a somewhat informal enquiry regarding a more efficient distribution in the upcoming smart grid.

above is a great illustration i found on fb, and i totally dig it - but lately i have been more puzzled by https://da.wikipedia.org/wiki/faradays_induktionslov, which indicate an induced voltage of tesla * no of turns in a circuit * circumference in meter * sinus ø / seconds for magnet to pass area of circuit, even if the distance of a passing magnet to a given circuit is not mentioned, since faradays cookie jar was loaded by a touching swing wheel. most apparently, all them tesla in the magnets, are to multiply by the relative permeability of atmospheric air / distance r^2, in meter.

so, if difference is voltage, current is amp, also measured in coulumb, or amp per sec = amp/sec, and resistance in the wire is ohm, or ohm per m3 in a given material, then according to ohms law of v=a*o, or a=v/o, a given generator with 96 spoles & 64 magnets on either side, spinning at some 50 rpm, churning a difference of 50 volt & 69 amp at 50 hertz from every evil eye ball, could serial connect & make a perfect 3 phase sine curve wave with a peak difference of + to - 1.600 voltage, or an average line phase of 1 kv, into all 3 wires down the circuit - with the much fabled difference between any 2 given wires being max line voltage / 1,6077 * square root of 3.

this is all neat, and i worked on it real hard for a long time, so i am not that smart here in lilly dk - but my inquiry relates to differentiate & resistance in a given wire. cupper has a resistance of 1,724*10^-8 ohm per m2 per meter, and alumin 2,640*10^-8 ohm per m2 per m, which is quite easy on the intuition to most people - since the electrons (e-) in the atomic grid of the cupper are more loosely bound, than the electrons in the alumin, the first is quite the better conductor.

now you dont want to toast your wires in just about any configuration, so even if you dont run an undercover russian spy network, you check the chart American_wire_gauge over on wikipedia, to see if you are passing more amps through any given wire than it can handle -- which lead to my out-standing question; will a differentiate in voltage, say ac 1 kv at 50 hertz, run more amps back & forth a very thick line, in relation to a circuit with a somewhat more thin line?

a 100 km cupper wire, with a width area of (ø/2) ^ 2 * pi = 2.500 mm2, considered a 115 kv line by some, has a resistance of about 0,69 ohms, since 1,724*10^-8 ohm/m3 * 2.500 mm2 / 1.000.000 mm2 per m2 * 100.000 m equals 0,69 ohm, whereas a line of similar dimensions made of alumin, has a resistance of 1,06 ohm. so according to ohms law, more amps will flow in a thick, irt to a thin circuit, made of cupper or alumini, since a = v / o, right?

i.e. more efficient & cheap amps down the line, for the same voltage or differentiate churned in.

on a similar note, given the above, going from 230 to 120 voltage over at the consumer side, should save us almost 50% on the power production, here on the poorly resourced continent...

ps: the slide from facebook implies a certain confusion between current (amps) and effect (watt), where watt = voltage * ampere, which in si units may also be expressed in joules per second.

a certain differentiate (voltage) over the entire circuit of the 100 kilometer line with 0,69 ohm in resistance, would create a wave, or a current, of x voltage divided with 0,69 ohm = y ampere.

this current will hardly leave any effect within the wire, unless the differentiate or voltage turn so high, that the entire line would begin to heat up & glow like edisons lamp - or your oven bar.

however, before the potential kva from mains, meet the load w at the consumer side, the voltage is stepped down in a trafo with primary & secondary windings, to match a local requirement.

now how will amps, drawn from rho + secondary windings, effect kva over on primary & mains?

. . .

since ameri-joe is keen to shoot first, plough the fields afterwards, and rarely measure the results, they might nevertheless take an interest in considering the fuze level, in fundamental charges.

https://en.wikipedia.org/wiki/American_wire_gauge

from the more pragmatic chart of measurements above, the following correalance does appear;

a cupper wire with ø 10 mm, has a fuze level of 2.700 amps, one with ø 1 mm goes toast at 90 amps, while a cupper string with a diameter of ø 0,1 mm, would call it quits at some 3 ampere.

by some odd law of nature, this is an ampacity of factor 30, every time ø subdivides by 10 - but more interestingly, you can measure the fuze level, in fundamental charges, instead of ampere.

since the fundamental charge, ev, or e-, of a single electron is 1,602176634*10^-19 coulomb, or amps per second, you may divide the fuze level of an ø 1 mm cupper wire, with the above a/s,

to realize an ø l mm cupper wire, has a fuze level of 5,617358167*10^20 fundamental charges.

. . .

of course my intention is to spin loads of the cheapest ev's available into the grid, and get paid a premium for it - but if the consumer goes broke, I will be met by a hoarde of angry off-takers.

so; how many amps do you prefer to have jiffin down the line for all them monkey voltage?

with somewhat impatient regards,

mr gulbenkian 2.0 from the gutter.