30 A.

In a shunt motor the flux is constant (field current depends only on V and R_f), so N ∝ E_b.

Equivalent circuit

 +--[Ra=0.2]--(Eb)--+
 |       Ia          |
 +--- Rf = 115 ohm --+--- V = 230 V

First condition

  • I_sh = 230 / 115 = 2 A
  • I_a1 = 60 − 2 = 58 A
  • E_b1 = 230 − 58 × 0.2 = 230 − 11.6 = 218.4 V
  • N₁ = 600 rpm

Second condition (I_a = 30 A)

E_b2 = 230 − 30 × 0.2 = 224 V

Speed

N₂ = N₁ × (E_b2 / E_b1) = 600 × 224 / 218.4 = 615.4 rpm

(The speed rises slightly because the armature drop is smaller at the lighter load.)

Answer: speed ≈ 615 rpm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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