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

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  $m^{2}$