Question

(a) Write a short note on various types of DC Motors. (6)

(b) A 440V shunt motor takes an armature current of 30A at 700 rev/min. The armature resistance is 0.7ohm. If the flux is suddenly reduced 20 per cent, to what value will the armature current rise momentarily? Assuming unchanged resisting torque to motion, what will be the new steady values of speed and armature current? Sketch graphs showing armature current and speed as functions of time during the transition from initial to final, steady-state conditions.

Answer 1

Part (a)

Short note on the various types of D.C. motors:

Shunt motor: the field winding is connected in parallel with the armature. It has a nearly constant speed characteristic and moderate starting torque. Used for constant-speed drives such as fans, blowers, and machine tools.
Series motor: the field winding is in series with the armature. It has a very high starting torque and a falling speed characteristic (constant power). It must never be run without load (overspeed). Used for cranes, hoists, winches, and traction.
Compound motor: has both a shunt and a series field. Cumulative compound gives high starting torque with a stable speed (used for deck machinery, presses, and loads requiring high starting torque). Differential compound gives a nearly constant speed and is rarely used.
Permanent magnet motor: uses permanent magnets for the field. Compact and efficient, used for small servo and control applications.

Part (b)

440 V shunt motor, armature current 30 A at 700 rev/min, armature resistance 0.7 ohm. Flux suddenly reduced 20%.

Back e.m.f. E1 = V - Ia Ra = 440 - 30 x 0.7 = 440 - 21 = 419 V.
Flux reduced by 20%: phi2 = 0.8 phi1.
Momentarily, the speed cannot change instantly, so the back e.m.f. falls to E2 = 0.8 x 419 = 335.2 V.
The armature current momentarily rises to Ia2 = (V - E2)/Ra = (440 - 335.2)/0.7 = 104.8/0.7 = 149.7 A.
New steady state: torque constant (resisting torque unchanged), so phi1 Ia1 = phi2 Ia2, giving Ia2 = Ia1 (phi1/phi2) = 30/0.8 = 37.5 A.
New back e.m.f. E2 = V - Ia2 Ra = 440 - 37.5 x 0.7 = 440 - 26.25 = 413.75 V.
New speed: N proportional to E/phi. N2 = N1 x (E2/E1) x (phi1/phi2) = 700 x (413.75/419) x (1/0.8) = 700 x 0.9875 x 1.25 = 864 rev/min.
So the armature current momentarily rises to about 150 A, then settles at 37.5 A, and the new steady speed is about 864 rev/min.
Sketch: the armature current shows a sharp spike to 150 A at the instant of flux reduction, then falls to the new steady value of 37.5 A. The speed rises smoothly from 700 to 864 rev/min as the motor accelerates to the new steady state.

✓ Verified Model Answer (Text Solution)

Short note on the various types of D.C. motors:

440 V shunt motor, armature current 30 A at 700 rev/min, armature resistance 0.7 ohm. Flux suddenly reduced 20%.