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Shunt motor: resistance for 1200 rpm

Short answer

Field unchanged, flux constant. At constant load torque, armature current remains 37.8A. E_b at 1400 rpm = 219.244V, at 1200 rpm = 187.923V. Required series resistance R = (220–187.923)/37.8 – 0.02 ≈ 0.83Ω.

A 220 V dc shunt motor draws a current of 40 A at full load and runs with speed of 1400 rpm. Calculate the value of resistance required to be inserted in the armature circuit so that speed drops to 1200 rpm at constant load. Given that Ra = 0.02 ohm and Rf = 100 ohms.

At constant load torque and constant flux (shunt field), the armature current remains unchanged, and the speed falls by the reduction of back emf.

Equivalent circuit

 +--[Ra=0.02]--[R ?]--(Eb)--+
 |                           |
 +---- Rf = 100 ohm ---------+--- V = 220 V

Initial state

  • I_sh = 220 / 100 = 2.2 A
  • I_a = 40 − 2.2 = 37.8 A
  • E_b1 = 220 − 37.8 × 0.02 = 219.244 V at 1400 rpm

New state (1200 rpm, same I_a)

E_b2 = 219.244 × (1200 / 1400) = 187.923 V

Resistance

220 = 187.923 + 37.8 (0.02 + R)

0.02 + R = 32.077 / 37.8 = 0.8486 Ω

R = 0.8486 − 0.02 = 0.829 Ω

Answer: about 0.83 Ω must be inserted in series with the armature.

✓ Answer checked by Claude Opus 5.5 (high effort)

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