The solar panels on the roof of a house measure \(4.0 \mathrm{m}\) by $6.0 \mathrm{m} .\( Assume they convert \)12 \%$ of the incident EM wave's energy to electric energy. (a) What average power do the panels supply when the incident intensity is \(1.0 \mathrm{kW} / \mathrm{m}^{2}\) and the panels are perpendicular to the incident light? (b) What average power do the panels supply when the incident intensity is \(0.80 \mathrm{kW} / \mathrm{m}^{2}\) and the light is incident at an angle of \(60.0^{\circ}\) from the normal? (W) tutorial: solar collector) (c) Take the average daytime power requirement of a house to be about 2 kW. How do your answers to (a) and (b) compare? What are the implications for the use of solar panels?

First, let's calculate the area (A) of the solar panels, which is the product of their dimensions: \(A = 4.0\,\text{m} \times 6.0\,\text{m} = 24.0\,\text{m}^2\)

The efficiency of the solar panels is 12%, meaning the panels convert 12% of the incident energy to electric energy. For the first scenario, the incident intensity is \(1.0\,\text{kW/m}^2\). To find the average power supplied by the panels, we'll multiply the incident intensity, efficiency, and area of the panels: \(P_1 = (1.0\,\text{kW/m}^2) \times (0.12) \times (24.0\,\text{m}^2) = 2.88\,\text{kW}\)

Since the incident light is not perpendicular to the panels in the second scenario, we need to consider the angle of the incident light. The angle factor is the cosine of the angle: \(\text{Angle factor} = \cos(60^{\circ}) = \frac{1}{2}\)

Now, we can calculate the average power for the second scenario, considering the incident intensity (\(0.80\,\text{kW/m}^2\)), efficiency (12%), area of the panels (\(24.0\,\text{m}^2\)), and the angle factor: \(P_2 = (0.80\,\text{kW/m}^2) \times (0.12) \times (24.0\,\text{m}^2) \times \frac{1}{2} = 1.15\,\text{kW}\)

Now, we can compare the average power for the two scenarios with the average daytime power requirement of a house, which is about 2 kW: (a) The average power supplied by the solar panels in the first scenario is \(2.88\,\text{kW}\), which is greater than the power requirement of 2 kW. (b) The average power supplied by the solar panels in the second scenario is \(1.15\,\text{kW}\), which is less than the power requirement of 2 kW. The implication is that solar panels can supply enough power if incident light is perpendicular to the panels and has a certain intensity. However, their efficiency decreases significantly when the angle of incident light changes. This highlights the importance of installing solar panels at the optimal angle to maximize their efficiency and usage in meeting the power requirements of a household. Another solution is to use more solar panels or improve solar panel efficiency while considering the variation in the angle of the incident light throughout the day.