Question

A single phase, 2000 V alternator has armature resistance and reactance of 0.8 $\Omega$and 4 $\Omega$ respectively. The voltage regulation of alternator at 100 A load at 0.8 leading power factor is______%.

• A. -6.96

Answer 1

The correct option is A -6.96
$E_f^2=(V_{t}cos\varphi +I_a{r}_a{)}^2+(V_{t}sin\varphi \pm I_a{X}_a{)}^2$

$=(1600+80{)}^2+(1200-400{)}^2$

$\text{(Negative sign due to leading p.f.)}$

$E_f=1860.752V$

$\%V.R.=\frac{1860.752-2000}{2000}\times 100=-6.96\%$