Question

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

Answer 1

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%.