Reactance diagram: three generators, 30 MVA 6.6 kV base

Base values

• $S_b = 30$ MVA, and $V_b = 6.6$ kV in the G1 circuit.
• Line section (through T1, 6.6/115 kV): $V_b = 115$ kV, so $Z_b = 115^2/30 = 440.83\ \Omega$.
• G2 circuit (through T2, 115/6.6 kV): $V_b = 6.6$ kV.
• T3 is a Y–Y bank of single-phase 69/6.9 kV units. Its line ratio is $\sqrt3\times69 = 119.5$ kV / $\sqrt3\times6.9 = 11.95$ kV, and its rating is $3\times10 = 30$ MVA.
• G3 circuit: $V_b = 115\times\dfrac{11.95}{119.5} = 11.5$ kV.

Per-unit reactances

$$\begin{aligned} G_1&: 0.2\times\frac{30}{25} = 0.24\\ G_2&: 0.15\times\frac{30}{15} = 0.30\\ G_3&: 0.15\times\frac{30}{30}\left(\frac{13.2}{11.5}\right)^2 = 0.1976\\ T_1&: 0.1\times\frac{30}{30} = 0.10\\ T_2&: 0.1\times\frac{30}{15} = 0.20\\ T_3&: 0.1\times\frac{30}{30}\left(\frac{119.5}{115}\right)^2 = 0.1080\\ \text{Line } j120&: \frac{120}{440.83} = 0.2722\\ \text{Line } j90&: \frac{90}{440.83} = 0.2042 \end{aligned}$$

Reactance diagram

 j0.24  j0.10   j0.2722        j0.2042  j0.108  j0.1976
+-/\/\--/\/\----/\/\----+(J)+---/\/\----/\/\----/\/\--+
|                        |                            |
(E1)                   j0.20 (T2)                   (E3)
|                        |                            |
|                      j0.30 (G2)                     |
|                        |                            |
|                      (E2)                           |
+------------------- reference -----------------------+

Here J is the junction bus. Winding connections and earthing do not affect these positive-sequence reactances.