Question

A 22 kV, 100 MVA alternator is provided with differential protection. The percentage of winding to be protected against phase to ground fault is 90%. The relay operates when there is 20% out of balance current. The value of resistance required to be placed in neutral to ground connection will be ________$\Omega$.

• A. 2.42

Answer 1

The correct option is A 2.42
Given
$V_{line}=22kV$
MVA rating = 100 MVA
Primary earth fault current for relay operation
$=\frac{100\times {10}^3}{22\times \sqrt{3}}\times \frac{20}{100}=521.86A$

Let x% of winding remains unprotected and $R_a$ be resistance connected to neutral
So, x = 200 - 90 = 10%
$Faultcurrent=\frac{x}{100}\times \frac{22\times {10}^3}{\sqrt{3}{R}_n}=\frac{10}{100}\times \frac{22\times {10}^3}{\sqrt{3}{R}_n}$
$524.86=\frac{10}{100}\times \frac{22\times 1063}{\sqrt{3}\times R_n}$
$R_n=2.42\Omega$