> VOLTAGE is to electricity what pressure is to water: the more you have of it the more oomph you get. That is why electrical power lines work at high voltage.
No. Losses in transmission lines are proportional to current squared [1]. High voltage is used because it results in lower current (for the same power), so lower losses.
Idk, that's how I learned it growing up before I learned about stranded vs solid core wire, skin effect, resistance, etc.
Power delivered by the battery is this equation:
P = I * V
So you can have (1 volt @ 100 amps), or (100 V @ 1 amp) and still get (100 watts). If both are the same, wouldn't the low voltage solution be safer, and more desirable?
Massaging the same equation a little bit, we can model our losses along the power lines. Substituting V = IR you get...
P = (I^2)R
It's difficult to make conductors less resistive without using expensive metals or unconventional techniques (like cryogenically cooled wires). So to cut our losses, we'll go with high voltage, low current.
> VOLTAGE is to electricity what pressure is to water: the more you have of it the more oomph you get.
... or about the claim that follows:
> That is why electrical power lines work at high voltage.
The former sentence (aka the hydraulic analogy, [1]) is a very well-established thing. It's not a perfect analogy, but it's quite reasonable. The latter (high voltage power lines) is demonstrably not correct, as the two of you have pointed out.
Extending your power formula to the hydraulic world and simplifying the calculus to steady-state algebra:
Power = (Volumetric flow rate) * (pressure)
Though, to be fair, the analogy breaks down a little there; the Poiseuille equation [2] (the fluid equivalent of V = IR) involves 6 variables instead of 3. But the point is, you still end up with a final equation of similar form to P = (I^2)R.
[1] https://en.wikipedia.org/wiki/Hydraulic_analogy
[2] https://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equat...