Question

A 11 kV, 100 MVA alternator is provided with differential protection. The percentage of winding to be protected against phase to ground fault is 85%. The relay is set to operate when there is 20% out of balance current. The value of resistance to be placed in the neutral to ground connection is ______ $\Omega$.

• A. 0.91

Answer 1

The correct option is A 0.91
Primary earth fault current at which relay operates

$=\frac{100\times {10}^3}{\sqrt{3}\times 11}\times \frac{20}{100}=1049.73A$
Percentage of winding which remains unprotected is

$\varphi =100-85=15$

The fault current $=\frac{\varphi }{100}\times \frac{11\times {10}^3}{\sqrt{3}{R}_n}$

$\therefore \frac{15}{100}\times \frac{11\times {10}^3}{\sqrt{3}{R}_n}=1049.73$

$\Rightarrow R_n=0.91\Omega$