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Anton [14]
3 years ago
6

A sunny day has a light intensity of 1200 w/m^2. The efficiency of the cells is 9.1%. What total area must the cells be to gener

ate 6.7 kW of power while the light is shining? a. 61.4 m^2 b. 4.71 m^2 c. .047 m^2 d. .061 m^2 e. None of the above and the answer is ___________
Physics
1 answer:
Art [367]3 years ago
5 0

Answer:

Option a.

Total area of the cells to generate 6.7 kW, A = 61.4 m^{2}

Given:

The efficiency of the cell, \eta _{cell} = 9.1%

Power generated by the cells, P = 6.7 kW

Intensity of light, I = 1200 W/ m^{2}

Solution:

Using the formula for intensity of light, I = \frac{P}{\eta _{cell}A}

rearranging the above formula for Area, A:

Area, A = \frac{P}{I\eta _{cell}}

Now, putting values in the above formula:

Area, A = \frac{6.7\times 10^{3}}{1200\times 0.091}}

Area, A = 61.355 ≈ 61.4  m^{2}

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