Question

The Positive, Negative and Zero sequence per unit impedances of two generators connected in parallel are X₁ = 0.12, X₂ = 0.096 and X₀ = 0.036 pu. For an L.G fault at generator terminals (with 1 pu voltage) the positive sequence current will be:
1. 7.936 pu
2. 11.936 pu
3. 10.936 pu
4. 8.936 pu

Answer 1

Correct Answer - Option 1 : 7.936 pu

Concept:

For LG fault at generator terminals, the positive sequence current is given by

\({I_{R1}} = \frac{E}{{\left( {{X_1}\; + \;{X_2}\; + \;{X_0}} \right)}}\)

Where X₁ is the positive sequence reactance

X₂ is the negative sequence reactance

X₀ is the zero-sequence reactance

Calculation:

Given that, X₁ = 0.12 pu, X₂ = 0.096pu, X₀ = 0.036pu, V = 1pu

As two generators connected in parallel,

∴ X'1 = X1/2, X'2 = X2/2, and  X'0 = X0/2

Now \({I_{R1}} = \frac{V}{{\left( {\frac{{{X_1}}}{2} + \frac{{{X_2}}}{2} + \frac{{{X_0}}}{2}} \right)}}\)

\({I_{R1}} = \frac{1}{0.126}\)

I_R1 = 7.936pu