Q6 (10 Marks) Electrical Circuits & Calculations
MET • Written Exam

(a) What are factors on which the speed of a motor depends? Discuss them for series and shunt motors. (6)

(b) Three equal resistors are connected to a three-phase system, If one resistor is removed find the reduction in load if they are connected in (a) Star, (b) Delta. (10)

Appeared In: Nov 2023

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready
Part (a)

Factors on which the speed of a motor depends, for series and shunt motors:

  • The speed of a d.c. motor is given by N = (V - Ia Ra)/(k phi), so it depends on the applied voltage V, the armature current Ia, the armature resistance Ra, and the field flux phi.
  • Shunt motor: the field flux is nearly constant (the field is connected across the supply). The speed falls only slightly as the load (armature current) increases, because the armature resistance drop (Ia Ra) increases. The shunt motor has a nearly constant speed characteristic.
  • Series motor: the field winding is in series with the armature, so the flux is proportional to the armature current (until saturation). As the load increases, the flux increases, so the speed falls sharply. The series motor has a falling speed characteristic and must never be run without load (it would overspeed). Its speed depends strongly on the load.
  • The speed can also be varied by changing the applied voltage (armature voltage control) or the field flux (field weakening).
Part (b)

Three equal resistors on a three-phase system, one removed:

Let each resistor be R and the line voltage be V.

(i) Star:

  • Three resistors in star: each phase voltage = V/root 3. Power per resistor = (V/root3)^2/R = V^2/(3R). Total P3 = 3 x V^2/(3R) = V^2/R.
  • Two resistors remaining: P2 = 2 x V^2/(3R).
  • Reduction = V^2/R - 2V^2/(3R) = V^2/(3R). Fractional reduction = 1/3 = 33.3%.

(ii) Delta:

  • Three resistors in delta: each phase voltage = V. Power per resistor = V^2/R. Total P3 = 3V^2/R.
  • Two resistors remaining: P2 = 2V^2/R.
  • Reduction = 3V^2/R - 2V^2/R = V^2/R. Fractional reduction = 1/3 = 33.3%.

So in both cases the load is reduced by one third (33.3%).

← Back to MET Question Bank Upload Recent Question Paper →