It is a long question, but better than a short one, as you've shown your own research.
1) Solar cells. If you're stacking your own ones, stack 9 of them and get the 4.5V of the original circuit.
2) Battery charging. Batteries are the only thing you've left out of your spec. This is an area where the circuit design relies on cutting a lot of corners. In theory it might be out of spec, if you were to put 4.5V at 280ma through AA NiMH cells indefinitely. In practice, you don't get full sun all day, you'll be using it indoors, and you're not going to get optimal power transfer from the cells, so this isn't going to cause problems.
3) Diode. It's just a regular diode, not a zener. Current through it is actually determined by the battery and right hand side circuit, not the solar panel - the transistor is off when the panel is generating electricity. The original 1N914 will be fine. 1N4004 will also be fine.
4) Resistors: not a precision component here, use whatever meets your cost constraint. 5.1k for 5k is fine.
5) Wire: not critical. Your ebay link looks suitable. Thinner is better for the toroid.
6) Transistors: stick with the exact part numbers. Design may rely on specific parameters.
7) LED: again, this circuit relies on cheating. Normally a white LED won't run from two NiMH cells. The joule thief part provides a boost converter that gives small pulses of higher voltage. It doesn't have the capacity to provide a lot of current at that voltage. In combination with the pulsing this means there should be no risk of damaging it.
(A proper analysis of this circuit would be good, if nobody else supplies one I'll do it in a few days).
You seem to think that you have created a problem, but not in this case. The 1A will flow through current source, there will be 1V across the voltage source. The current source will have the voltage over it that makes this happen, likewise the voltage source will have the current through it to make this happen (which must be 1A).
1A flows through the resistor too. Let's (arbitrarily) define its lefts side as 0V, then its right side must be at +100V, hence the voltage over the current source is 99V.
If you invert the current source the right side of the resistor is at -100V, hence the voltage over the current source is -101V.
Note that voltage sources and current sources are theoretical constructs, just like a line, circle and point. You can create 'impossible' diagrams with them, for instance a shorted voltage source or an open current source, or voltage sources for different voltages in parallel. It makes no sense to ask 'what would happen in such a case' because those idealized components do not exist. We can calculate with them within certain restrictions, outside those restrictions we can't calculate with them. That's all.
Best Answer
In a 2 wire system the only thing you can do is make the total wire so long that it either acts inductively or resistively against your system. In the case of resistive it makes absolutely no difference where the wire is and in the case of inductive the total length or discrepancy therein is of much smaller interest than the path those wires take. For there to be a mile-long difference, you're likely coiling one up, which is much more interesting in that respect.
For three phase systems that directly use the phase synchronicity, such as fixed armature three phase motors, you can disrupt it all if you can delay one of the phases by enough.
Again, however, the speed of light being 3*10^8 m/s and the speed of
electronselectricity [credit:Andrew] in free hanging copper being approximately 95% of that, you would need a huge deal of length to get enough displacement, say 10%-ish (rough guesstimate - too lazy to really think heavily on that number) of the 60Hz wave (or in my case 50Hz), to start noticing it. 10% of a 60Hz wave would take about1.68μs1.68ms [credit: Jon].In
1.68μs1.68ms the wave in copper would have travelled (approximate speed of electricity in copper) * (time) = (3*10^8 * 0.95)m/s * 1.68ms =478m478km.Seems
feasible, right?quite ridiculous already...Well, to go on... Let's, for the ease of it, assume that my 10% was a correct guesstimate and it isn't 30%. Let's also assume that the other wires are all connected directly to the source and thus have no losses at all (good luck with that one), then still:
A 3 phase system usually needs at least 10A per phase (and that's tiny!), if you'd want to limit the total resistive losses to 5%, you'd need a resistance smaller than, if we assume an RMS voltage of 240V for your 120VAC 60Hz situation, of less than 12V/10A = 1.2Ohm. And losing 12V of your RMS is quite a lot to power engineers! To get 1.2Ohm at 478km of wire, and I won't throw around another formula, as when looking for the exact resistivity of copper I found this, which shows me for 47897000cm (= the value I got a bit more exactly, which I rounded to 478km) I'd need a wire diameter of 92.335mm. That's an area of (0.92335 / 2)^2 * pi = 0.67 dm2. I'm using dm, or 1/10th of a meter, for a reason. 47897000cm = 4789700dm. So the volume of copper you need is 0.67 dm2 * 4789700dm ~= 3207235 dm3.
1dm3 is the same as 1 liter. So that's 3207235 liters of copper. Copper weighs about 8.93kg per liter, so that's about 28640616 kg of copper. At currently about €5.30 per kg, that comes to € 152 million!! With only about €10 you can buy an inductor to screw things up just as bad.
Which brings me to the next problem. 478m of copper wire, diameter 2.9201mm, laid out in a perfect loop, without coiling it, from anice site with a calculatorwhich I did not verify as I am still lazy, is close to 1mH. So in the most efficient way you could place that wire you already get 1mH, assuming the calculator is right. Placed on a normal type of floor which is sort-of-grounded I suspect it to be higher. This may, depending on the motor, already be enough of an inductor to put things out of whack. But if you then want to put that wire in an orientation so that you don't need a HUGE amount of space to make a perfect circle, the inductance will increase significantly.The calculator stops working at about 5km, but with the numbers I entered I'd say it's safe to say we get in the range of actual Henris, not mili or micro.
At the time of writing €1 is approximately $1.08
Note: I made a mistake in my simplification steps to get 5% of RMS 240V in a three phase means 1.2 Ohm, but since it's all blue-sky I'm not going to re-do it.
Another note: Just calculated "back-of-envelope"-style, because my brain wouldn't let it go, that if you'd want the wire to be coiled such that it could only just stand up in my living room
you'd need 77 windings, making an inductor of 74mH, assuming no metals interfere. (And it's a largely concrete living room).you'd need many of my living rooms for the immense length the coil would take, making that a futile exercise.