From the above figure, calculate the short circuit kVA feed to the symmetrical fault between phases if it occurs.

(i) at the end of transmission line.

(ii) at the high voltage terminals of the transformer.

% reactance of alternator on base kVA,

$\%X_A=\frac{10000}{10\times {10}^3}\times 10=10\%$

% reactance of transformer on base kVA,

$\%X_T= \frac{10000}{5\times {10}^3}\times 5=10\%$

% reactance of transmission line is

$\%X_L=\frac{\left(kVA\right) \times reac\tan ce in \Omega }{10 {\left(kV\right)}^2}$

$=\frac{10,000 \times 4}{10\times {\left(10\right)}^2}= 40\%$

% age resistance of transmission line,

$\%R_L =\frac{10,000 \times 1}{10\times (10{)}^2}= 10\%$

(i) Total % reactance = %X_A+%X_T+%X_L

= 10 + 10 + 40 = 60%

% resistance = 10%

∴ % impedance from generator neutral upto fault point F₂

$=\sqrt{(60{)}^2+(10{)}^2}$ = 60.83%

∴ Short-circuit kVA $= 10000\times \frac{100}{60.83}= 16 ,440 kVA$

(ii) For a fault at the high voltage terminals of the transformer (point F₁),

Total % reactance from generator neutral upto fault point F₁

$=\%X\mathrm{<em></em>}T=10+10=20\%$

∴ Short-circuit kVA$= 10000\times \frac{100}{20}$

= 50000kVA