Given that,

Machine A: V_{A} = 11 kV, X_{SA} = 1 Ω

Machine B: V_{B} = 11 kV, X_{SB} = 3 Ω

Grid: 11 kV, 50 MVAR

And I_{A} = I_{B}

The reactive power supplied by both machines are equal.

\({Q_A} = {Q_B} = \frac{{50}}{2} = 25\;MVAR\)

\({I_B} = {I_A} = \frac{{{Q_A}}}{{\sqrt 3 \times V}} = \frac{{25 \times {{10}^6}}}{{\sqrt 3 \times 11 \times {{10}^3}}} = 1312.16\;A\)

E_{A} = V_{A} - jI_{A}X_{SA}

\(= \frac{{\left( {11 \times {{10}^3}} \right)}}{{\sqrt 3 }} - j\left( {1312.16\;\angle 90} \right)\left( 1 \right)\)

⇒ E_{A} = 7663.05 A

E_{B} = V_{B} – jI_{B}X_{SB}

\(= \frac{{11 \times {{10}^3}}}{{\sqrt 3 }} - j\left( {1312.16\;\angle 90} \right)\left( 3 \right)\)

⇒ E_{B} = 10287.33 V

We know that,

E ∝ I_{f}

\(\Rightarrow \frac{{{I_{fA}}}}{{{I_{fB}}}} = \frac{{{E_A}}}{{{E_B}}} = \frac{{7663.05}}{{10287.33}} = 0.745\)