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? }

• A. $200~m^{2}$
• B. $40~m^{2}$
• C. $5~m^{2}$
• D. 800 m²

Hint:

Efficiency indicates what fraction of available energy is actually usable.

Answer 1

The Correct Option is A

Step 1: Concept
Useful Power = Total Incident Power $\times$ Efficiency.
Step 2: Analysis
Target Power = 8 kW = 8000 W. Incident Intensity = 200 W/$m^2$. Efficiency = 20% = 0.20.
Step 3: Calculation
Useful power per $m^2$ = $200 \times 0.20 = 40$ W/$m^2$. Total Area needed = Total Power / Useful Power per $m^2$ = $8000 / 40 = 200$ $m^2$.
Step 4: Conclusion
Hence, an area of 200 $m^2$ is required.
Final Answer:(A)