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Shunt motor speed with 50 Ω field resistance (1200 rpm)

Short answer

Field current drops to 1.4286 A (flux ratio 0.7143). Armature current rises to 38.64 A, back emf 234.54 V. Speed = 1200 × (234.54/240.8) / 0.7143 ≈ 1636 rpm.

A 250 V dc shunt motor has armature-winding resistance of 0.4 ohm and field winding resistance of 125 ohms. It draws a current of 25 amp at half load and the corresponding speed is 1200 rpm. If a resistance of 50 ohm is connected in series with the field winding and load torque is increased by 20%, calculate the new speed.

Given: V=250V = 250 V, Ra=0.4 ΩR_a = 0.4\ \Omega, Rsh=125 ΩR_{sh} = 125\ \Omega, IL1=25I_{L1} = 25 A at half load, N1=1200N_1 = 1200 rpm; 50 Ω added in field; load torque +20%.

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=250125=2 A,Ia1=25−2=23 AEb1=250−23×0.4=240.8 V\begin{aligned} I_{sh1} &= \frac{250}{125} = 2\ \text{A}, \quad I_{a1} = 25 - 2 = 23\ \text{A} \\ E_{b1} &= 250 - 23 \times 0.4 = 240.8\ \text{V} \end{aligned}

New field

Ish2=250125+50=1.4286 A,ϕ2ϕ1=125175=0.7143I_{sh2} = \frac{250}{125 + 50} = 1.4286\ \text{A}, \qquad \frac{\phi_2}{\phi_1} = \frac{125}{175} = 0.7143

New armature current (T2=1.2 T1T_2 = 1.2\,T_1)

Ia2=1.2×230.7143=38.64 AI_{a2} = \frac{1.2 \times 23}{0.7143} = 38.64\ \text{A}

New speed

Eb2=250−38.64×0.4=234.54 VN2=1200×234.54240.8×10.7143=1636.4 rpm\begin{aligned} E_{b2} &= 250 - 38.64 \times 0.4 = 234.54\ \text{V} \\ N_2 &= 1200 \times \frac{234.54}{240.8} \times \frac{1}{0.7143} = 1636.4\ \text{rpm} \end{aligned}

Answer: New speed ≈1636\approx 1636 rpm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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