SSE JE Electrical Engineering 24 March 2021 Shift 1 Paper

Rating of alternator = 500 KVA
Terminal voltage of alternator ${\mathrm{V}}_{\mathrm{t}\mathrm{L}-\mathrm{L}}$ $=3.3\mathrm{KV}=3300\mathrm{V}$
Short circuit current ${\mathrm{I}}_{\mathrm{s}\mathrm{c}}$ $=110\sqrt{3}\mathrm{A}$
Winding Resistance $=1\mathrm{\Omega }$ Per phase Open circuit voltage ${\mathrm{V}}_{\mathrm{o}\mathrm{c}}=\frac{3300}{\sqrt{3}}\mathrm{V}$
Per phase short circuit current ${\mathrm{I}}_{\mathrm{s}\mathrm{c}}=110\sqrt{3}\mathrm{A}$ (by default we assume star connected)
Therefore,
per phase ${\mathrm{Z}}_{\mathrm{s}}=\frac{\frac{3300}{\sqrt{3}}}{110\sqrt{3}}$ $=10\mathrm{\Omega }$
${\mathrm{Z}}_{\mathrm{s}}=\sqrt{\overline{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{s}}^2}}$
$10=\sqrt{1^2+{\mathrm{X}}_{\mathrm{s}}^2}$
$100=1+{\mathrm{X}}_{\mathrm{s}}^2$
${\mathrm{X}}_{\mathrm{s}}=\sqrt{99}\mathrm{\Omega }$