Reactive power flow between machines via j10 Ω

Per phase, with I flowing from A to B through $Z = j10\ \Omega$. Then $S_A = V_AI^*$ is the power supplied by A and $S_B = V_BI^*$ is the power received by B.

Current

$$\begin{aligned} V_A &= 200\angle10^\circ = 196.96 + j34.73\ \text{V}\\ V_B &= 200\angle30^\circ = 173.21 + j100.00\ \text{V}\\ I &= \frac{V_A - V_B}{j10} = \frac{23.76 - j65.27}{j10} = -6.527 - j2.376\\ &= 6.946\angle-160^\circ\ \text{A} \end{aligned}$$

Powers

$$\begin{aligned} S_A &= V_AI^* = (200\angle10^\circ)(6.946\angle160^\circ) = -1368.08 + j241.23\ \text{VA}\\ S_B &= V_BI^* = (200\angle30^\circ)(6.946\angle160^\circ) = -1368.08 - j241.23\ \text{VA}\\ Q_{loss} &= |I|^2X = 48.246\times10 = 482.46\ \text{var} \end{aligned}$$

Check: $Q_{loss} = \frac{|V_A - V_B|^2}{X} = \frac{(2\times200\sin10^\circ)^2}{10} = \frac{69.46^2}{10} = 482.46$ var.

Direction of reactive power

• Machine A supplies +241.23 var into the link.
• Machine B receives −241.23 var, so it also supplies 241.23 var into the link.
• So reactive power flows from both ends into the link. There is no net transfer of vars from one machine to the other, because the voltage magnitudes are equal (200 V each). Each machine supplies half of the link's $I^2X$.
• Real power: 1368.08 W per phase flows from B (leading, 30°) to A (lagging, 10°).

Answer: Reactive power flows from both machines into the link, 241.2 var each per phase. The reactive loss in the link is 482.5 var per phase (1447.4 var for three phases).