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Series resistance to halve series motor speed

Short answer

Torque is proportional to I², so halving torque gives I2=40/√2=28.28 A. With Eb1=242 V and N2/N1=0.5, Eb2=85.56 V. Required series resistance R=(250−85.56)/28.28−0.2≈5.61 Ω.

A dc series motor running at 600 rpm on 250 V mains draws a current of 40 A. The total resistance of armature and series field of the machine is 0.20 ohm. Calculate the value of resistance connected in series with the machine which will reduce the speed to 300 rpm, assume the load torque being half of the previous value.

For a series motor (unsaturated), ϕ∝I\phi \propto I, so T∝I2T \propto I^2 and N∝Eb/IN \propto E_b / I.

Given: V=250V = 250 V, N1=600N_1 = 600 rpm, I1=40I_1 = 40 A, Ra+Rse=0.2 ΩR_a + R_{se} = 0.2\ \Omega, N2=300N_2 = 300 rpm, T2=0.5 T1T_2 = 0.5\,T_1.

Step 1: New current

T2T1=(I2I1)2=0.5  ⇒  I2=402=28.28 A\frac{T_2}{T_1} = \left(\frac{I_2}{I_1}\right)^2 = 0.5 \;\Rightarrow\; I_2 = \frac{40}{\sqrt 2} = 28.28\ \text{A}

Step 2: Back emfs

Eb1=250−40×0.2=242 VE_{b1} = 250 - 40 \times 0.2 = 242\ \text{V} N2N1=Eb2Eb1×I1I2Eb2=Eb1×N2N1×I2I1=242×300600×28.2840=85.56 V\begin{aligned} \frac{N_2}{N_1} &= \frac{E_{b2}}{E_{b1}} \times \frac{I_1}{I_2} \\ E_{b2} &= E_{b1} \times \frac{N_2}{N_1} \times \frac{I_2}{I_1} \\ &= 242 \times \frac{300}{600} \times \frac{28.28}{40} = 85.56\ \text{V} \end{aligned}

Step 3: Series resistance

Eb2=V−I2(0.2+R)0.2+R=250−85.5628.28=5.814 ΩR=5.814−0.2=5.61 Ω\begin{aligned} E_{b2} &= V - I_2 (0.2 + R) \\ 0.2 + R &= \frac{250 - 85.56}{28.28} = 5.814\ \Omega \\ R &= 5.814 - 0.2 = 5.61\ \Omega \end{aligned}

Answer: Extra series resistance R≈5.61 ΩR \approx 5.61\ \Omega (current falls to 28.28 A).

✓ Answer checked by Claude Opus 5.5 (high effort)

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