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?

Short Answer

Expert verified
Answer: To optimize the efficiency of solar panels in terms of incident light angles, it is essential to install solar panels at the optimal angle that maximizes the absorption of sunlight throughout the day. Another solution is to use more solar panels or improve solar panel efficiency, considering the variation in the angle of the incident light throughout the day.

Step by step solution

01

Calculate the area of the solar panels

First, let's calculate the area (A) of the solar panels, which is the product of their dimensions: \(A = 4.0\,\text{m} \times 6.0\,\text{m} = 24.0\,\text{m}^2\)
02

Calculate the average power in the first scenario

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: \(P_1 = (1.0\,\text{kW/m}^2) \times (0.12) \times (24.0\,\text{m}^2) = 2.88\,\text{kW}\)
03

Calculate the angle factor for the incident light in the second scenario

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: \(\text{Angle factor} = \cos(60^{\circ}) = \frac{1}{2}\)
04

Calculate the average power in the second scenario

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: \(P_2 = (0.80\,\text{kW/m}^2) \times (0.12) \times (24.0\,\text{m}^2) \times \frac{1}{2} = 1.15\,\text{kW}\)
05

Compare the average power and discuss implications

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In order to study the structure of a crystalline solid, you want to illuminate it with EM radiation whose wavelength is the same as the spacing of the atoms in the crystal \((0.20 \mathrm{nm}) .\) (a) What is the frequency of the EM radiation? (b) In what part of the EM spectrum (radio, visible, etc. \()\) does it lie?
The radio telescope in Arecibo, Puerto Rico, has a diameter of $305 \mathrm{m}$. It can detect radio waves from space with intensities as small as \(10^{-26} \mathrm{W} / \mathrm{m}^{2} .\) (a) What is the average power incident on the telescope due to a wave at normal incidence with intensity \(1.0 \times 10^{-26} \mathrm{W} / \mathrm{m}^{2} ?\) (b) What is the average power incident on Earth's surface? (c) What are the rms electric and magnetic fields?
Two sheets of polarizing material are placed with their transmission axes at right angles to one another. A third polarizing sheet is placed between them with its transmission axis at \(45^{\circ}\) to the axes of the other two. (a) If unpolarized light of intensity \(I_{0}\) is incident on the system, what is the intensity of the transmitted light?(b) What is the intensity of the transmitted light when
How far does a beam of light travel in 1 ns?
In astronomy it is common to expose a photographic plate to a particular portion of the night sky for quite some time in order to gather plenty of light. Before leaving a plate exposed to the night sky, Matt decides to test his technique by exposing two photographic plates in his lab to light coming through several pinholes. The source of light is \(1.8 \mathrm{m}\) from one photographic plate and the exposure time is \(1.0 \mathrm{h}\). For how long should Matt expose a second plate located \(4.7 \mathrm{m}\) from the source if the second plate is to have equal exposure (that is, the same energy collected)?
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free