A two-slit Fraunhofer interference-diffraction pattern is observed with light of wavelength 672 nm. The slits have widths of 0.0
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Answer:
You will need 450 cells (3 cm each) to meet the voltage/current requirement.
The panel must be 3 cells in one side, by 150 cell in another side. 1350 cm^2 or 0.135 m^2. They must be connected 3 in row in parallel (to add current), then each of the former group must be connected in series to meet the voltage, so it would be 150 rows of connected in series.
The panel can be optimized using a voltage inverter, to convert current to voltage. In this way, less cells can be used achieving the same output specs.
Explanation:
To meet the voltage:
120 [v] required voltage
0.8 [v] voltage of each cell
So we need 150 cells in series for the voltage.
To meet the current
1.0 [A] Required current
350[mA]=0.35[A] cell current
1/0.35=3 cell So we need 3 cells in parallel to add the currents and meet the requirement.
See the attached figure
