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In the simplest case, and some unmentioned assumptions, the minimum charging time would be given by the formula already mentioned: $$T = \frac{V \times Ah}{W}$$ this gives 12x150/25 = 72 hours. However, this case makes at least two major assumptions, the battery is fully discharged and there are no losses of any kind! In the real world, the battery ...

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There would be real problems if as you said " run a really long wire from the array to the ground" if you did that you would have alot of voltage drop. There is a significant voltage drop even for a run of wire of 100 ft. But if you put a pv system on say a mountain top with few sun obstructions and gathered the power there. But you would have to live ...

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I come across this cause I am working on this problem. I see long talk and they don't know the answer. Yes, the energy is lost, when all systems are fully charged and outside the sun is still shining. No meter PWM or MPPT charging system. You just cant put more into batteries when its full. Full means full. Solution is ultracapcitors. Assemble a capacitor ...

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Light gets "trapped" in an optical fiber when the light travels such that the angle of incidence when hitting the surface is low enough that the optical index change causes the light to "bend" back into the fiber. This is known as total internal reflection. Basically with larger fibers you'll get light bleeding out. Thin fibers work over long distance ...

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