#### Question

A family uses 8 kW of power. (a) 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? (b) Compare this area to that of the roof of a typical house.

#### Solution 1

**Power used by the family,**

*P*= 8 kW = 8 × 10

^{3}W

Solar energy received per square metre = 200 W

Efficiency of conversion from solar to electricity energy = 20 %

Area required to generate the desired electricity = *A*

As per the information given in the question, we have:

8 × 10^{3} = 20% × (*A* × 200)

= (20 /100) × *A* × 200

∴ *A* = 8 × 10^{3} / 40 = 200 m^{2}

(b)** **The area of a solar plate required to generate 8 kW of electricity is almost equivalent to the area of the roof of a building having dimensions 14 m × 14 m. (≈ `sqrt200`).

#### Solution 2

(a) Power used by family, p = 8 KW = 8000 W

As only 20% of solar energy can be converted to useful electrical energy, hence, power

8000 W to be supplied by solar energy = 8000 W/20 = 40000 W

As solar energy is incident at a rate of 200 Wm^{-2}, hence the area needed

A=4000 W/200 Wm^{-2} =200 m^{2}

(b) The area needed is camparable to roof area of a large sized house.