# Energy and Joule heating

1. Mar 18, 2011

### BillJ3986

1. The problem statement, all variables and given/known data
A solar panel (an assemblage of solar cells) measures 58 cm x 53 cm. When facing the sun, this panel generates 2.7A at 14V. Sunlight delivers an energy of 1.0 x 10^3 W/m^2 to an area facing it. What is the efficiency of this panel, that is, what fraction of the energy in sunlight is converted into electric energy?

2. Relevant equations
Am I doing this problem correctly

3. The attempt at a solution
The power delivered by the panel is P(delivered)= I x V= 2.7A x 14V= 37.8 W
For the power delivered by sunlight I multiplied 1.0 x 10^3 W/m^2 by the area of .58m x .53m:
so the area equals .307m^2.
1.0 x 10^3 W/m^2 x .307m^2= 307.4 W delivered by the sun.

So I needed to find the fraction of energy in sunlight that is converted into electric energy. I did that by dividing 37.8W/307.4. The fraction of energy converted is .123 or 12.3%.

Or is the answer 307.4/37.8= 8.13?

I'm not sure which is the right answer or if it I did it correctly please let me know.

2. Mar 18, 2011

### Staff: Mentor

Looks good to me.

No, you want the fraction of the energy in the sunlight that is converted to electrical energy. Your first answer was correct. (Efficiency could never be greater that 1 or 100%,)

3. Mar 18, 2011

Thank you.