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Shunt motor speed with 0.1 Ω armature resistance added

A 220 V dc shunt motor draws a current of 20 A and runs at 1400 rpm with certain load on the shaft. The armature winding resistance and field winding resistance are 0.2 Ω and 220 Ω respectively. If a resistance of 0.1 Ω is added in series with armature and the load torque is increased by 10%, calculate the new speed.

Adding resistance in the armature circuit does not change the field current, so flux is constant. Then T∝IaT \propto I_a and N∝EbN \propto E_b.

Given: V=220V = 220 V, IL1=20I_{L1} = 20 A, N1=1400N_1 = 1400 rpm, Ra=0.2 ΩR_a = 0.2\ \Omega, Rsh=220 ΩR_{sh} = 220\ \Omega, extra 0.1 Ω in armature, torque +10%.

Initial condition

Ish=220220=1 A,Ia1=20−1=19 AEb1=220−19×0.2=216.2 V\begin{aligned} I_{sh} &= \frac{220}{220} = 1\ \text{A}, \quad I_{a1} = 20 - 1 = 19\ \text{A} \\ E_{b1} &= 220 - 19 \times 0.2 = 216.2\ \text{V} \end{aligned}

New condition

Flux constant, torque up 10%:

Ia2=1.1×19=20.9 AEb2=220−20.9×(0.2+0.1)=220−6.27=213.73 V\begin{aligned} I_{a2} &= 1.1 \times 19 = 20.9\ \text{A} \\ E_{b2} &= 220 - 20.9 \times (0.2 + 0.1) = 220 - 6.27 = 213.73\ \text{V} \end{aligned}

New speed

N2=N1×Eb2Eb1=1400×213.73216.2=1384.0 rpmN_2 = N_1 \times \frac{E_{b2}}{E_{b1}} = 1400 \times \frac{213.73}{216.2} = 1384.0\ \text{rpm}

Answer: New speed ≈1384\approx 1384 rpm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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