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Series motor speed with 6 Ω in series

Short answer

With same current and supply voltage, flux is constant, so speed is proportional to back emf. Eb drops from 230 V to 110 V due to added 6 Ω resistance, giving speed ≈ 478 rpm.

A dc series motor of resistance 1 Ω between terminals runs at 1,000 RPM at 250 V with a current of 20 A. Find the speed at which it will run when connected in series with a 6 Ω resistance and taking the same current at the same supply voltage.

For a series motor with the same current and the same supply, the flux is unchanged (Φ ∝ I), so the speed is proportional to the back emf.

Equivalent circuit

 Case 1: V --[1 ohm (Ra+Rse)]--(Eb1)       I = 20 A
 Case 2: V --[1 ohm]--[6 ohm ext]--(Eb2)   I = 20 A

Case 1

E_b1 = V − I R = 250 − 20 × 1 = 230 V at N₁ = 1000 rpm

Case 2 (extra 6 Ω, same current 20 A)

E_b2 = 250 − 20 × (1 + 6) = 250 − 140 = 110 V

Speed

N₂ / N₁ = E_b2 / E_b1 (same Φ)

N₂ = 1000 × 110 / 230 = 478.3 rpm

Answer: the motor will run at about 478 rpm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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