Reactance diagram 1000 kVA base and generator currents

No figure is printed, so the usual arrangement is assumed: G1–Tr1 and G2–Tr2 feed a common 66 kV bus; from it the line TL and Tr3 (66/400 kV) supply the load. The two generator emfs are taken as equal, so the load current divides between the two generator branches in inverse ratio of their impedances. The 1200 A load current is taken as referred to the 11 kV level, since its voltage level is not stated.

 G1--Tr1(11/66)--+
                 |66 kV bus
 G2--Tr2(12.5/75)+--TL(10 ohm)--Tr3(66/400)--Load

Base values ($S_b$ = 1000 kVA)

• G1 side: 11 kV (given). 66 kV bus: $11\times\frac{66}{11} = 66$ kV.
• G2 side through Tr2: $66\times\frac{12.5}{75} = 11$ kV.
• Load side: $66\times\frac{400}{66} = 400$ kV.
• $Z_b$ at 66 kV $= \frac{66^2}{1} = 4356\ \Omega$. $I_b$ at 11 kV $= \frac{1000}{\sqrt3\times11} = 52.49$ A.

Per-unit reactances

$$\begin{aligned} X_{G1} &= 0.025\ \text{pu}\\ X_{G2} &= 0.001\times\frac{1000}{500}\times\left(\frac{11}{11}\right)^2 = 0.002\ \text{pu}\\ X_{Tr1} &= 0.02\ \text{pu}\\ X_{Tr2} &= 0.02\times\frac{1000}{500}\times\left(\frac{12.5}{11}\right)^2 = 0.0517\ \text{pu}\\ X_{TL} &= \frac{10}{4356} = 0.0023\ \text{pu}\\ X_{Tr3} &= 0.02\times\frac{1000}{1500} = 0.0133\ \text{pu} \end{aligned}$$

Reactance diagram

 j0.025  j0.02
+-/\/\/--/\/\/--+
|  G1     Tr1   |   j0.0023  j0.0133
(E1)            +---/\/\/----/\/\/---> Load
|  G2     Tr2   |     TL      Tr3
+-/\/\/--/\/\/--+
 j0.002  j0.0517
(E1, E2 return to the common reference bus)

Current shared by G1 and G2

Branch impedances: $Z_1 = 0.025 + 0.02 = 0.045$ pu and $Z_2 = 0.002 + 0.0517 = 0.0537$ pu.

$$\begin{aligned} I_{G1} &= I_L\frac{Z_2}{Z_1+Z_2} = 1200\times\frac{0.0537}{0.0987} = 1200\times0.5439 = 652.6\ \text{A}\\ I_{G2} &= I_L\frac{Z_1}{Z_1+Z_2} = 1200\times0.4561 = 547.4\ \text{A} \end{aligned}$$

(In per unit the load current is $1200/52.49 = 22.86$ pu. G1 carries 12.43 pu and G2 carries 10.43 pu.)

Answer: G1 supplies about 652.6 A and G2 about 547.4 A (54.4 % and 45.6 % of the load current).

Note: if the G2 reactance of 0.1 % is a misprint for 10 %, then $X_{G2} = 0.2$ pu, $Z_2 = 0.2517$ pu, and the shares become 1018.0 A (G1) and 182.0 A (G2).