If lightning were to strike power lines, wouldn't you expect it to increase the power, and hence your room lights get brighter?
Why do they go dim for 1 second at a time?
Electronic – Why does power dip during a lightning storm
powerpower electronicspower-engineeringpower-generation
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Yes it surely is. People divert lightning all the time using fine wires dragged up into the sky, as user30997 suggested.
The Doc already knew approximately when the tower was going to be hit, so all he had to do was to give the lightning a nice length of wire to follow.
The mistake they made though was trying to get the car to touch the wire at the exact moment the lightning struck. As you pointed out, they'd have to get the timing superhumanly accurate.
What they should have done was have a spool of fine wire in the car, paying it out as they drove away from the clock tower. That way they'd only have to make sure that the car was travelling at 88mph when the lightning struck. This would still have required timing accurate to a few seconds, and I don't know if the clock had a second hand.
Personally, I think he should have just carved out a nice little niche for himself as the inventor of Rock and Roll.
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.
Best Answer
Janka comes close, but there are are several more details. (Note, recalling from EE classes about 45 years ago.)
On many high voltage lines there are arc electrodes at various points. When lightning strikes the line, the increased voltage causes an arc to form across the electrodes. This helps to dissipate the voltage of the lightning strike.
But the spacing of the electrodes is such that, once the arc has formed, the normal voltage on the line is sufficient to keep it going. So the electrodes are formed in the classical V shape, close together at the bottom and wider at the top.
Heat causes the arc to rise (see "Jacob's ladder") and so it gets longer. Eventually, when it gets to the top, the voltage will (hopefully) no longer sustain the arc.
If the arc does not extinguish itself, eventually a nearby over-current detector (fuse) will trip and cut off the power.
But to save the lineman a trip to reset it, the over-current detector is often designed with a timer so that it resets itself after a few seconds. But usually there's a limit to how many times it will reset (in case the over-current condition is due to, eg, a downed power line).