## College Physics (4th Edition)

(a) The panels supply an average power of $2880~W$ (b) The panels supply an average power of $1152~W$ (c) The situation in part (a) supplies an average power that is 2.5 times more than the average power supplied in part (b). The implication for the use of solar panels is that the angle of incidence is a big factor, so we should try to make the angle of incidence as close to perpendicular as possible.
(a) We can find the power absorbed by the solar panels: $P = (I~cos~\theta)~A$ $P = (1000~W/m^2)~(cos~0^{\circ})~(4.0~m)(6.0~m)$ $P = 24,000~W$ The panels can supply 12% of this power. We can find the power $P_s$ supplied by the panels: $P_s = (0.12)(24,000~W) = 2880~W$ The panels supply an average power of $2880~W$ (b) We can find the power absorbed by the solar panels: $P = (I~cos~\theta)~A$ $P = (800~W/m^2)~(cos~60^{\circ})~(4.0~m)(6.0~m)$ $P = 9600~W$ The panels can supply 12% of this power. We can find the power $P_s$ supplied by the panels: $P_s = (0.12)(9600~W) = 1152~W$ The panels supply an average power of $1152~W$ (c) In part (a), the panels supply an average power of $2880~W$ which is more than the required $2~kW$. In part (b), the panels supply an average power of $1152~W$ which is less than the required $2~kW$. We can find the ratio of the two values: $\frac{2880~W}{1152~W} = 2.5$ The situation in part (a) supplies an average power that is 2.5 times more than the average power supplied in part (b). The implication for the use of solar panels is that the angle of incidence is a big factor, so we should try to make the angle of incidence as close to perpendicular as possible.