Electronic – Series and Parallel together for 3 Solar Panels

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).

Since in electrical the above equation P= V*I is possible mathematically, is it also possible practically ??

Basically yes. That equation exists because it matches real world observations. However, you must also account for other equations that match real world observations. Here's one that is especially relevant in this context:

$$ P = I^2 R $$

This is Joule's first law. If you have 1205A, R better be very small. R here is the resistance of the wire carrying your 1205A. Let's say it's just \$1m\Omega\$. Then:

$$ P = (1205A)^2 \cdot 0.001\Omega \approx 1452W $$

That's an awful lot of power to be wasted just in the wire. It's probably getting pretty hot.

It's usually easier and more economical to make wires with strong insulation that won't break down under high voltage than it is to make wires with low resistance, since this requires large amounts of expensive metals. Thus, when people have a lot of electrical energy to transfer, as a power plant does, they tend to use high voltage rather than high current.

Regarding the calculation of how much power you can get out of some number of solar panels, you need not make the math so complex. If one panel gets you 250W, then two gives you twice that much. I'm not sure how you are "combining in series and parallel", but to get 518750 watts, you need, one way or another:

$$ \frac{518750W}{250W} = 2075 \text{ panels} $$

If you connect the panels in series, then their voltages add, but the current remains the same. If you connect them in parallel, then the currents add, but the voltage remains the same.

And after converting direct current to alternating current will there be a power change ?

There is nothing that makes AC or DC inherently have more or less power than the other. However, any way you might realize such a conversion will involve some losses, so the output electrical power will be somewhat less than the input power, with the difference likely converted to heat in your conversion device.