This website explains the basics of FACTS technologies and how they can be used to exchange active and reactive power with the transmission system, namely:

  • SSSCs - can exchange active and reactive power
  • STATCOM - can exchange reactive power only

But why is this desirable? Why do we want to add or remove reactive power onto the transmission system?

  • 1
    \$\begingroup\$ The short answer is voltage control. By injecting or removing reactive power the voltage can be controlled and kept at an acceptable level. I'll come back and write a thorough explanation, but I'm way too busy at the moment. In a few days hopefully. Good question though! =) \$\endgroup\$ Nov 28, 2014 at 23:14

2 Answers 2


The main reason for reactive power compensation is to regulate the voltage magnitude. Note that the compensation might be both positive and negative (reactive power in, or reactive power out). In a transmission system, there is a strong correlation between reactive power and voltage magnitude, whereas the active power is mainly dependent on the voltage angle. Have a look here for a bit more information.

In the transmission system, a branch might have impedance Z = R + jX, where the reactive X is about 10 times the purely resistive R.

I'm assuming you're familiar with the per unit system. Let me know if you're not and I'll explain it closer.

Let's just review a few basic relationships first:

\begin{align*} S = V \cdot I^*\\ => I = (S/ V)^*\\ \Delta V = I^2\cdot Z\\ Z = (R + jX) \end{align*}

Let's assume we have a very simple power system that looks like this:

G ---|------------------|------------------|----->
     3   Z = R + jX     2   Z = R + jX     1   Load               
  • G is the generator
  • The verticals lines are buses, labeled 1 - 3
  • The load is at the end of the radial.
  • The voltage at bus 1 is assumed to be 1pu with angle 0 degrees.
  • The load is (1 + j0.2) pu. (If S_base = 100MVA, this would be equal to 100MW + 20MVAr)
  • Z = 0.01 + j0.1

The current necessary to supply the load is given by:

\begin{align*} I = (S/ V)^* =((1 + j0.2) / 1)^* = 1 - j0.2\\ \end{align*}

No compesation:

The voltage at bus 2 is give by the voltage at bus 1 plus the voltage rise over the cable (seen from 1 to 2):

\begin{align*} V_2 = V_1 + I^2 \cdot Z = (1-j0.2)\cdot(0.01 + j0.1) = 1.054\angle 5.01 ^{\circ}\;\text{pu} \end{align*}

This means the power injection into the cable between 1 and 2 is:

\begin{align*} S_2 = V_2 \cdot I^* = (1.031 + j0.302) \;\text{pu} \end{align*}

The voltage at V3 is:

\begin{align*} V_3 = V_2 + I^2 \cdot Z = 1.11\angle 9.50^{\circ}\;\text{pu} \end{align*}

Now we can find the power output from the generator by using the first equation:

\begin{align*} S_{Gen} = V_3 \cdot I^* = (1.062 + j0.404) \;\text{pu} \end{align*}

With compensation:

Let us add a capacitor that injects 0.3pu reactive power at bus 2.

The voltage at bus 2 is still given by the voltage at bus 1 and the voltage rise over the cable, so it's still \$\underline{1.054 \angle5.01^{\circ}\;\text{pu}}\$.

Now, the reactive power injection of 0.3pu will give a current injection of:

\begin{align*} I_{inj} = (Q / V_2)^* = 0.285 \angle{-85.0}^{\circ}\;\text{pu} \end{align*}

The current through cable 1-2 is equal to the current through cable 2-3 plus the current injection, so:

\begin{align*} I_3 = I_2 - I_{inj} = 0.979\angle 4.90^{\circ}\;\text{pu} \end{align*}

You see that the current magnitude is lower than it was without compensation. So, let's have a look at the voltage at bus 3:

\begin{align*} V_3 = V_2 + {I_3}^2 \cdot Z = 1.06\angle10.22^{\circ}\;\text{pu} \end{align*}

Now we can find the power output from the generator by using the first equation:

\begin{align*} S_{Gen} = V_3 \cdot I_3^* = (1.037 + j0.096)\;\text{pu} \end{align*}

So, to summarize:

       W/O comp:   W comp:
|V1|   1.000       1.000
|V2|   1.054       1.054
|V3|   1.115       1.060

       W/O comp:         W comp:
Gen    1.062 + j0.404    1.033 + j0.096

As you can see from the above results, the voltage is much more stable with compensation. The current gets lower through the cable resulting in lower active losses.

The reason why the reactive power is needed in the first place is because it accounts for the magnetization of the equipment. If there's no reactive power, transformers, generator rotors/stators, machines etc. have no magnetic field. With no magnetic field, there is no torque, no magnetic coupling in the transformer etc. So, a lot of equipment have to consume reactive power in order to work. If there's too little reactive power available the equipment will try to draw more current to compensate. This will lead to higher voltage drops, which in the end might cause voltage collapse.

As Andy points out, it can also be used as power factor correction for large industrial loads. However when we're talking about reactive power compensation it's most often because of what I've described above.

In a meshed grid it can also be used to control power flow. This works because the active power flow through a cable is mainly given by the voltage angle difference over it. If you inject reactive power, the voltage and currents angles will change, thus it will affect the power flow. If you inject the right amount at the right place you can redistribute the power flow the way you want (but only to a small extent).

Hope this answers your question!

  • 1
    \$\begingroup\$ This is a really great answer! Thank you very much! \$\endgroup\$
    – Blue7
    Nov 30, 2014 at 13:37
  • \$\begingroup\$ @Blue7, glad I could help =) \$\endgroup\$ Nov 30, 2014 at 14:02
  • \$\begingroup\$ Can I just ask, when you say "the voltage is much more stable with compensation", are you referring to the fact that the buses now all have similar voltages, whereas without compensation the buses decrease in voltage as you move away from the generator? I'm asking because in the past I have been taught that stability usually refers to disturbance rejection, steady state tracking of a desired value etc \$\endgroup\$
    – Blue7
    Dec 2, 2014 at 1:25
  • \$\begingroup\$ You are correct, it refers to disturbance rejection too. I forgot to mention it in the answer (was kind of in a rush). I can expand on it later, but I won't be able to do so until mid-December probably. Good comment though. By the way: if you figure it out in the mean time and post it as an answer, I will definitely vote it up, because it's important... \$\endgroup\$ Dec 2, 2014 at 6:01
  • \$\begingroup\$ I lost you at this equation: \$\Delta V = I^2\cdot Z\\\$. Can you explain where this comes from? \$\endgroup\$
    – LShaver
    Feb 26, 2023 at 3:40

I see it like this...

A power transmission system may be connected to a large 100kW load that has a significant reactive element i.e. the power factor is not perfect.

If it were a purely resistive load lets say the voltage is 1000 volts and the current is 100 amps. That equals 100kW. If the power factor were not unity the current taken down the transmission line would be more than 100 amps even though (in a perfect world where only power is metred) the customer is billed for 100kW consumption. Because of this the electricity companies would be somewhat aggrevated by having to lose a little more \$I^2R\$ loss in the cable without call to extract extra money from the bad-power-factor consumer.

If an opposing reactive injection were made, the current can be largely returned to 100 amps and everyone is satisfied.

  • \$\begingroup\$ Thanks for your answer. So you're saying that we inject reactive power to bring the power factor towards unity, and this reduces the current needed to provide the same amount of power to the end user ? \$\endgroup\$
    – Blue7
    Nov 27, 2014 at 23:54
  • 1
    \$\begingroup\$ What you describe is power factor correction. There is more to the story than just saving a little money on I2R losses. I am not an expert, but I believe an important application is controlling power flow, i.e. transferring load between transmission lines without physical switching. \$\endgroup\$ Nov 28, 2014 at 4:41

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