# Voltage regulation in alternator

A single phase, 2000 V alternator has armature resistance and reactance of 0.8 Ω and 4 Ω respectively. The voltage regulation of alternator at 100 A load at 0.8 leading power factor is _______?

My approach

$$V_{nl} = 2000 \text{ V}$$ When loaded $$V_{l} = 2000 - 100 \angle \cos^{-1}0.8 (0.8 +j4) = 2206.9 \text{ V}$$ In alternator $V_{reg}$ is defined as $$V_{reg} = \frac{|V_{nl}| - |V_l|}{|V_l|} * 100 = -9.38 \%$$ Am I right ? Because when I saw the answer it was given -6.96%.

There is a sign error in your calculation. For unloaded operation, you have : V = E. So: $$E_{nl} = 2000V$$
$$E_l = V + R \times I + j \times X \times I$$ $$E_l = 2000 + (0.8 + j \times 4) \times I = 1824 + 368j$$ $$\left | E_l \right | = 1861 V$$
Note that El is lower than Enl because of the leading power factor. Finally : $${{1861 - 2000} \over {2000}} = -6.96\%$$
• Haven't you change the formula used for calculating the voltage regulation. It's alternator so the reference taken should be $E_l$ ? – Ansh Kumar Feb 23 '17 at 13:19