Perhaps it would be better to say "direct current is beneficial to hydroelectricity plants".
Wikipedia shows several hydroelectric power plants use HVDC to transmit power, including
Volga Hydroelectric Station,
Tianshengqiao-I Hydropower Station,
Itaipu Dam,
Cahora Bassa,
Inga Dam,
Tianshengqiao-I Hydropower Station,
the various hydro plants that feed into the Radisson Substation,
the various hydro plants that feed into the Nelson River Bipole,
Benmore Dam,
Sakuma Dam,
Three Gorges Dam,
etc.
Why would using direct current at a plant be beneficial?
Wikipedia: HVDC has a good explanation of why HVDC (sometimes) is better than AC. In summary:
With a given long-distance transmission line, HVDC has fewer losses than high-voltage AC. Both have roughly the same parasitic resistance losses, but only AC has losses due to parasitic inductance and parasitic capacitance.
With a given long-distance transmission line, HVDC can transmit more energy per minute. HVDC can transmit at maximum power all the time (the maximum voltage and maximum current supported by the line), while AC has zero crossings where a line is not transmitting any power.
HVDC is immune to some kinds of failures that occasionally hit AC systems, such as loss of synchronization.
The main disadvantage of HVDC is the cost of inverters to convert power to 60 Hz AC.
So these HVDC systems are point-to-point, with inverters only at the two ends, rather than a heavily branching structure which would require many inverters, one at each endpoint.
As the cost of inverters continues to drop, I expect HVDC to be used more often.
As far as I can tell, all these advantages would also apply to pumped-storage hydroelectricity facilities.
Your maths is wrong.
100 Watts generated by solar power at 12V.
Your 8.33 amps is the correct calculation.
The mistake you have made is assuming that this current stays the same value when added to the 110V source. You cannot just connect the systems. To feed the 12V (DC) into the 110V (AC) you need to convert them to the same thing - usually the 110V AC. This means using some form of converter (basically a transformer with the DC turned into AC by transistor switches). The POWER OUT will always be less than the POWER in so that at 110V the current will be less than 0.909 Amps NOT 8.33 Amps. (Vin * Iin >= Vout * Iout)
Best Answer
In an AC system the mean voltage and the mean current are both zero but just because the mean of a quantity is zero doesn't mean that the quantity doesn't exist.
As an analogy consider two hydralic cylinders connected together by a pipe. I push one cylinder in and pull it out again. There was no net flow in the pipe but there was clearly flow when I was pushing the cylinder in and also flow in the opposite direction when I was pulling it out. If I place a load on one cylinder and move the other up and down i'm clearly using the flow of water to transfer power even though there is no net water flow.
Getting back to electricty consider a resistor connected to a sinewave AC power supply. For simplicity lets assume that our voltage source has a peak voltage of 1 unit and our resistor has a resistance of one unit. We can write equations for the instantaneous voltage, instantaneous current and instantaneous power.
If we calculate the mean values of those functions we quickly see that V and I have a mean of 0 but P doesn't, it has a mean of 0.5.
So if our mean voltage and mean current are zero how do we meaningfully measure voltage and current? We use something called the "root mean square" value. That is we take the voltage or current waveform, square it, take the mean and then take the square root. Voltages and currents in AC power system are nearly always measured and discussed in RMS values.
We can use these RMS voltage and current values in the equations P=I2R and P=V2/R equations to calculate average power delivered to a resistive load.
It's tempting to also think we can apply RMS values to the P=IV equation but it is important to remember that this only applies if the voltage and current waveforms are proportional.
Further reading: https://en.wikipedia.org/wiki/Root_mean_square#Average_electrical_power