Skip to main content

Shunt motor speed with 250 ohm in field circuit

Short answer

Inserting 250 Ω in the field circuit halves the field current to 0.5 A, halving flux. To maintain constant torque, armature current doubles to 40 A, back emf becomes 240 V, and the new speed is 1500 × (240/245) × 2 ≈ 2939 rpm.

A 250 V dc shunt motor draws an armature current of 20 A and runs with a speed of 1500 rpm. If a resistance of 250 Ω250\ \Omega is inserted in series with field winding keeping the load torque constant, find the new speed. Given that armature winding resistance is 0.25 Ω0.25\ \Omega and field winding resistance is 250 Ω250\ \Omega.

Assume the flux is proportional to field current. The load torque is constant, so ΦIa\Phi I_a is constant. Field weakening therefore increases the armature current.

Initial:

If1=250250=1 A,Eb1=250−20(0.25)=245 VI_{f1} = \frac{250}{250} = 1\ \text{A}, \qquad E_{b1} = 250 - 20(0.25) = 245\ \text{V}

After inserting 250 Ω in the field:

If2=250250+250=0.5 A,Φ2Φ1=0.51=0.5I_{f2} = \frac{250}{250 + 250} = 0.5\ \text{A}, \qquad \frac{\Phi_2}{\Phi_1} = \frac{0.5}{1} = 0.5

Constant torque (T∝ΦIaT\propto\Phi I_a):

Ia2=Ia1×Φ1Φ2=20×2=40 AI_{a2} = I_{a1}\times\frac{\Phi_1}{\Phi_2} = 20\times 2 = 40\ \text{A} Eb2=250−40(0.25)=240 VE_{b2} = 250 - 40(0.25) = 240\ \text{V}

Speed (from Eb∝ΦNE_b\propto\Phi N):

N2=N1×Eb2Eb1×Φ1Φ2=1500×240245×2=2939 rpmN_2 = N_1\times\frac{E_{b2}}{E_{b1}}\times\frac{\Phi_1}{\Phi_2} = 1500\times\frac{240}{245}\times 2 = 2939\ \text{rpm}

Answer: new speed ≈2939 rpm\approx 2939\ \text{rpm}.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Machines chapter 4 questions →