Question

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?

Options

• 200 m²

• 40 m²

• 5 m²

• 800 m²

Answer 1

200 m² 

Explanation:

Let area needed is x m².

Therefore as per question, `(200/100) xx x xx 20/100` = 8

`therefore x = (8 xx 100 xx 1000)/(200 xx 20)` = 200 m²