Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
A farmer plans a solar panel for a small water pump. Each cell has an area of . In sunlight of , a cell has a short-circuit current of A and an open-circuit voltage of V, with a fill factor of . The cell works best at V. The pump needs V. The cells are joined in series. Find the greatest power of one cell, its efficiency, the number of cells in series that give V at the best working point and the power of the whole panel. The pump runs for h in the sunlight of . Find the energy the panel gives in watt-hour. Treat every cell as the same.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. One cell. W.
Step 2. The efficiency. The light on a cell is W. So .
Step 3. The cells in series. Each cell gives V at its best point. cells.
Step 4. The panel. The current at the best point is the same in every cell, A. The power is W.
Step 5. The energy. .
Answer: W, , cells, W and Wh, which is kWh.
Check. The panel is . The light on it is W, and of W is W. It agrees to rounding.
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An LED gives light of wavelength , with , and needs a series resistor . A photodiode with a share of photons counted gives for light of power and photon energy in eV. A solar cell gives and has the efficiency .