How large a solar panel do you need to drive a television (which needs 100 W to run) on a sunny day, assuming that the solar panel operates at 15 % efficiciency?
Assuming that you have the full S = 1 .36 kW m^{−2} at your disposal, the area needed is
\frac{100 W}{0.15\times 1.36\times 10^{3} W m^{−2} } ≈ 0.5 m^{2} . (37.6)