Asunny 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

Option a.

Total area of the cells to generate 6.7 kW, A = 61.4

Given:

The efficiency of the cell, = 9.1%

Power generated by the cells, P = 6.7 kW

Intensity of light, I = 1200 W/

Solution:

Using the formula for intensity of light, I =

rearranging the above formula for Area, A:

Area, A =

Now, putting values in the above formula:

Area, A =

Area, A = 61.355 ≈ 61.4  

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