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Shunt motor speed with 0.1 Ω armature resistance, less load

Short answer

Fixed field current gives constant flux. Initial back emf 238.8 V; with extra 0.1 Ω and reduced armature current (2.4 A), back emf stays 238.8 V, so speed remains 1500 rpm.

A 240 V dc shunt motor has armature resistance of 0.4 Ω and field winding resistance of 120 Ω. It runs at 1500 rpm and draws a current of 5 A with certain load on its shaft. A resistance of 0.1 Ω is connected in series with armature winding and the load on the shaft is reduced by 20%, calculate the new speed of the motor.

Adding resistance in series with the armature leaves the field current unchanged, so flux is constant; then T∝IaT \propto I_a and N∝EbN \propto E_b.

Given: V=240V = 240 V, Ra=0.4 ΩR_a = 0.4\ \Omega, Rsh=120 ΩR_{sh} = 120\ \Omega, N1=1500N_1 = 1500 rpm, IL1=5I_{L1} = 5 A; extra 0.1 Ω in armature; load torque reduced by 20%.

Initial condition

Ish=240120=2 A,Ia1=5−2=3 AEb1=240−3×0.4=238.8 V\begin{aligned} I_{sh} &= \frac{240}{120} = 2\ \text{A}, \quad I_{a1} = 5 - 2 = 3\ \text{A} \\ E_{b1} &= 240 - 3 \times 0.4 = 238.8\ \text{V} \end{aligned}

New condition

Ia2=0.8×3=2.4 AEb2=240−2.4×(0.4+0.1)=240−1.2=238.8 V\begin{aligned} I_{a2} &= 0.8 \times 3 = 2.4\ \text{A} \\ E_{b2} &= 240 - 2.4 \times (0.4 + 0.1) = 240 - 1.2 = 238.8\ \text{V} \end{aligned}

New speed

N2=1500×238.8238.8=1500 rpmN_2 = 1500 \times \frac{238.8}{238.8} = 1500\ \text{rpm}

The extra drop in the added resistance exactly balances the reduced drop from the smaller current (3×0.4=2.4×0.5=1.23 \times 0.4 = 2.4 \times 0.5 = 1.2 V), so the speed is unchanged.

Answer: New speed =1500= 1500 rpm (no change).

✓ Answer checked by Claude Opus 5.5 (high effort)

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