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Series motor speed with 5 Ω series resistance

Short answer

Back emf without added resistance is 185 V, with 5 Ω it drops to 110 V. Constant current keeps flux unchanged, so speed is proportional to back emf: 800 × 110/185 ≈ 476 rpm.

A 200 V d.c. series motor runs at 800 rpm when taking a line current of 15 A. The armature and field resistances are 0.6 Ω and 0.4 Ω respectively. Find the speed at which it will run when connected in series with a 5 Ω resistance and taking the same current at the same voltage.

For a series motor, the same current means the same flux, so speed is directly proportional to back emf: N2N1=Eb2Eb1\dfrac{N_2}{N_1} = \dfrac{E_{b2}}{E_{b1}}.

Given: V=200V = 200 V, I=15I = 15 A, N1=800N_1 = 800 rpm, Ra=0.6 ΩR_a = 0.6\ \Omega, Rse=0.4 ΩR_{se} = 0.4\ \Omega, extra series resistance 5 Ω.

Back emfs

Eb1=V−I(Ra+Rse)=200−15×1.0=185 VEb2=200−15×(1.0+5)=200−90=110 V\begin{aligned} E_{b1} &= V - I (R_a + R_{se}) = 200 - 15 \times 1.0 = 185\ \text{V} \\ E_{b2} &= 200 - 15 \times (1.0 + 5) = 200 - 90 = 110\ \text{V} \end{aligned}

New speed

N2=800×110185=475.7 rpmN_2 = 800 \times \frac{110}{185} = 475.7\ \text{rpm}

Answer: Speed with 5 Ω in series ≈476\approx 476 rpm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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