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Shunt motor speed with 50 Ω field resistance and lower torque

Short answer

Field current drops from 2 A to 1.375 A, flux ratio 0.6875. Torque reduced 10% gives I_a = (0.9×23)/0.6875 ≈ 30.11 A. E_b = 220 – 30.11×0.1 = 216.99 V. New speed = 1200 × (216.99/217.7) / 0.6875 ≈ 1740 rpm.

A 220 V dc shunt motor draws a current of 25 A and runs at 1200 rpm with certain load on its shaft. The armature winding resistance and field winding resistances are 0.1 Ω and 110 Ω respectively. If a resistance of 50 Ω is connected in series with the field winding and load torque on the shaft is reduced by 10%, calculate the new speed.

Given: V=220V = 220 V, IL1=25I_{L1} = 25 A, N1=1200N_1 = 1200 rpm, Ra=0.1 ΩR_a = 0.1\ \Omega, Rsh=110 ΩR_{sh} = 110\ \Omega; 50 Ω added in field; load torque reduced by 10%.

For a shunt motor, take flux proportional to field current (no saturation). Then

T∝ϕIa,N∝Ebϕ,Eb=V−IaRaT \propto \phi I_a, \qquad N \propto \frac{E_b}{\phi}, \qquad E_b = V - I_a R_a

Initial condition

Ish1=220110=2 A,Ia1=25−2=23 AEb1=220−23×0.1=217.7 V\begin{aligned} I_{sh1} &= \frac{220}{110} = 2\ \text{A}, \quad I_{a1} = 25 - 2 = 23\ \text{A} \\ E_{b1} &= 220 - 23 \times 0.1 = 217.7\ \text{V} \end{aligned}

New field

Ish2=220110+50=1.375 A,ϕ2ϕ1=1.3752=0.6875I_{sh2} = \frac{220}{110 + 50} = 1.375\ \text{A}, \qquad \frac{\phi_2}{\phi_1} = \frac{1.375}{2} = 0.6875

New armature current (T2=0.9 T1T_2 = 0.9\,T_1)

Ia2=0.9×230.6875=30.11 AI_{a2} = \frac{0.9 \times 23}{0.6875} = 30.11\ \text{A}

New speed

Eb2=220−30.11×0.1=216.99 VN2=1200×216.99217.7×10.6875=1739.8 rpm\begin{aligned} E_{b2} &= 220 - 30.11 \times 0.1 = 216.99\ \text{V} \\ N_2 &= 1200 \times \frac{216.99}{217.7} \times \frac{1}{0.6875} = 1739.8\ \text{rpm} \end{aligned}

Answer: New speed ≈1740\approx 1740 rpm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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