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 X_{1} is the positive sequence reactance

X_{2} is the negative sequence reactance

X_{0} is the zero-sequence reactance

**Calculation:**

Given that, X_{1} = 0.12 pu, X_{2} = 0.096pu, X_{0} = 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