# Q23 (a)   A family uses 8 kW of power. Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?

D Devendra Khairwa

It is given that the efficiency of energy conversion is 20 per cent.

According to question, we can write (equating power used by family) :

$8\times 10^3\ =\ \frac{20}{100}\times A\times 200$                           (Here A is the area required.)

or                                                   $A\ =\ \frac{8\times 10^3}{40}$

or                                                          $=\ 200\ m^2$

Thus required area is 200 m2.

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