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Series motor speed with 6 ohm series resistance

Short answer

With same current, flux remains constant. At 1000 rpm, Eb = 230 V; with added 6 Ω total resistance 7 Ω, Eb = 110 V. Speed = 478.3 rpm.

A DC series motor of resistance 1 Ω1\ \Omega between terminals runs at 1,000 RPM at 250 V with a current of 20 A. Find the speed at which it will run when connected in series with a 6 Ω6\ \Omega resistance and taking the same current at the same supply voltage.

The current is the same in both cases, so the flux is the same (Φ∝I\Phi\propto I). Hence N∝EbN\propto E_b.

Case 1: 1000 rpm, 250 V, 20 A, R=1 ΩR = 1\ \Omega (armature plus series field).

Eb1=V−IR=250−20(1)=230 VE_{b1} = V - IR = 250 - 20(1) = 230\ \text{V}

Case 2: extra series resistance 6 Ω, so total R=1+6=7 ΩR = 1 + 6 = 7\ \Omega and the same I=20I = 20 A.

Eb2=250−20(7)=110 VE_{b2} = 250 - 20(7) = 110\ \text{V} N2=N1Eb2Eb1=1000×110230=478.3 rpmN_2 = N_1\frac{E_{b2}}{E_{b1}} = 1000\times\frac{110}{230} = 478.3\ \text{rpm}

Answer: the motor runs at 478.3 rpm478.3\ \text{rpm}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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