Solved Examples
Problem: A 4 pole generator supplies a current of 143 A. It has 492 conductors lap connected and delivering full load, brushes are given a shift of lead to 10°. calculate demagnetizing ampere turns per pole? The field winding is shunt connected and takes 10 A. Find no. of extra shunt turns necessary to neutralize this demagnetization.
Solution: Current in armature, Ia = 143 + 10 = 153 A
ꞵelect = P/2 * 10° = 4/2 * 10° = 20°
Z = 492
A = P = 4, (A=P for lap winding)
Far (demagnetizing) = {(Z/2) / P} * {(2ꞵelect) / 180°} * {Ia / A}
= {492 / (2*9)} * {(2*20°) / 180°} * {153 / 4}
= 522.75 AT/pole
Extra field turns = 522.75 / 10 = 53 turns
Problem: Neglecting all losses, the developed torque (T) of a d.c. separately excited motor, operating under constant terminal voltage, is related to its output power (P) as?
Solution: Power developed, P = Eb.Ia = 2𝜋NT / 60
P ∝ NT ⇒ T ∝ P / N
For a given Torque, Speed is constant.
∴ T ∝ P .
Problem: A 4 pole dynamo with wave wound armature has 51 slots containing 20 conductors in each slot. The induced emf is 357 volts and the speed is 8500 rpm. The flux per pole will be?
Solution: For wave winding, A =2
P=4, S=51
Number of conductors, Z= 51*20 =1020
N=8500 rpm
Induced emf, Ea=357 V
Ea = ⲫZNP / 60A
⇒ ⲫ = 60EaA / ZNP = (60*357*2) / (1020*8500*4) = 1.23 mWb
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