Impedance diagram: three generators, j100 and j130 Ω lines

Common base: 30 MVA, 6.6 kV in the G1 circuit. Base voltages in other zones follow the line-to-line ratios.

Base voltages

• G1 zone: 6.6 kV
• Line zone (T1 6.6/115 kV): 115 kV, $Z_b = \frac{115^2}{30} = 440.83\ \Omega$
• G2 zone (T2 115/6.6 kV): 6.6 kV
• G3 zone: T3 is a Y–Y bank of 69/6.9 kV units, so the line-to-line ratio is 119.5/11.95 kV (10 : 1). Then $V_b = 115/10 = 11.5$ kV

Per-unit values (30 MVA)

$$\begin{aligned} X_{G1} &= 0.2\times\frac{30}{25} = 0.24\ \text{pu}\\ X_{G2} &= 0.15\times\frac{30}{15} = 0.30\ \text{pu}\\ X_{G3} &= 0.15\times\frac{30}{30}\times\left(\frac{13.2}{11.5}\right)^2 = 0.1976\ \text{pu}\\ X_{T1} &= 0.1\times\frac{30}{30} = 0.10\ \text{pu}\\ X_{T2} &= 0.1\times\frac{30}{15} = 0.20\ \text{pu}\\ X_{T3} &= 0.4\times\frac{30}{3\times10}\times\left(\frac{11.95}{11.5}\right)^2 = 0.432\ \text{pu}\\ X_{L1} &= \frac{100}{440.83} = 0.2268\ \text{pu}\\ X_{L2} &= \frac{130}{440.83} = 0.2949\ \text{pu} \end{aligned}$$

Impedance diagram

    j0.10   j0.2268       j0.2949   j0.432
 +--/\/\/---/\/\/----+----/\/\/-----/\/\/---+
 |   T1     line 1   |    line 2     T3     |
j0.24              j0.20 (T2)            j0.1976
 |                   |                      |
(E1)               j0.30 (G2)             (E3)
 |                   |                      |
 |                 (E2)                     |
 +-------------------+----------------------+
           reference (neutral) bus

Answer: $X_{G1}=0.24$, $X_{G2}=0.30$, $X_{G3}=0.1976$, $X_{T1}=0.10$, $X_{T2}=0.20$, $X_{T3}=0.432$, $X_{L1}=0.2268$, $X_{L2}=0.2949$ pu (base 30 MVA; 6.6 / 115 / 6.6 / 11.5 kV).