- 2072 Magh
Series motor speed at reduced load
Short answer
At 19 A, speed is about 1277 rpm. At 1 A, theoretical speed is about 25,660 rpm, a dangerously high runaway speed.
A 200 V DC series motor draws full-load line current of 38 A at the rated speed of 600 rpm. The motor has armature resistance of 0.4 Ω and the series field resistance is 0.2 Ω find: i) The speed of the motor when the load current drops to 19 A ii) The speed on removal of load when the motor takes only 1 A from supply
For a series motor, Φ ∝ I (assumed unsaturated), and N ∝ E_b / Φ ∝ E_b / I.
Equivalent circuit
+--[Ra=0.4]--[Rse=0.2]--(Eb)--+
| I |
+------------ V = 200 V --------+
Total resistance R = 0.4 + 0.2 = 0.6 Ω.
Full load (38 A, 600 rpm)
E_b1 = 200 − 38 × 0.6 = 200 − 22.8 = 177.2 V
i) Load current = 19 A
E_b2 = 200 − 19 × 0.6 = 188.6 V
N₂ = N₁ × (E_b2/E_b1) × (I₁/I₂) = 600 × (188.6/177.2) × (38/19)
= 600 × 1.0643 × 2 = 1277 rpm
ii) Current = 1 A (no load)
E_b3 = 200 − 1 × 0.6 = 199.4 V
N₃ = 600 × (199.4/177.2) × (38/1) = 600 × 1.1253 × 38 = ≈ 25 660 rpm
| Current (A) | E_b (V) | Speed (rpm) |
|---|---|---|
| 38 | 177.2 | 600 |
| 19 | 188.6 | 1277 |
| 1 | 199.4 | about 25 660 |
The second result is theoretical: with flux ∝ I, the speed becomes extremely high at light load. In practice, residual flux, saturation, friction and windage limit it, but the speed is still dangerously high. This is why a series motor must never run on no load.
Answer: (i) about 1277 rpm; (ii) about 25 660 rpm (a runaway speed).
✓ Answer checked by Claude Opus 5.5 (high effort)
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