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Why does reactive power influence the voltage? Suppose you have a (weak) power system with a large reactive load. If you suddenly disconnect the load, you would experience a peak in the voltage.

Is there a good explanation why this happens?

Edit: Unfortunately I'm no longer in academia, and therefore don't have access to all the same sources as I used to have. I work with this on a daily basis and wrote my Master's about a closely related subject and I'm therefore quite certain my statements are correct. I see that this might not be common knowledge for electronics engineer, but I believe people in the power transmission industry are familiar with what I'm describing.

For those interested in why voltage level and reactive power closely related from a reliable source, here is the original paper describing the Fast Decoupled Load Flow algorithm (you need access to IEEE):

"Stott and O. Alsac, “Fast decoupled load flow” IEEE Trans. on PAS, vol. 93, no. 3, pp. 859-869, May/June 1974"

See also page 91 in this textbook by Wood / Wollenberg on

A quote from the Roger C Dugan, the author of this textbook on Electrical Power Systems:

Reactive power (vars) is required to maintain the voltage to deliver active power (watts) through transmission lines. Motor loads and other loads require reactive power to convert the flow of electrons into useful work. When there is not enough reactive power, the voltage sags down and it is not possible to push the power demanded by loads through the lines.

I believe the edit history might be interesting for anyone wondering what the edit and all the comments are about.

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As a power electrical engineer, this is a valid and interesting question. (Admittedly, I do not know the answer of the top of my head, and I will have to do some research.) – Li-aung Yip Jul 3 '14 at 14:22
Related: static VAR compensators (devices which inject or consume reactive power at substations, in order to control transmission line voltage) and the general concept of reactive power compensation. – Li-aung Yip Jul 3 '14 at 14:30
I have cleaned out the comments here because I believe the major complaints have been addressed. – W5VO Jul 3 '14 at 14:43

2 Answers 2

up vote 4 down vote accepted

Why does reactive power influence the voltage? Suppose you have a (weak) power system with a large reactive load. If you suddenly disconnect the load, you would experience a peak in the voltage.

First, we need to define what exactly is being asked. Now that you have stated this is regarding a utility-scale power system, not the output of a opamp or something, we know what "reactive power" means. This is a shortcut used in the electric power industry. Ideally the load on the system would be resistive, but in reality is is partially inductive. They separate this load into the pure resistive and pure inductive components and refer to what is delivered to the resistance as "real power" and what is delivered to the inductance as "reactive power".

This gives rise to some interesting things, like that a capacitor accross a transmission line is a reative power generator. Yes, that sounds funny, but if you follow the definition of reactive power above, this is all consistant and no physics is violated. In fact, capacitors are sometimes used to "generate" reactive power.

The actual current coming out of a generator is lagging the voltage by a small phase angle. Instead of thinking of this as a magnitude and phase angle, it is thought of as two separate components with separate magnitudes, one at 0 phase and the other lagging at 90° phase. The former is the current that causes real power and the latter reactive power. The two ways of describing the overall current with respect to the voltage are mathematically equivalent (each can be unambiguously converted to the other).

So the question comes down to why does generator current that is lagging the voltage by 90° cause the voltage to go down? I think there are two answers to this.

First, any current, regardless of phase, still causes a voltage drop accross the inevitable resistance in the system. This current crosses 0 at the peak of the voltage, so you might say it shouldn't effect the voltage peak. However, the current is negative right before the voltage peak. This can actually cause a little higher apparent (after the voltage drop on the series resistance) voltage peak immediately before the open-circuit voltage peak. Put another way, due to non-zero source resistance, the apparent output voltage has a different peak in a different place than the open-circuit voltage does.

I think the real answer has to do with unstated assumptions built into the question, which is a control system around the generator. What you are really seeing the reaction to by removing reactive load is not that of the bare generator, but that of the generator with its control system compensating for the change in load. Again, the inevitable resistance in the system times the reactive current causes real losses. Note that some of that "resistance" may not be direct electrical resistance, but mechanical issues projected to the electrical system. Those real losses are going to add to the real load on the generator, so removing the reactive load still relieves some real load.

This mechanism gets more substantial the wider the "system" is that is producing the reactive power. If the system includes a transmission line, then the reactive current is still causing real I2R losses in the transmission line, which cause a real load on the generator.

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Thanks Olin! I didn't know the distinction between active and reactive power was uncommon for others than power system engineers. You are absolutely right in your answer, and I guess you answered my question. However, in the case I'm looking at, the load shedding is faster than the generator regulators. When the control system reacts, the voltage is reduced to its nominal value again. Anyway, +1 =) – Stewie Griffin Jul 3 '14 at 14:31
@Robert: This is exactly the kind of assumption that is missing in your question, which is why writing a answer can be a waste of time. Earlier you had several more implied assumptions. I tried to answer when you eliminated some of them. See how assumptions can waste everyone's time, and why questions relying on them should be closed? – Olin Lathrop Jul 3 '14 at 14:43
My apologies for being unclear in the question. I tried to phrase it as well as I could, and include the necessary information. I agree that all the comments going back and forth might have been a waste of time. However, I have to disagree with you if you think writing this answer was a waste of your time.I believe you just provided a very high quality answer that explains the mechanisms in a great way. I believe you just provided a lot of information that IMO belongs on a site like EE.SE. It benefits me, but more importantly future visitors which, if I'm right is the purpose of the site! =) – Stewie Griffin Jul 3 '14 at 14:51
I think Olin is essentially correct - the transmission line has an inductance, and Ohm's Law says that there will be a voltage drop across such an inductance. The wording about 'reactive power' is really talking about this voltage drop. You can counteract the inductance by adding some capacitance, which is essentially what a static VAR compensator does. Note: I have only researched this to a shallow level and will need to check some resources at work (though we are very busy right now, so don't hold your breath.) – Li-aung Yip Jul 3 '14 at 15:03
@Yip: Ohm's law states that there will be a voltage drop across a resistance proportional to the current through it. I believe it was Faraday and Henry who worked out the particulars for capacitance and inductance under the influence of AC. (The capacitors and inductors, not Henry and Faraday) – EM Fields Jul 3 '14 at 17:11

"If you suddenly disconnect the load, you would experience a peak in the voltage." I suggest you look up the Ferranti effect. When you remove the load you are essentially creating a lightly loaded line.

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Can you elaborate on this a bit? Link-only answers are discouraged. – Adam Haun Oct 19 at 5:50
This isn't even a link-only answer... – Null Oct 19 at 13:34
@Null: link added. – Dave Tweed Oct 19 at 14:27

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