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Shunt motor speed with field resistance added (250 V, 1400 rpm)

Short answer

With field resistance increased, flux ratio is 0.7143. Armature current becomes 38.64 A, back emf 242.27 V. Speed = 1400 × (242.27/245.4) / 0.7143 ≈ 1935 rpm.

A 250 V dc shunt motor has armature-winding resistance of 0.2 ohm and field winding resistance of 125 ohms. It draws a current of 25 amp at half load and the corresponding speed is 1400 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=0.2 ΩR_a = 0.2\ \Omega, Rsh=125 ΩR_{sh} = 125\ \Omega, IL1=25I_{L1} = 25 A at half load, N1=1400N_1 = 1400 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=250−23×0.2=245.4 V\begin{aligned} I_{sh1} &= \frac{250}{125} = 2\ \text{A}, \quad I_{a1} = 25 - 2 = 23\ \text{A} \\ E_{b1} &= 250 - 23 \times 0.2 = 245.4\ \text{V} \end{aligned}

New field

Ish2=250175=1.4286 A,ϕ2ϕ1=125175=0.7143I_{sh2} = \frac{250}{175} = 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=250−38.64×0.2=242.27 VN2=1400×242.27245.4×10.7143=1935.0 rpm\begin{aligned} E_{b2} &= 250 - 38.64 \times 0.2 = 242.27\ \text{V} \\ N_2 &= 1400 \times \frac{242.27}{245.4} \times \frac{1}{0.7143} = 1935.0\ \text{rpm} \end{aligned}

Answer: New speed ≈1935\approx 1935 rpm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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