Power through a line
The key idea
Everything a power flow computes reduces to one picture: two voltage phasors with an impedance between them. The angle between the phasors decides how much active power crosses, and in which direction. The difference in their lengths decides the reactive power. And whatever the impedance keeps for itself is the loss.
The idea
Every branch in a network has two ends: the sending end at voltage Vs and the receiving end at Vr. Each voltage is a phasor, with a magnitude and an angle, and the branch impedance R + jX sits between them. That is the entire problem.
The difference between the two ends is what forces the current: I = (Vs − Vr) / (R + jX). Two identical phasors give no flow at all. Every number a study reports — the flow, the loading, the loss — comes out of that one subtraction.
Now, there are two ways to make the phasors differ, and each has its own effect. You can rotate one relative to the other, opening an angle difference δ, or you can make one longer, opening a magnitude difference. On a transmission line the two act almost independently, and that independence is the most useful fact in this lesson.
Take the angle first. On a transmission line the reactance dwarfs the resistance, and active power comes out as P ≈ VsVr·sin δ / X. The magnitudes barely matter: both are near 1.0 per unit, multiplied together. Active power follows the angle, and the leading end sends it. Tip the sending end forward by 10° and megawatts appear; tip it backward and the same line imports instead. Nothing physical about the line has changed — only the order of the two angles.
Now the magnitudes. Reactive power comes out as Q ≈ (Vs² − VsVr·cos δ) / X, and for small angles the cosine is nearly 1, so this is mostly Vs(Vs − Vr) / X. Reactive power follows the magnitudes, and the higher-voltage end sends it. Raise a generator's terminal voltage by one percent and it exports vars; its megawatts hardly move.
So the angle and the magnitude are two separate controls with almost no crosstalk. Engineers call this decoupling, and it runs deep: it is why voltage control and dispatch are separate problems, and why a solver can exploit the split to converge faster. An operator who needs MW moves the governor; an operator who needs Mvar moves the excitation.
The impedance takes its share on the way through. The resistance turns |I|²R into heat, so less active power arrives than left. The reactance absorbs |I|²X of reactive power, and that absorption grows faster than the loading does. A heavily loaded line therefore drags the voltage down at both ends, and the reactive flows at its two ends often point in opposite directions: both ends are feeding the reactive power the line itself absorbs.
Distribution breaks the clean split. On a low-voltage feeder R and X are comparable (R is sometimes the larger), and the two couplings mix. Active power now moves the voltage noticeably: a street of rooftop solar exporting on a sunny afternoon pushes the local voltage up, where the same export on a transmission line would be almost invisible.
Try it
Start with equal magnitudes and δ at +10°. Drag the angle through zero and watch both power arrows turn around. Then set δ back to zero and move the sending voltage instead.
every value in per unit
P sent
1.736
P received
1.736
P loss
0.000
Q sent
0.152
Q received
−0.152
Q absorbed
0.304
Active power crosses from the sending bus to the receiving bus. Reactive power enters the line from both ends, and the line keeps it.
This is a single-phase per-unit model. The receiving end holds 1.00 pu ∠0°, and X is 0.1 pu. The model has no shunt capacitance, so nothing on this line generates vars. Move the angle, and P moves. Move the sending voltage, and Q moves. Turn up R/X to model a distribution feeder, and the two controls are no longer separate.
Why it matters
- Flow direction is a consequence, not a setting. Nobody tells the power which way to go. Dispatch and voltage set the angles, and the angles set the direction, which is exactly what a power flow study solves for.
- Voltage support is a reactive-power problem. A shortage of vars collapses the magnitude gradient and the voltage sags. Capacitors, generator excitation and tap changers all work on the magnitude difference, not the angle.
- Losses grow with the square of the current. Double the flow and the loss quadruples. That is why the same MW is cheaper to move at a higher voltage, where the per-unit current is smaller.
- On a feeder, the couplings mix. With R comparable to X, active power exports raise the voltage, and the voltage limits then decide how much generation a feeder can host.
The math, if you want itOptional — the page reads completely without it
Nothing here is an approximation to start with. The whole branch is Ohm's law and S = V·I*:
the exact relations
I = Vs − VrR + jX · Ss = Vs·I* · Sr = Vr·I*
Now set R to zero, a fair approximation on transmission, where X is many times R. The two flows separate into the classic pair:
active power, R ≈ 0
P ≈ Vs·Vr·sin δX
reactive power at the sending end, R ≈ 0
Qs ≈ Vs² − Vs·Vr·cos δX
P depends on δ through a sine: roughly proportional to the angle when it is small, and reversing sign with it. Q depends on δ through a cosine, which is flat near zero, so small angle changes barely move Q, and the magnitudes control it. That is the decoupling, written out.
The impedance keeps exactly the difference between the two ends, because Ss − Sr = (Vs − Vr)·I* = |I|²·Z:
what the impedance keeps
Ploss = |I|²·R · Qloss = |I|²·X
Both are positive whenever current flows, so a series branch can only consume reactive power. Real lines also carry shunt capacitance, which generates reactive power in proportion to the voltage squared: a lightly loaded transmission line is actually a net var source. This model has no shunt branch, so it cannot show that effect.
See it in Phasor
In Phasor, every branch result is this calculation. The study reports P and Q at both ends of each branch, plus the loss between them: a flow that leaves one bus at 12 MW arrives at the next at 11.8 MW, and the difference is exactly what the impedance took. A sign change tells you the angle order flipped.