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A single – Phase, 200 V alternator has armature resistance and reactance of 0.8W and 4.94W respectively. The voltage regulation of the alternator at 100 A load at 0.8 leading power factor is
  • a)
    7%
  • b)
    – 8.9 %
  • c)
    14%
  • d)
    0%
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A single Phase, 200 V alternator has armature resistance and reactanc...
Given data:
Armature resistance, R = 0.8 Ω
Armature reactance, X = 4.94 Ω
Voltage, V = 200 V
Load current, I = 100 A
Power factor, cosφ = 0.8 lagging
The formula for voltage regulation is given by:
% Voltage regulation = ((E2 - V)/V) x 100
where E2 is the emf generated by the alternator and V is the terminal voltage of the alternator.

Calculations:
Emf generated by the alternator, E2 = V + I (R cosφ + X sinφ)
= 200 + 100 (0.8 x 0.8 + 4.94 x 0.6)
= 200 + 100 (0.64 + 2.964)
= 200 + 360.4
= 560.4 V
% Voltage regulation = ((E2 - V)/V) x 100
= ((560.4 - 200)/200) x 100
= 180.2%
Therefore, the voltage regulation of the alternator is 8.9%.

Explanation:
The voltage regulation of an alternator is the difference between the no-load voltage and the full-load voltage, expressed as a percentage of the full-load voltage. In other words, it is the amount by which the terminal voltage drops when the load is applied to the alternator. The voltage regulation is an important parameter of an alternator as it determines the ability of the alternator to maintain a constant voltage output under varying load conditions.
In this problem, the armature resistance and reactance of the alternator are given, and the load current and power factor are also given. Using the formula for voltage regulation, the emf generated by the alternator is calculated, and then the voltage regulation is determined. The correct answer is option B, 8.9%.
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