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Shunt motor speed with 0.05 Ω armature resistance added

Short answer

Field current 2 A, flux constant. Initial back emf 218.16 V; after adding 0.05 Ω, armature current 27.6 A, back emf 216.41 V. Speed = 1500 × 216.41/218.16 ≈ 1488 rpm.

A dc shunt motor supplied by 220 V dc draws a current of 25 A and runs at a speed of 1500 rpm. The armature and field winding resistances are 0.08 Ω and 110 Ω respectively. A resistance of 0.05 Ω is added in series with armature and load torque is increased by 20%, calculate the new speed.

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

Given: V=220V = 220 V, IL1=25I_{L1} = 25 A, N1=1500N_1 = 1500 rpm, Ra=0.08 ΩR_a = 0.08\ \Omega, Rsh=110 ΩR_{sh} = 110\ \Omega; extra 0.05 Ω in armature; load torque +20%.

Initial condition

Ish=220110=2 A,Ia1=25−2=23 AEb1=220−23×0.08=218.16 V\begin{aligned} I_{sh} &= \frac{220}{110} = 2\ \text{A}, \quad I_{a1} = 25 - 2 = 23\ \text{A} \\ E_{b1} &= 220 - 23 \times 0.08 = 218.16\ \text{V} \end{aligned}

New condition

Ia2=1.2×23=27.6 AEb2=220−27.6×(0.08+0.05)=220−3.588=216.41 V\begin{aligned} I_{a2} &= 1.2 \times 23 = 27.6\ \text{A} \\ E_{b2} &= 220 - 27.6 \times (0.08 + 0.05) = 220 - 3.588 = 216.41\ \text{V} \end{aligned}

New speed

N2=1500×216.41218.16=1488.0 rpmN_2 = 1500 \times \frac{216.41}{218.16} = 1488.0\ \text{rpm}

Answer: New speed ≈1488\approx 1488 rpm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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