A solar panel (an assemblage of solar cells) measures 58 cm x 53 cm. When facing the Sun, this panel generates 2.7 A at 14 V. Sunlight delivers an energy of 1.0×10^3 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?
Power out = V I = 2.7*14 = 37.8 W
Solar Power in = (Area)*(Irradiance)
= 0.58 * 0.53 * 1000 = 307.4 W
Efficiency = (Power out)/(Power in)
= ___
To find the efficiency of the solar panel, we need to calculate the power output and power input.
1. Power Output: The power output is the electric power generated by the solar panel. It is given by:
Power Output = Current (I) x Voltage (V)
Power Output = 2.7 A x 14 V
Power Output = 37.8 W
2. Power Input: The power input is the power received from sunlight, which is determined by the area of the panel and the energy delivered by sunlight per unit area. The area of the panel is given as 58 cm x 53 cm, we need to convert it to square meters by dividing by 100:
Area = (58 cm / 100) x (53 cm / 100)
Area = 0.58 m x 0.53 m
Area = 0.3074 m²
Power Input = Area x Energy Delivered by Sunlight
Power Input = 0.3074 m² x 1.0 × 10^3 W/m²
Power Input = 307.4 W
Now that we have the power output and power input, we can calculate the efficiency.
Efficiency = (Power Output / Power Input) x 100%
Efficiency = (37.8 W / 307.4 W) x 100%
Efficiency = 0.123 x 100%
Efficiency = 12.3%
Therefore, the efficiency of this solar panel is 12.3%.