Two sources via 0 - j5 Ω: power flow

The line impedance $Z = 0 - j5\ \Omega$ is a pure capacitive reactance. Let I flow from machine 1 to machine 2. Then $S_1 = E_1I^*$ is the power supplied by machine 1 and $S_2 = E_2I^*$ is the power received by machine 2.

Current

$$\begin{aligned} E_1 - E_2 &= 100 - (86.60 + j50) = 13.40 - j50\\ I &= \frac{13.40 - j50}{-j5} = 10 + j2.679 = 10.353\angle15^\circ\ \text{A} \end{aligned}$$

Powers

$$\begin{aligned} S_1 &= E_1I^* = 100(10 - j2.679) = 1000 - j267.95\ \text{VA}\\ S_2 &= E_2I^* = (100\angle30^\circ)(10.353\angle-15^\circ) = 1000 + j267.95\ \text{VA}\\ S_Z &= |I|^2Z = 107.18(-j5) = -j535.90\ \text{VA} \end{aligned}$$

Check with $P = \frac{|E_1||E_2|}{X}\sin(\delta_1 - \delta_2)$ using $X = -5$: $P = \frac{100\times100}{-5}\sin(-30^\circ) = +1000$ W, from 1 to 2.

Magnitude and direction

Note: with an inductive link, real power flows from the leading machine (2) to the lagging one. A capacitive link reverses this, so power flows from machine 1 to machine 2 even though $E_2$ leads. The line capacitance supplies reactive power equally to both ends, because the voltage magnitudes are equal.

Answer: P = 1000 W flows from machine 1 to machine 2. Each machine absorbs 267.95 var, all supplied by the capacitive line (535.9 var).