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Shunt motor speed with added resistance

Short answer

Field current constant, flux constant, torque ∝ I_a. Initial I_a=18.18A, E_b=198.545V. After 30% torque rise, I_a=23.64A, total armature resistance 0.10Ω, E_b=197.636V. New speed = 1500 × (197.636/198.545) ≈ 1493 rpm.

A dc shunt motor is supplied by a source of 200 V. It draws a current of 20 A and runs at speed of 1500 rpm. The armature and field winding resistance are 0.08 Ω and 110 Ω respectively. A resistance of 0.02 Ω is added in series with armature and load torque is increased by 30%, calculate new speed.

In a shunt motor the field is connected across the supply, so flux Φ is constant. Torque ∝ I_a, and speed N ∝ E_b.

Equivalent circuit

 +--[Ra=0.08]--[Rext=0.02]--(Eb)--+
 |         Ia                      |
 +---- Rf = 110 ohm ---------------+--- V = 200 V

Initial condition

  • I_sh = 200 / 110 = 1.818 A
  • I_a1 = 20 − 1.818 = 18.18 A
  • E_b1 = V − I_a1R_a = 200 − 18.18 × 0.08 = 198.545 V
  • N₁ = 1500 rpm

New condition

  • Torque rises by 30 % with Φ constant, so I_a2 = 1.3 × 18.18 = 23.64 A
  • Total armature circuit resistance = 0.08 + 0.02 = 0.10 Ω
  • E_b2 = 200 − 23.64 × 0.10 = 197.636 V

New speed

N₂ / N₁ = E_b2 / E_b1 (Φ unchanged)

N₂ = 1500 × 197.636 / 198.545 = 1493.1 rpm

QuantityBeforeAfter
I_a (A)18.1823.64
R in armature (Ω)0.080.10
E_b (V)198.545197.636
N (rpm)15001493.1

Answer: new speed ≈ 1493 rpm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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