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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%.

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There is a sign error in your calculation. For unloaded operation, you have : V = E. So: $$E_{nl} = 2000V$$

You want to calculate the variation of E you need to have in order to ensure that V stay constant when the alternator is loaded :

$$ E_l = V + R \times I + j \times X \times I $$ $$ E_l = 2000 + (0.8 + j \times 4) \times I = 1824 + 368j $$ $$ \left | E_l \right | = 1861 V $$

Note that El is lower than Enl because of the leading power factor. Finally : $$ {{1861 - 2000} \over {2000}} = -6.96\% $$

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  • \$\begingroup\$ Haven't you change the formula used for calculating the voltage regulation. It's alternator so the reference taken should be \$E_l\$ ? \$\endgroup\$ – Ansh Kumar Feb 23 '17 at 13:19
  • \$\begingroup\$ Basically, I used the formula which is given here : electricalengineeringinfo.com/2014/11/… . It corresponds to the difference (EMF(loaded)-EMF(unloaded)/EMF(unloaded) = (EMF(loaded)-V)/V. This gives the amount of change you need to apply to the EMF in order to get a constant output voltage (V). \$\endgroup\$ – Charles JOUBERT Feb 23 '17 at 14:47
  • \$\begingroup\$ The formula in the link simply relates terminal and internal voltage of an alternator. Check nptel.ac.in/courses/108106071/pdfs/1_9.pdf \$\endgroup\$ – Ansh Kumar Feb 23 '17 at 15:02
  • \$\begingroup\$ The discrepancy is due to defining 2000V as the terminal voltage at the specified load versus at no load. The problem statement is a bit ambiguous in this regard. \$\endgroup\$ – user28910 Feb 23 '17 at 15:39

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