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Nonamiya [84]
3 years ago
8

if 500 watts hit the surface of a solar panel every second and it generates 150 watts every second, how efficient is it?

Physics
1 answer:
alukav5142 [94]3 years ago
4 0
The electrical efficiency is defined as the ratio between the useful power output and the total power input:
\epsilon =  \frac{P_{out}}{P_{in}}

In our problem, P_{out} = 150 W (power generated by the solar panel) and P_{in} = 500 W (power hitting the surface of the solar panel), therefore the efficiency of the panel is
\epsilon =  \frac{150 W}{500 W}=0.3
which corresponds to 30%.
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