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Series motor resistance for 800 rpm

Short answer

Constant-torque load keeps armature current at 100 A. Back emf at 1200 rpm: 220 − 100×(0.2+0.05) = 195 V. At 800 rpm: 195 × (800/1200) = 130 V. Total circuit resistance needed: (220−130)/100 = 0.9 Ω. Extra resistance = 0.9 − 0.25 = 0.65 Ω.

A 220 V, DC series motor draws 100 A current and runs at 1200 RPM. What is the value of armature resistance required to run the motor at 800 RPM keeping load torque constant. Given armature resistance = 0.2 ohm, field winding = 0.05 ohm.

For a series motor the flux is proportional to the armature current (below saturation). At constant load torque, T ∝ Φ I_a ∝ I_a², so the current stays the same (100 A) after the speed is reduced, and the flux is unchanged too.

Equivalent circuit

 +--[Ra=0.2]--[Rse=0.05]--[R extra]--(Eb)--+
 |            I = 100 A                     |
 +---------------- V = 220 V ---------------+

Step 1: back emf at 1200 rpm

E_b1 = V − I(R_a + R_se) = 220 − 100 × (0.2 + 0.05) = 220 − 25 = 195 V

Step 2: back emf at 800 rpm

E_b ∝ Φ N, and Φ is the same (current is same), so E_b2 = E_b1 × (N₂/N₁) = 195 × (800/1200) = 130 V

Step 3: total resistance needed

V = E_b2 + I (R_a + R_se + R)

220 = 130 + 100 × (0.25 + R)

0.25 + R = 90/100 = 0.9 Ω

R = 0.9 − 0.25 = 0.65 Ω

ConditionSpeed (rpm)E_b (V)Total series R (Ω)
Normal12001950.25
Reduced8001300.90

Answer: an extra resistance of 0.65 Ω must be connected in series with the armature.

✓ Answer checked by Claude Opus 5.5 (high effort)

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