Electrical – What happens to the Real and Reactive power when I increase the excitation current in a 3-Phase Synchronous Generator

"Load" is a term for how hard the device has to work to power whatever is attached to it. "High load" means the output power is high, "low load" means the output power is low. You can increase the load by, say, connecting a second device to the output of the generator. Assuming the voltage is constant, the current will be higher (since there are now two deices powered by the generator). That means that the equivalent resistance of both deices in parallel will be lower than that of a single device.

So, increasing the load means decreasing the resistance of the load. This may be counter intuitive, but usually devices are not rated in resistance - they are rated in input current or power, so the "load" follows that. If I have a 1kW load attached to the power supply and now increase it to 2kW, the current will increase (assuming the voltage stays constant).

Electricity doesn't travel infinitely fast so, at some distance down a cable, the voltage will be delayed by a small amount with the delay getting bigger as distance increases. A fixed delay = some amount of phase shift.

I know this is for much higher frequencies but the dielecric material in a cable affects velocity of propagation of electricity at any frequency but, it's more noticable at higher frequencies in a lab using an oscilloscope: -

Basically it's telling you that the wavefront of a signal travels down a cable at speeds slower than the speed of light when the relative dielectic constant is greater than 1.

For a very long power cable you will see what appear to be phase shift along the cable but this is due to the velocity of propagation not being infinite. Here are the simple formulas that describe the speed of light and velocity of a wave down a cable: -

This is a useful site for learning.