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Series motor speed at half torque

Short answer

For an unsaturated series motor, torque is proportional to the square of armature current, so halving torque reduces current to 70.71 A. The new back emf is 212.93 V, and speed becomes 800 × (212.93/210) × (100/70.71) ≈ 1147 rpm.

A 220 V series motor is running at a speed of 800 rpm and draws 100 A. Calculate at what speed the motor will run developing half the torque. Total resistance of the armature and field is 0.1 Ω0.1\ \Omega. Assume that the magnetic circuit is unsaturated.

For an unsaturated series motor Φ∝I\Phi\propto I, so the torque T∝ΦI∝I2T\propto\Phi I\propto I^2, and the speed N∝Eb/Φ∝Eb/IN\propto E_b/\Phi\propto E_b/I.

Case 1: 800 rpm, I1=100I_1 = 100 A, R=0.1 ΩR = 0.1\ \Omega.

Eb1=220−100(0.1)=210 VE_{b1} = 220 - 100(0.1) = 210\ \text{V}

Case 2: torque T2=T1/2T_2 = T_1/2.

T2T1=(I2I1)2=12  ⇒  I2=1002=70.71 A\frac{T_2}{T_1} = \left(\frac{I_2}{I_1}\right)^2 = \frac12 \;\Rightarrow\; I_2 = \frac{100}{\sqrt{2}} = 70.71\ \text{A} Eb2=220−I2R=220−(70.71)(0.1)=212.93 VE_{b2} = 220 - I_2R = 220 - (70.71)(0.1) = 212.93\ \text{V} N2N1=Eb2Eb1×I1I2  ⇒  N2=800×212.93210×10070.71=1147 rpm\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}}\times\frac{I_1}{I_2} \;\Rightarrow\; N_2 = 800\times\frac{212.93}{210}\times\frac{100}{70.71} = 1147\ \text{rpm}

Answer: the motor runs at about 1147 rpm1147\ \text{rpm} when developing half the torque.

✓ Answer checked by Claude Opus 5.5 (high effort)

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