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.
The electrical technician has suggested installing a 3 phase capacitor
bank to compensate for the “Reactive power” produced by the motors
because if not, I may be penalized economically by the local
electrical company.
That's all fine and dandy if you are running your appliances all together 24 7 but, in the dead of night, ask yourself, what is that bank of capacitors doing?
Answer - they are taking reactive current that they are hoping would balance the reactive current taken by the motors BUT the motors are not switched on SO, if your utility company can bill you extra for having a poor lagging power factor then they surely will bill you for having a poor leading power factor for all those hours that the laundry is shut.
Maybe the electrician has in mind some clever caps that automatically switch in and out or maybe he means installing several sets so that each only activates when their particular motors start turning? Either way, seek clarification from him and if he falls down at the first hurdle then fire him.
Best Answer
I'll offer an intuitive explanation instead. The calculations would seem to depend on specifics – maybe if you sketch a circuit that illustrates your situation it would be easier.
You have a single phase power system from the utility. You want to dump PV energy energy onto the power line but also regulate it to standard line voltage.
How do you regulate voltage? Well, you can either dissipate energy by dumping power into a resistor, or store the energy in a reactive load. Clearly energy storage will be more efficient than dissipating the excess as heat, but the problem is that it takes time to store energy (e.g. to charge a capacitor).
So if there is a sharp transient you may not have enough to store the excess energy quickly enough. On the other hand, you can start dissipating energy in the time it takes to switch a transistor.
If there is a transient and the X/R ratio is low (resistive) then energy can be stored. If it is high (reactive) then energy storage of the transmission line will create a DC offset, so you need to dissipate real power.