Skip to main content

Shunt motor: resistance to reduce speed to 1200 rpm

Short answer

Field current is 2.2 A, armature current 37.8 A. Initial back emf is 219.244 V; at 1200 rpm back emf is 187.923 V. Required total armature resistance is 0.8486 ohm, so inserted resistance is 0.8286 ohm.

A 220 V dc shunt motor draws a current of 40 A at full load and runs with speed of 1400 rpm. Calculate the value of resistance required to be inserted in the armature circuit so that speed drops to 1200 rpm at constant load. Given that Ra=0.02 ΩR_a = 0.02\ \Omega and Rf=100 ΩR_f = 100\ \Omega.

The field is across the supply, so the flux is constant and N∝EbN\propto E_b. Constant load torque means constant armature current.

Initial condition:

If=220100=2.2 A,Ia=40−2.2=37.8 AI_f = \frac{220}{100} = 2.2\ \text{A}, \qquad I_a = 40 - 2.2 = 37.8\ \text{A} Eb1=220−37.8(0.02)=219.244 VE_{b1} = 220 - 37.8(0.02) = 219.244\ \text{V}

At 1200 rpm:

Eb2=Eb1×12001400=187.923 VE_{b2} = E_{b1}\times\frac{1200}{1400} = 187.923\ \text{V} Ra+Rx=V−Eb2Ia=220−187.92337.8=0.8486 ΩR_a + R_x = \frac{V - E_{b2}}{I_a} = \frac{220 - 187.923}{37.8} = 0.8486\ \Omega Rx=0.8486−0.02=0.8286 ΩR_x = 0.8486 - 0.02 = 0.8286\ \Omega

Answer: resistance to be inserted in the armature circuit Rx=0.8286 ΩR_x = 0.8286\ \Omega.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Machines chapter 4 questions →