A 4-pole 500V shunt motor has 720 wave connected conductors on its arm...
Given data:
Number of poles (P) = 4
Voltage rating (V) = 500V
Armature conductors (N) = 720
Full-load armature current (Ia) = 60A
Flux per pole (Φ) = 0.03 Wb
Armature resistance (Ra) = 0.2Ω
Contact drop per brush (Vc) = 1V
To find:
Back emf (Eb)
Solution:
1. Calculation of back emf (Eb)
The back emf of a DC shunt motor is given by the expression:
Eb = V - Ia(Ra + Rb)
where,
V is the applied voltage
Ia is the armature current
Ra is the armature resistance
Rb is the brush contact resistance
In this case, the brush contact resistance is not given, but we can assume it to be negligible compared to the armature resistance. Therefore,
Rb = 0
Substituting the given values,
Eb = V - Ia(Ra + Rb)
Eb = V - IaRa
Eb = ΦPN/60A
Eb = 0.03 × 4 × 720 / 60
Eb = 0.144V
Therefore, the back emf of the motor is 0.144V.
2. Conversion of back emf to actual voltage
The back emf is a theoretical value that is always less than the actual voltage of the motor. The actual voltage can be found by adding the voltage drop due to the armature resistance and the brush contact drop to the back emf.
Vactual = Eb + IaRa + 2Vc
Substituting the given values,
Vactual = 0.144 + (60 × 0.2) + 2
Vactual = 12.24V
Therefore, the actual voltage of the motor is 500 - 12.24 = 487.76V.
Rounding off to the nearest whole number, we get:
Vactual ≈ 488V
Hence, the correct answer is option 'D', 486V.
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