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?

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 79 in this textbook by Wood / Wollenberg on books.google.

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.

  • 4
    \$\begingroup\$ 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.) \$\endgroup\$ Jul 3, 2014 at 14:22
  • \$\begingroup\$ 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. \$\endgroup\$ Jul 3, 2014 at 14:30
  • \$\begingroup\$ A much more accurate answer is that a motor produces reactive current from excitation of coils or +VAR's and someone must compensate with -VAR's (e.g. cap series or shunt) to reduce apparent power increase. tinyurl.com/y9zmovut \$\endgroup\$ Apr 8, 2018 at 19:30

3 Answers 3


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.

  • \$\begingroup\$ @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? \$\endgroup\$ Jul 3, 2014 at 14:43
  • \$\begingroup\$ 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.) \$\endgroup\$ Jul 3, 2014 at 15:03
  • \$\begingroup\$ @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) \$\endgroup\$
    – EM Fields
    Jul 3, 2014 at 17:11
  • \$\begingroup\$ @EMFields: we work with some simplifying assumptions in power engineering. We assume constant frequency (ω = 50Hz or 60Hz) in which case a given inductance in Henries converts into a given number of ohms, given as X [Ω] = j × ω × L. The impedance of a transmission line then becomes an imaginary number of ohms (i.e. Z = j10 Ω) and you can do Ohm's Law using complex numbers, to determine the complex-number voltage drop - V = I* × Z. (I neglect the resistive part of the impedance which is much smaller than the inductive reactance.) This seems wacky, but it accurately models what we observe. \$\endgroup\$ Jul 4, 2014 at 18:55

Consider the source impedance of the weak power system has both a resistive and reactive component (i.e. an "ideal" voltage source in series with an RL combination). Just as a resistive load will form a "voltage divider" with the source, a reactive load will do the same. By applying the standard voltage divder rules to complex impedances, the reason for the observed result (greater voltage drop with inductive loads than with purely resistive) becomes clear.

To put it another way, there are two ways to get more current out of a reactive source impedance - one is to increase the voltage drop, the second is to increase the phase shift across the inductive component. Adding a reactive load with the same "sign" of complex impedance reduces that phase shift (as the resulting AC current in the system produces a voltage at the load more in-phase with that of the "ideal" component of the source), so the voltage drop across the source impedance must increase to deliver the same load current.

The other interpretation i make of the question relates to transients, when a large current passing though an inductor (all wiring has an inductive property) is interrupted, the collapsing magnetic field induces a voltage rise in the inductor proportional to di/dt. This creates a transient peak at the load for a fraction of a cycle, however if there is significant capacitance in the system, ringing (oscillation) can occur which spreads out the transient over a few cycles. These transients make the switching of heavy inductive loads a design challenge.


"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.

  • 1
    \$\begingroup\$ Can you elaborate on this a bit? Link-only answers are discouraged. \$\endgroup\$
    – Adam Haun
    Oct 19, 2015 at 5:50
  • \$\begingroup\$ This isn't even a link-only answer... \$\endgroup\$
    – Null
    Oct 19, 2015 at 13:34
  • \$\begingroup\$ @Null: link added. \$\endgroup\$
    – Dave Tweed
    Oct 19, 2015 at 14:27
  • \$\begingroup\$ Some people have no learning , or surfing skills \$\endgroup\$ Apr 8, 2018 at 19:25

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