Question

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 generate 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 ___________

Answer 1

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 =
`(P)/(\eta _(cell)A)`

rearranging the above formula for Area, A:

Area, A =
`(P)/(I\eta _(cell))`

Now, putting values in the above formula:

Area, A =
`(6.7* 10^(3))/(1200* 0.091)}`

Area, A = 61.355 ≈ 61.4
`m^(2)`