ELECTRICAL ENERGY IN PHYSICS Karen Porter-Davis

Physics Teacher

Chamblee Charter High School, DeKalb Public Schools, Georgia

2018 Georgia Tech STEP-UP Program

Advised by

Kevin Caravati, Senior Research Scientist

Matt Swarts, Senior Research Scientist

Sam Kim, Research Assistant

Georgia Tech Research Institute

GridEd K-12 Outreach

Problem: The median age for employees in the energy industry is over 50 years old. Most of these employees will be retiring in the next five to ten years – that is over 500,000 employees. Not only will the industry need to replace these workers, but will need to hire even more due to technological advances and demand. That means more power plant construction, more fuel sources being needed, more infrastructure being created and more research to find reliable and efficient renewable energy sources. There are skill sets and educational requirements to be successful in the energy industry, however public perception especially with the younger generations see this pathway as not as lucrative and lower skilled than other careers. This means that qualified workers are not taking the opportunities that the energy field possesses. One solution that many states, universities and companies are looking at is developing curriculum and pathways for high school students and even starting these programs at the middle school level.

Abstract: Globally and locally we are in an energy crisis. This is not just due to the loss of non-renewable fuel sources that have been projected to be depleted in the next 50 – 100 years, but the lack of an upcoming utility workforce. One possible solution that many states, universities and companies are looking to is the development of energy curriculum/pathways for high/middle school students. GridEd is one such group. It is a collaborative group of universities, educators, industry stakeholders and the Electric Power Research Institute to create energy lessons for K-12 students in effort to teach about the importance of energy production and encourage the next generation to enter into the energy sector. GridEd has developed high school level lessons for Unit 1: Introduction to Energy and mini units for middle school aged students. The lessons have been constructed to allow schools and educators to pick and choose lessons, PowerPoints and activities to match their framework and time line. For example, some schools might have a dedicated year- long Energy class, have break-out/after school programs a couple of times a month, or even yet science and engineer teachers can adopt these items within their classes. We wanted to make the material as flexible as possible to reach as many students as we can. The next steps are to continue additions/revisions for existing lessons, complete programs for the other six units in the created course outline and to design useful professional development for educators to bring these lessons back to their classroom.

National Standards:

As a result of their activities in grades 9-12, all students should develop an understanding of

PS 1a: Matter is made of minute particles called atoms, and atoms are composed of even smaller components. These components have measurable properties, such as mass and electrical charge. Each atom has a positively charged nucleus surrounded by negatively charged electrons. The electric force between the nucleus and electrons holds the atom together.

PS 2b: Atoms interact with one another by transferring or sharing electrons that are furthest from the nucleus. These outer electrons govern the chemical properties of the element. Bonds between atoms are created when electrons are paired up by being transferred or shared.

PS 3a: Chemical reactions occur all around us, for example in health care, cooking, cosmetics, and automobiles. Chemical reactions may release or consume energy. Some reactions such as the burning of fossil fuels release large amounts of energy by losing heat and by emitting light.

PS 3b: A large number of important reactions involve the transfer of either electrons (oxidation/reduction reactions) or hydrogen ions (acid/base reactions) between reacting ions, molecules, or atoms. In other reactions, chemical bonds are broken by heat or light to form very reactive radicals with electrons ready to form new bonds.

PS 4c: The electric force is a universal force that exists between any two charged objects. Opposite charges attract while like charges repel. The strength of the force is proportional to the charges, and, as with gravitation, inversely proportional to the square of the distance between them.

PS 4d: Between any two charged particles, electric force is vastly greater than the gravitational force. Most observable forces such as those exerted by a coiled spring or friction may be traced to electric forces acting between atoms and molecules.

PS 5a: The total energy of the universe is constant. Energy can be transferred by collisions in chemical and nuclear reactions, by light waves and other radiations, and in many other ways. However, it can never be destroyed. As these transfers occur, the matter involved becomes steadily less ordered.

PS 5b: All energy can be considered to be either kinetic energy, which is the energy of motion; potential energy, which depends on relative position; or energy contained by a field, such as electromagnetic waves.

PS 6d: In some materials, such as metals, electrons flow easily, whereas in insulating materials such as glass they can hardly flow at all. Semiconducting materials have intermediate behavior. At low temperatures some materials become superconductors and offer no resistance to the flow of electrons

GPS STANDARDS FOR PHYSICS

SCSh1. Students will identify and investigate problems scientifically. a. Suggest reasonable hypotheses for identified problems. b. Develop procedures for solving scientific problems. c. Collect, organize and record appropriate data. d. Graphically compare and analyze data points and/or summary statistics. e. Develop reasonable conclusions based on data collected. f. Evaluate whether conclusions are reasonable by reviewing the process and checking

against other available information.

SCSh2. Students will use standard safety practices for all classroom laboratory and field investigations.

a. Follow correct procedures for use of scientific apparatus. b. Demonstrate appropriate technique in all laboratory suations. c. Follow correct protocol for identifying and reporting safety problems and

violations.

SCSh3. Students will identify and investigate problems scientifically. Suggest reasonable hypotheses for identified problems.

a. Develop procedures for solving scientific problems. b. Collect, organize and record appropriate data. c. Graphically compare and analyze data points and/or summary statistics. d. Develop reasonable conclusions based on data collected. e. Evaluate whether conclusions are reasonable by reviewing the process and checking

against other available information.

SCSh4. Students use tools and instruments for observing, measuring, and manipulating scientific equipment and materials.

a. Develop and use systematic procedures for recording and organizing information. b. Use technology to produce tables and graphs. c. Use technology to develop, test, and revise experimental or mathematical models.

SCSh6. Students will communicate scientific investigations and information clearly.

a. Write clear, coherent laboratory reports related to scientific investigations. Write clear, coherent accounts of current scientific issues, including possible alternative interpretations of the data.

b. Use data as evidence to support scientific arguments and claims in written or oral presentations.

c. Participate in group discussions of scientific investigation and current scientific issues.

SCSh9. Students will enhance reading in all curriculum areas by:

a. Reading in All Curriculum Areas • Read technical texts related to various subject areas

b. Discussing books • Examine author’s purpose in writing.

• Recognize the features of disciplinary texts.

c. Building vocabulary knowledge

• Demonstrate an understanding of contextual vocabulary in various subjects.

• Use content vocabulary in writing and speaking.

• Explore understanding of new words found in subject area texts.

d. Establishing context

• Explore life experiences related to subject area content.

• Discuss in both writing and speaking how certain words are subject area related.

• Determine strategies for finding content and contextual meaning for unknown words.

SP3. Students will evaluate the forms and transformations of energy.

a. Analyze, evaluate, and apply the principle of conservation of energy and measure the components of work-energy theorem by • describing total energy in a closed system. • identifying different types of potential energy. • relating transformations between potential and kinetic energy.

b. Explain the relationship between matter and energy. c. Analyze and measure power.

SP5. Students will evaluate relationships between electrical and magnetic forces.

a. Describe the transformation of mechanical energy into electrical energy and the transmission of electrical energy.

b. Determine the relationship among potential difference, current, and resistance in a direct current circuit.

c. Determine equivalent resistances in series and parallel circuits.

Objectives: 1) Blur the lines between the fields of science. Physics, chemistry, and biology are not

independent of each other and should be integrated. Examples of this would be through technological uses and material science.

2) To tie in more real-world experiences. Sometimes the concepts taught are very abstract to the students and they do not connect to them.

3) To create more inquiry based labs. I want to require my students to use high order thinking skills and to become comfortable with scientific investigations.

4) To vary my teaching techniques. I need to be sure to use many teaching styles to reach the most students.

5) To inspire more students to go into the fields of engineering and applied sciences. This is crucial for the United States to stay competitive in our technology driven world and the great need for young talent to enter the utilities industry.

I want to take the “scariness” out of physics. I want my students to move past their insecurities of math and science and begin to understand the importance and relevance of the physical sciences. I want my students to be excited about coming to class and excited to tell their friends, parent/guardian, etc. about what they have discovered in class. I want my students to become interested in the content discussed and have them research and explore on their own. I want my students to feel comfortable to ask questions and question concepts (this means to actively participate).

To accomplish this I also have to be an active and creative participant. I will need to know my students and their learning styles. Due to time constraints and student personalities this is not always an easy task. Personal reflection and the willingness to try new methods are always crucial to teaching. Also, identifying students with special needs is very important so I can modify lessons and offer additional support.

Another big concern within a science classroom is financial. Public school budgets are usually tight so it is important to be creative when creating activities. Thanks to the internet many demonstrations can be shown to the class without expensive equipment.

Anticipated Learner Outcomes:

Students should demonstrate mastery in the following tasks:

• Accurate measurement of quantitative and qualitative data to support conclusions. • Formulate and prove/disprove hypotheses with relevant supporting data collected from

experiments or activities. • Application of quantitative data into graphic representations of data. • Correlation of learned content to factors involved in electrical storage. • Application of Ohm’s Law and the creation of electrical circuits. • Creation of a series and parallel circuit. • Demonstration of mastery of subject through successful completion of various assessments

(test, homework check, lab practicums, modeling, reports). • Continual interest in electrical energy outside of the classroom.

Follow safety and clean-up protocols for laboratory activities.

Students will be using electrical components and should be reminded of the following safety considerations.

• Review of fire drills and placement of fire extinguishers. • Review of proper lab dress code and rules. • Electrical Safety Guidelines:

1. Be sure that equipment is in the “off” position before you plug it in.

2. Never use an electrical appliance around water or with wet hands or clothing. 3. Be sure to turn off and unplug electrical equipment when you are finished using it. 4. Do not work with any batteries, electrical devices, or magnets other than those

provided by your teacher.

Use of sharp dissecting tools (razors, knives, and scalpels) may lead to cutting injuries. This can be minimized through supervision of students.

Assessment/Rubrics: My lesson plans includes the following:

1) Collection of teacher created worksheets 2) Investigation Lab. 3) Class Lab and Teacher Guide. 4) Lab Practicum and Teacher Guide. 5) Alternative Labs (I took the main lab and divided it into three separate labs to use alone)

and Teacher Guide. 6) Journal Project and Rubric 7) Electromagnetism Project and Rubric 8) Pioneers in Electricity Project and Rubric 9) Technology and Design Project and Rubric 10) PowerPoints (including instructional PPT for Lab Practicum and Team Trivia)

Background: Electrical basics – From howstuffworks: How Light Bulbs Work Harris, T., Garden, H. and Appliances, H. (2018). How Light Bulbs Work. [online] HowStuffWorks. Available at: https://home.howstuffworks.com/light-bulb.htm [Accessed 24 Jul. 2018].

In an electrical system, increasing either the current or the voltage will result in higher power. Let's say you have a system with a 6 V light bulb hooked up to a 6 V battery. The power output of the light bulb is 100 watts. Using the equation above, we can calculate how much current in amps would be required to get 100 watts out of this 6 V bulb.

You know that P = 100 W, and V = 6 V. So you can rearrange the equation to solve for I and substitute in the numbers.

I = P/V = 100 W / 6 V = 16.66 amps

What would happen if you use a 12-volt battery and a 12-volt light bulb to get 100 watts of power?

100 W / 12 V = 8.33 amps

So this system produces the same power, but with half the current. There is an advantage that comes from using less current to make the same amount of power. The resistance in electrical wires consumes power, and the power consumed increases as the current going through the wires increases. You can see how this happens by doing a little rearranging of the two equations. What you need is an equation for power in terms of resistance and current. Let's rearrange the first equation:

I = V / R can be restated as V = I R

Now you can substitute the equation for V into the other equation:

P = V I substituting for V we get P = IR I, or P = I2R

What this equation tells you is that the power consumed by the wires increases if the resistance of the wires increases (for instance, if the wires get smaller or are made of a less conductive material). But it increases dramatically if the current going through the wires increases. So using a higher voltage to reduce the current can make electrical systems more efficient. The efficiency of electric motors also improves at higher voltages.

This improvement in efficiency is what is driving the automobile industry to adopt a higher voltage standard. Car makers are moving toward a 42-volt electrical system from the current 12-volt electrical systems. The electrical demand on cars has been steadily increasing since the first cars were made. The first cars didn't even have electrical headlights; they used oil lanterns. Today cars have thousands of electrical circuits, and future cars will demand even more power. The

change to 42 volts will help cars meet the greater electrical demand placed on them without having to increase the size of wires and generators to handle the greater current.

http://www.qrg.northwestern.edu/projects/vss/docs/Power/2-whats-electron-flow.html

Electrochemical Cells- From Qrg.northwestern.edu. (2018). What's electron flow?. [online] Available at: http://www.qrg.northwestern.edu/projects/vss/docs/Power/2-whats-electron-flow.html [Accessed 24 Jul. 2018].

And Nave, C. (2018). Electrochemical Cells. [online] Hyperphysics.phy-astr.gsu.edu. Available at: http://hyperphysics.phy-astr.gsu.edu/hbase/Chemical/electrochem.html [Accessed 24 Jul. 2018]. Oxidation and reduction reactions are used to provide useful electrical energy in batteries. A simple electrochemical cell can be made from copper and zinc metals with solutions of their sulfates. In the process of the reaction, electrons can be transferred from the zinc to the copper through an electrically conducting path as a useful electric current. An electrochemical cell can be created by placing metallic electrodes into an electrolyte where a chemical reaction either uses or generates an electric current. Electrochemical cells which generate an electric current are called voltaic cells or galvanic cells, and common batteries consist of one or more such cells. In other electrochemical cells an externally supplied electric current is used to drive a chemical reaction which would not occur spontaneously. Such cells are called electrolytic cells. The metals used in a dry-cell battery are usually copper and zinc. These materials are called the electrodes. The current is generated because a chemical reaction causes the copper electrode to develop a shortage of electrons. At the same time the zinc electrode develops an over-supply of electrons. When the two are connected, a flow of electrons from the zinc to the copper electrode results. Over a period of time, the zinc electrode will dissolve and increase the concentration of the zinc ion solution. The copper ion will plate onto the copper electrode and thus the concentration of the copper ion in solution will decrease. As this happens, the reaction slows down and the voltage decreases.

Electro-potential Table Understanding how batteries actually work requires a knowledge of chemistry. The most important factor in battery design is the electrical relationship between the two metals used in the battery. Some metals give electrons away while other metals accept extra electrons. This difference is exploited in a battery to create a flow of electrons. Electronegativity is a measure of the magnitude of the force by which an atom or molecule is able to acquire an extra electron, thereby becoming a negatively-charged ion. Differing electronegativies is a major reason why different substances must be used as the electrodes in an electrochemical cell. For example,

consider Substance A, which has a higher electronegativity than Substance B: If these are used as electrode materials, then electrons will flow from B to A, because A has the greater greed for extra electrons. This means that B will be the anode and A will be the cathode of that particular cell. From the preceding we may conclude that if we want to use copper as an electrode, and we want it to be the anode, then we must select a substance for the second electrode that has a higher electronegativity than copper. Now it is widely known that if copper and zinc are inserted into a lemon, the citric acid of the lemon will work as an electrolyte, and a small voltage and current can be produced for a short time. However, it happens that zinc has a lower electronegativity than copper; this means that in a 'lemon' cell the copper electrode is the cathode, and the zinc electrode is the anode. The lemon cell is peculiar in that both oxidation and reduction take place at the same electrode. The anode metals become oxidized (Zn to Zn+2, Mg to Mg+2) and the hydrogen ions in the lemon are reduced to hydrogen gas, in part, at the zinc and magnesium electrodes. In fact, hydrogen gas can be seen vigorously bubbling out from around the magnesium electrode. The copper electrode is simply an auxiliary electrode; it merely acts as an electron shunt, where reduction of hydrogen ions to hydrogen gas also takes place.

The table can be used to calculate theoretical voltages for various metal combinations.

Metal Potential, Volts Metal Potential, Volts

Calcium +2.20 Hydrogen 0.000

Magnesium +1.87 Antimony -0.190

Aluminum +1.30 Arsenic -0.320

Manganese +1.07 Bismuth -0.330

Zinc +0.758 Copper -0.345

Chromium +0.600 Mercury -0.799

Lemon Battery Activity

Iron +0.441 Silver -0.800

Cadmium +0.398 Platinum -0.863

Nickel +0.220 Gold -1.100

Chart Source: http://hilaroad.com/camp/projects/lemon/electric_potential.html

We've had some students do this project and then try to use the lemon "battery" to light a small flashlight's light bulb. The lemons did not work. Why? The reason is that the lemons produce only a very small current (about one milliamp). This is not enough electric current to light the bulb. Even with multiple lemons, the amount of current flowing through the wire is not enough. Though the voltage is high enough (1.5 volts with two lemons), the current is too weak. But it was a great experiment! Even if an experiment doesn't work, it helps us to understand why.

Oxidation Reduction Reactions Many definitions can be given to oxidation and reduction reactions. In terms of electrochemistry, the following definition is most appropriate, because it let us see how the electrons perform their roles in the chemistry of batteries.

Loss of electrons is oxidation, and gain of electrons is reduction.

Copper-Zinc Voltaic Cells As an introduction to electrochemistry let us take a look of a simple Voltaic cell or a galvanic cell.

http://hyperphysics.phy-astr.gsu.edu/hbase/Chemical/electrochem.html

When a stick of zinc (Zn) is inserted in a salt solution, there is a tendency for Zn to loose electron according to the reaction,

Zn = Zn2+ + 2 e-.

The arrangement of a Zn electrode in a solution containing Zn2+ ions is a half cell, which is usually represented by the notation:

Zn | Zn2+,

Zinc metal and Zn2+ ion form a redox couple, Zn2+ being the oxidant, and Zn the reductant. The same notation was used to designate a redox couple earlier.

Similarly, when a stick of copper (Cu) is inserted in a copper salt solution, there is also a tendency for Cu to loose electron according to the reaction,

Cu = Cu2+ + 2 e-.

This is another half-cell or redox couple: Cu | Cu2+.

However, the tendency for Zn to loose electron is stronger than that for copper. When the two cells are connected by a salt bridge and an electric conductor as shown to form a closed circuit for electrons and ions to flow, copper ions (Cu2+) actually gains electron to become copper metal. The reaction and the redox couple are respectively represented below,

Cu2+ + 2 e- = Cu, Cu2+ | Cu.

This arrangement is called a galvanic cell or battery as shown here. In a text form, this battery is represented by,

Zn | Zn2+ || Cu2+ | Cu,

in which the two vertical lines ( || ) represent a salt bridge, and a single vertical line ( | ) represents the boundary between the two phases (metal and solution). Electrons flow through the electric conductors connecting the electrodes and ions flow through the salt bridge. When

[Zn2+] = [Cu2+] = 1.0 M,

the voltage between the two terminals has been measured to be 1.100 V for this battery.

A battery is a package of one or more galvanic cells used for the production and storage of electric energy. The simplest battery consists of two half cells, a reduction half cell and an oxidation half cell.

Oxidation and Reduction Reactions -- a review

The overall reaction of the galvanic cell is

Zn + Cu2+ = Zn2+ + Cu

Note that this redox reaction does not involve oxygen at all. For a review, note the following:

Oxidant + n e- = Reductant Example: Cu2+ + 2 e = Cu Cu2+ is the oxidizing agent and Cu the reducing agent. Reductant = n e- + Oxidant Example: Zn = Zn2+ + 2 e-. Zn is the reducing agent, and Zn2+ the oxidizing agent.

Theoretically, any redox couple may form a half cell, and any two half cells may combine to give a battery, but we have considerable technical difficulty in making some couples into a half cell.

A simple electrochemical cell can be made from two test tubes connected with a third tube (the crossbar of the “H”), as shown in figure below. The hollow apparatus is filled by simultaneously pouring different solutions into the two test tubes, an aqueous solution (aq) of zinc sulfate into the left tube and a copper sulfate solution into the one on the right. Then a strip of zinc metal is dipped into the ZnSO4 solution, a piece of copper is inserted into the CuSO4 solution, and the two ends of the metal strips are connected by wires to a voltmeter. The lateral connecting tube allows ionic migration necessary for a closed electrical circuit. The voltmeter will show the electrical potential of 1.10 volts, which leads to the movement of electrons in the wire from the zinc electrode toward the copper electrode.

The Reaction of Metallic Zinc with Aqueous Copper(II) Ions in a Galvanic Cell. (a) A galvanic cell can be constructed by inserting a copper strip into a beaker that contains an aqueous 1 M solution

of Cu2+ ions and a zinc strip into a different beaker that contains an aqueous 1 M solution of Zn2+ ions. The two metal strips are connected by a wire that allows electricity to flow, and the

beakers are connected by a salt bridge. When the switch is closed to complete the circuit, the zinc electrode (the anode) is spontaneously oxidized to Zn2+ ions in the left compartment, while

Cu2+ ions are simultaneously reduced to copper metal at the copper electrode (the cathode). (b) As the reaction progresses, the Zn anode loses mass as it dissolves to give Zn2+(aq) ions, while the Cu cathode gains mass as Cu2+(aq) ions are reduced to copper metal that is deposited on the cathode.

Chemistry LibreTexts. (2018). Electrochemical Cells. [online] Available at: https://chem.libretexts.org/Textbook_Maps/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/

Electrochemistry/Basics_of_Electrochemistry/Electrochemical_Cells [Accessed 24 Jul. 2018].

The electric current is caused by a pair of redox reactions. At the zinc electrode, the metallic zinc is slowly being dissolved by an oxidation reaction:

An electrode at which oxidation occurs is called an anode; it strongly attracts negative ions in the solution, and such ions are consequently called anions.

Simultaneously, a reduction reaction at the copper cathode causes Cu2+ cations to be deposited onto the electrode as copper metal:

Because negatively charged electrons are flowing from the anode, it is the negative electrode. The cathode is the positive electrode.

Adding the reactions at the two electrodes gives

which is the overall redox reaction in the zinc-copper cell.

From:

CliffsNotes.com. Electrochemical Cells. 4 Jun 2009 <http://www.cliffsnotes.com/WileyCDA/CliffsReviewTopic/topicArticleId-21729,articleId-21712.html>.

Electrode Potential

The potential difference, which is measured in volts (v), depends upon the particular substances constituting the electrodes. For any electric cell, the total potential is the sum of those produced by the reactions at the two electrodes:

The EMF denotes electromotive force, another name for electrical potential.

Chemists have measured the voltages of a great variety of electrodes by connecting each in a cell with a standard hydrogen electrode, which is hydrogen gas at 1 atmosphere bubbling over a platinum wire immersed in 1 M H+ (aq). This standard electrode is arbitrarily assigned a potential of 0 volts, and measurement of the EMF of the complete cell allows the potential of the other electrode to be determined. Table 1 lists some standard potentials for electrodes at which reduction is occurring.

TABLE 1 Standard Electrode Potentials

Volts Reduction Half-Reaction

2.87 F2 (g) + 2e− → 2F− (aq)

1.36 Cl2 (g) + 2e− → 2Cl− (aq)

1.20 Pt2+ (aq) + 2e− → Pt (s)

0.92 Hg2+ (aq) + 2e− → Hg (l)

0.80 Ag+ (aq) + e− → Ag (s)

0.53 I2 (s) + 2e− → 2l− (aq)

0.34 Cu2+ (aq) + 2e− → Cu (s)

0 2H+ (aq) + 2e− → H2 (g)

−0.13 Pb2+ (aq) + 2e− → Pb (s)

−0.26 Ni2+ (aq) + 2e− → Ni (s)

−0.44 Fe2+ (aq) + 2e− → Fe (s)

−0.76 Zn2+ (aq) + 2e− → Zn (s)

−1.66 Al3+ (aq) + 3e− → Al (s)

−2.71 Na+ (aq) + e− → Na (s)

−2.87 Ca2+ (aq) + 2e− → Ca (s)

−2.91 K+ (aq) + e− → K (s)

−3.04 Li+ (aq) + e− → Li (s)

Near the middle of the list, you will see 0 volts arbitrarily assigned to the standard hydrogen electrode; all other potentials are relative to the hydrogen half-reaction. The voltages are given signs appropriate for a reduction reaction. For oxidation, the sign is reversed; thus, the oxidation half-reaction,

has an EMF of –1.20 volts, the opposite given in the Table 1 . Look this up in the chart to be sure that you understand.

Consider how these standard potentials are used to determine the voltage of an electric cell. In the zinc-copper cell described earlier, the two half-reactions must be added to determine the cell EMF. (See Table 2 .)

TABLE 2 Zinc-Copper Cell

Half-reaction Type Electrode Potential

Zn (s) → Zn2+ (aq) + 2e− Oxidation Anode 0.76 volts

Cu2+ (aq) + 2e− → Cu (s) Reduction Cathode 0.34 volts

The complete zinc-copper cell has a total potential of 1.10 volts (the sum of 0.76v and 0.34v). Notice that the sign of the potential of the zinc anode is the reverse of the sign given in the chart of standard electrode potentials because the reaction at the anode is oxidation.

In the chart of standard electrode potentials (Table 1 ), reactions are arranged in order of their tendency to occur. Reactions with a positive EMF occur more readily than those with a negative EMF. The zinc-copper cell has an overall EMF of +1.10 volts, so the solution of zinc and deposition of copper can proceed.

Calculate the total potential of a similar cell with zinc and aluminum electrodes. Table 3 shows the two pertinent half-reactions.

TABLE 3 Aluminum-Zinc Cell

Half-reaction Type Electrode Potential

2Al (s) → 2Al3+ (aq) + 6e− Oxidation Anode 1.66 volts

3Zn2+ (aq) + 6e− → 3Zn (s) Reduction Cathode −0.76 volts

Such a cell with zinc and aluminum electrodes would have an overall potential of +0.90 volts, with aluminum being dissolved and zinc metal being deposited out of solution.

If you select any two half-reactions from the chart of standard electrode potentials, the half-reaction higher on the list will proceed as a reduction, and the one lower on the list will proceed in the reverse direction, as an oxidation. Beware: Some references give standard electrode potentials for oxidation half-reactions, so you have to switch “higher” and “lower” in the rule stated in the preceding sentence, though this is not common.

The oxidation number rules for identifying the oxidant and reductant in a redox reaction seem complex at first but become quite simple with a small amount of practice and are an excellent tool for understanding the chemical reactions that occur all around us. Oxidation numbers are not charges even though many times they can be the same as ionic charges. They are an invented system for keeping track of electron loss and gain. We do not need to know anything about the bonding within the chemicals in order to be able to use these numbers effectively.

For each element or molecule that is involved in a reaction we need to follow these simple rules.

Rule 1. All pure substances have an oxidation number of zero. This applies to any pure substance whether it is a diatomic gas like O2 or a piece of pure metal like Iron (Fe). Examples of

Rule 2. In compounds, elements that usually have an ionic charge imparted by their position in a particular group have that same oxidation number. An example is Cl which is usually in the form Cl- in compounds; this will have an oxidation number of -1 in compounds.

Rule 3. When two or more usually negatively charged ions are involved in a compound, the one with the highest electronegativity value is given its ionic charge as the oxidation number; the others are worked out normally. An example is OF2. F is more electronegative, and so it is assigned the value of -1.

Rule 4. Oxygen in a compound always has an oxidation number of -2

Rule 5. Hydrogen in compounds always has an oxidation number of +1 except in the rare case of Metal Hydrides where it has a value of -1.

Rule 6. The oxidation numbers in the compound or molecule must total to the overall charge of that compound or molecule. For example CO2 has no overall charge and so the oxidation numbers must tally to zero. The sulfate ion, SO42-, has an overall charge of -2, so the oxidation numbers must tally to -2

Examples of Applying Oxidation Number Rules

These oxidation number rules only make sense when we start to use them. It is easiest to get the correct values if we break up molecules into their component atoms and line them up next to each other.

Example 1: Carbon Dioxide

Let's start with the very familiar Carbon Dioxide, CO2.

Carbon Dioxide contains one carbon atom and two oxygen atoms bound together in a discreet molecule. It is an uncharged molecule, meaning its overall charge value is zero.

First we draw each atom involved. Since there are two oxygen atoms in CO2, we put two of them in the drawing. We then make that collection of atoms equal to the overall charge which in this case is zero.

Next we go through the rules from 1 to 6, skipping any that are not relevant.

Rule 1: not relevant: CO2 is a compound and not a pure substance.

Rule 2: Oxygen is the most electronegative element and so it gets its standard charge of -2 for each oxygen atom.

Rule 3: not relevant: Carbon never forms negatively charged ions and even if it did it it is less electronegative than oxygen so what we have done is still good.

Rule 4: this has already been taken care of in Rule 2.

Rule 5: not relevant: there are no Hydrogen atoms involved.

Rule 6: In order to tally to zero, we MUST assign Carbon the oxidation number of +4. Therefore we can confidently state that after following the oxidation number rules, Carbon has an oxidation number of +4 in this compound.

Example 2: DichloroMethane

This unfamiliar chemical is also a discreet molecule that contains one Carbon atom, two Hydrogen atoms and two Chlorine atoms. Like Carbon Dioxide, it has an overall charge of zero.

Again, we go through the rules from 1 to 6:

Rule 1: not relevant: it is a compound.

Rule 2: Chlorine is the most electronegative element and so it gets its standard charge of -1 for each Chlorine atom.

Rule 3: not relevant: Carbon never forms negatively charged ions and even if it did it it is less electronegative than Chlorine so what we have done is still good.

Rule 4: not relevant: there are no Oxygen atoms involved.

Rule 5: both Hydrogen atoms are given an oxidation number of +1.

Rule 6: In order to tally to zero, we MUST assign Carbon the oxidation number of 0. Therefore we can confidently state that after following the oxidation number rules, Carbon has an oxidation number of 0 in this compound.

The steps taken to achieve this result are also shown in picture form below.

Example 3 + 4: More Challenging

Lastly, let's look at a couple of more challenging examples. The unnecessary steps have been omitted. See if you can follow the six rules to get the same result as those given here:

http://www.regentsprep.org/Regents/physics/phys03/apotdif/default.htm

Galvanic cell (also called voltaic cell) uses chemical reaction to produce electrical energy (flow of electrons).

When zinc metal placed in CuSO4 solution, following reaction take place:

Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)

Oxidation: Zn(s) Zn+2 + 2e-1

Reduction: Cu+2 + 2e-1 Cu

Overall: Zn(s) + Cu+2 Zn+2 + Cu(s)

Electrons will not flow in the following apparatus:

But if reaction carried out in the apparatus shown in Figure 21.2 (Cu/Ag system), electrons are transferred from Zn° to Cu+2 through a wire producing electrical energy.

The salt bridge is necessary to complete the circuit and maintain charge neutrality.

Zn+2 cathode SO4-2 anode

Anode Cathode

Oxidation occurs Reduction occurs

Eectrons produced Electrons are consumed

Anions migrate toward Cations migrate toward

Has negative sign Has positive sign

Shorthand Notation for:

Zn° + Cu+2 Zn+2 + Cu°

Write shorthand notation for: Fe(s) + 2Fe+3(aq) 3Fe+2(aq)

Fe° Fe+2 Fe+3 Fe+2

Write shorthand notation for: 2Ag+1(aq) + Ni(s) 2Ag(s) + Ni+2(aq)

Ni° Ni+2 Ag+1 Ag°

Capacitors

From Brain, M. and Bryant, C. (2018). How Capacitors Work. [online] HowStuffWorks. Available at: https://electronics.howstuffworks.com/capacitor.htm [Accessed 24 Jul. 2018].

In a way, a capacitor is a little like a battery. Although they work in completely different ways, capacitors and batteries both store electrical energy. If you have read a “How Batteries Work,” then you know that a battery has two terminals. Inside the battery, chemical reactions produce electrons on one terminal and absorb electrons on the other terminal. A capacitor is much simpler than a battery, as it can't produce new electrons -- it only stores them.

In this article, we'll learn exactly what a capacitor is, what it does and how it's used in electronics. We'll also look at the history of the capacitor and how several people helped shape its progress.

Inside the capacitor, the terminals connect to two metal plates separated by a non-conducting substance, or dielectric. You can easily make a capacitor from two pieces of aluminum foil and a piece of paper. It won't be a particularly good capacitor in terms of its storage capacity, but it will work.

In theory, the dielectric can be any non-conductive substance. However, for practical applications, specific materials are used that best suit the capacitor's function. Mica, ceramic, cellulose,

porcelain, Mylar, Teflon and even air are some of the non-conductive materials used. The dielectric dictates what kind of capacitor it is and for what it is best suited. Depending on the size and type of dielectric, some capacitors are better for high frequency uses, while some are better for high voltage applications. Capacitors can be manufactured to serve any purpose, from the smallest plastic capacitor in your calculator, to an ultra-capacitor that can power a commuter bus. NASA uses glass capacitors to help wake up the space shuttle's circuitry and help deploy space probes. Here are some of the various types of capacitors and how they are used.

• Air - Often used in radio tuning circuits • Mylar - Most commonly used for timer circuits like clocks, alarms and counters • Glass - Good for high voltage applications • Ceramic - Used for high frequency purposes like antennas, X-ray and MRI machines • Super capacitor – Powers electric and hybrid cars

Capacitor Circuit When you connect a capacitor to a battery, here's what happens:

• The plate on the capacitor that attaches to the negative terminal of the battery accepts electrons that the battery is producing.

• The plate on the capacitor that attaches to the positive terminal of the battery loses electrons to the battery.

Once it's charged, the capacitor has the same voltage as the battery (1.5 volts on the battery means 1.5 volts on the capacitor). For a small capacitor, the capacity is small. But large capacitors can hold quite a bit of charge. You can find capacitors as big as soda cans that hold enough charge to light a flashlight bulb for a minute or more.

Even nature shows the capacitor at work in the form of lightning. One plate is the cloud, the other plate is the ground and the lightning is the charge releasing between these two "plates." Obviously, in a capacitor that large, you can hold a huge amount of charge!

Let's say you hook up a capacitor like this:

Here you have a battery, a lightbulb and a capacitor. If the capacitor is pretty big, what you will notice is that, when you connect the battery, the light bulb will light up as current flows from the battery to the capacitor to charge it up. The bulb will get progressively dimmer and finally go out once the capacitor reaches its capacity. If you then remove the battery and replace it with a wire, current will flow from one plate of the capacitor to the other. The bulb will light initially and then dim as the capacitor discharges, until it is completely out.

Farad

A capacitor's storage potential, or capacitance, is measured in units called farads. A 1-farad capacitor can store one coulomb (coo-lomb) of charge at 1 volt. A coulomb is 6.25 E 18 (6.25 x 1018) or 6.25 billion billion) electrons. One amp represents a rate of electron flow of 1 coulomb of electrons per second, so a 1-farad capacitor can hold 1 amp-second of electrons at 1 volt.

A 1-farad capacitor would typically be pretty big. It might be as big as a can of tuna or a 1-liter soda bottle, depending on the voltage it can handle. For this reason, capacitors are typically measured in microfarads (millionths of a farad).

To get some perspective on how big a farad is, think about this:

• A standard alkaline AA battery holds about 2.8 amp-hours. • That means that an AA battery can produce 2.8 amps for an hour at 1.5 volts (about 4.2

watt-hours -- an AA battery can light a 4-watt bulb for a little more than an hour). • Let's call it 1 volt to make the math easier. To store one AA battery's energy in a capacitor,

you would need 3,600 * 2.8 = 10,080 farads to hold it, because an amp-hour is 3,600 amp-seconds.

If it takes something the size of a can of tuna to hold a farad, then 10,080 farads is going to take up a LOT more space than a single AA battery! Obviously, it's impractical to use capacitors to store any significant amount of power unless you do it at a high voltage.

Applications The difference between a capacitor and a battery is that a capacitor can dump its entire charge in a tiny fraction of a second, where a battery would take minutes to completely discharge. That's why the electronic flash on a Smartphone/camera uses a capacitor -- the battery charges up the

flash's capacitor over several seconds, and then the capacitor dumps the full charge into the flash tube almost instantly. This can make a large, charged capacitor extremely dangerous -- flash units and TVs have warnings about opening them up for this reason. They contain big capacitors that can, potentially, kill you with the charge they contain.

Capacitors are used in several different ways in electronic circuits:

• Sometimes, capacitors are used to store charge for high-speed use. That's what a flash does. Big lasers use this technique as well to get very bright, instantaneous flashes.

• Capacitors can also eliminate ripples. If a line carrying DC voltage has ripples or spikes in it, a big capacitor can even out the voltage by absorbing the peaks and filling in the valleys.

• A capacitor can block DC voltage. If you hook a small capacitor to a battery, then no current will flow between the poles of the battery once the capacitor charges. However, any alternating current (AC) signal flows through a capacitor unimpeded. That's because the capacitor will charge and discharge as the alternating current fluctuates, making it appear that the alternating current is flowing.

Capacitive Touch Screens

One of the more futuristic applications of capacitors is the capacitive touch screen. These are glass screens that have a very thin, transparent metallic coating. A built-in electrode pattern charges the screen so when touched, a current is drawn to the finger and creates a voltage drop. This exact location of the voltage drop is picked up by a controller and transmitted to a computer. These touch screens are commonly found in interactive building directories, Smartphones and Tablets.

History of the Capacitor

The invention of the capacitor varies somewhat depending on who you ask. There are records that indicate a German scientist named Ewald Georg von Kleist invented the capacitor in November 1745. Several months later Pieter van Musschenbroek, a Dutch professor at the University of Leyden came up with a very similar device in the form of the Leyden jar, which is typically credited as the first capacitor. Since Kleist didn't have detailed records and notes, nor the notoriety of his Dutch counterpart, he's often overlooked as a contributor to the capacitor's evolution. However, over the years, both have been given equal credit as it was established that their research was independent of each other and merely a scientific coincidence [source: Williams].

The Leyden jar was a very simple device. It consisted of a glass jar, half filled with water and lined inside and out with metal foil. The glass acted as the dielectric, although it was thought for a time that water was the key ingredient. There was usually a metal wire or chain driven through a cork in the top of the jar. The chain was then hooked to something that would deliver a charge, most likely a hand-cranked static generator. Once delivered, the jar would hold two equal but opposite charges in equilibrium until they were connected with a wire, producing a slight spark or shock [source: Williams].

Benjamin Franklin worked with the Leyden jar in his experiments with electricity and soon found that a flat piece of glass worked as well as the jar model, prompting him to develop the flat capacitor, or Franklin Square. Years later, English chemist Michael Faraday would pioneer the first practical applications for the capacitor in trying to store unused electrons from his experiments. This led to the first usable capacitor, made from large oil barrels. Faraday's progress with capacitors is what eventually enabled us to deliver electric power over great distances. As a result of Faraday's achievements in the field of electricity, the unit of measurement for capacitors, or capacitance, became known as the farad [source: Ramasamy].

Brain, Marshall, and Charles W. Bryant. "How Capacitors Work." 17 September 2007. HowStuffWorks.com. <http://electronics.howstuffworks.com/capacitor.htm> 04 June 2009.

The instructions to create a film canister Leyden Jar is from

Scrivs (2018). Instructions for Making a Film Canister Leyden Jar | ForeverGeek. [online] ForeverGeek. Available at: http://www.forevergeek.com/2005/04/instructions_for_making_a_film_canister_leyden_jar/ [Accessed 24 Jul. 2018].

“I wrote these plans after building the project pictured at Lenny R’s Web Page. This is a guide on how to make a Leyden Jar that makes awesome blue sparks with materials mostly just lying around your house. It’s cheap, it’s basically harmless, and it’s fun! First, here’s a list of the materials you will need:

• An empty film canister with its lid

• Multi-strand insulated wire

• Single conductor/solid uninsulated wire, about 1.5mm in diameter • Some aluminum or copper foil (NOTE: Any conductive foil will work. Copper foil is

thicker and easier to work with than aluminum foil, but aluminum foil works if that’s all you have.)

• A screw with a round head that is shorter than the film canister’s height • Two nuts that fit on the screw

• Scotch tape • optional. A PVC Pipe, 3/4″ wide and about 3 or 4 feet long. • optional. Fur, wool, or cotton fabric •

Now that you have all of the parts, let’s get to work. First, drill (or poke) a hole in the center of the lid of the film canister. The hole should be just wide enough to allow the screw to fit snugly inside it. Next, cut a rectangle of foil large enough to wrap around the outside of the canister, and about 2/3 of the height of the canister. Tape the foil to the canister, being sure to leave an open section for a loop of wire to go around the canister over the foil (you should just need to tape the foil’s edges to the outside of the canister.) This is roughly what your canister should look like with the

top on (sorry for the horrible MS Paint art.)

Make another piece of foil the same size as the first piece, and fasten it inside the canister. If you’re using heavy foil, you shouldn’t need to use tape; the tension of the foil being rolled up should be enough to keep it plastered to the inside of the canister. If you are using lighter foil, you’ll need to tape the edges of the foil to the inside of the canister in a similar fashion. It is very important that the foil touches the container all the way around the inside of the canister.

Now, it gets a little more complicated. Secure the multi-strand wire to the top of the

lid using the screw and a nut (see image below.) The multi-strand wire should be about 1î or 2 cm long, and should have about 1/4î or .75 cm of insulation stripped off both ends.

Make a hook on one end of the wire, and secure it in between a nut and the outside

of the lid. Fan/spread out the conductors on the other end of the wire. Secure the

solid wire to the bottom of the film canister in a similar fashion. Bend

the wire as shown below, making it poke out wider than the canister lid. This way,

the wire is guaranteed to touch the foil on the inside of the film

canister when the lid is put on. Make a loop at the end of the wire so that it won’t

tear through the foil on the inside of the canister and to make better a electrical contact. If this is

confusing, look at the drawing below (again, sorry about MS Paint.)

Place the lid on the container, ensuring that the wire loop touches the foil on the inside. If the loop doesn’t touch the foil on the inside, bend the loop farther outwards until it does.

Last, bend another piece of wire into the shape above, and insert it over the bottom of the canister onto the ‘open section’ of foil, so that it looks similar to the finished Leyden Jar

below. Make sure the top of the wire is at a MAXIMUM of 1/4 of an inch away from

the screw head. After the wire is in the correct position, cover the outside of the canister with scotch or electrical tape so that you don’t shock yourself while handling it (the canister below is covered with scotch tape, although it’s kind of hard to see.)

Image from Lenny R’s Web Page

You’re done! You have just built something called a Leyden Jar; in effect, it’s a very simple capacitor. To use it, simply wave the canister over the surface of a CRT TV screen or monitor, or

anything that makes static electricity(the multi-strand wire ‘receives’ the

static electricity and should therefore be closest to the static source.) Then gently

push down the solid wire on the outside so it reaches a little closer to the screw head (without touching) and ZAP! There’s your spark. Most of the time, the Jar will discharge itself with no extra help. If you’re having trouble charging the Jar, you can use the last two materials on the material list to build a simple but excellent static generator. Take the fabric and rub it along the pipe all the way up, then all the way down. Have someone else hold the Leyden jar; pressing the multi-strand pickup wires to the pipe, and discharge the Jar after a few seconds of rubbing.

Please note that the finished Leyden Jar holds high voltage, low current electricity, which is normally harmless. However, don’t take any chances! If you’re not sure if it’s charged or not and

you think it’s not safe, simply press the press on the outside wire against the screw to completely discharge it. Keep your finger away from the screw head area as much as possible while it’s charged. You don’t want an ‘accident’ (you’ll just get a static shock. No one likes getting a static shock.)

I would like to give special thanks to Lenny R for his ideas. As far as I know, the original “Film Can Leyden Jar” was his design, and although I’ve made some changes to the design and written this guide, I don’t want to risk getting sued or something for taking credit for the original design (editor’s note: smart thinking)

http://micro.magnet.fsu.edu/electromag/java/capacitor/index.html

DIODES, LEDs, AND LIGHTBULBS

From Harris, T. and Fenlon, W. (2018). How Light Emitting Diodes Work. [online] HowStuffWorks. Available at: https://electronics.howstuffworks.com/led2.htm [Accessed 24 Jul. 2018].

Basically, LEDs are just tiny light bulbs that fit easily into an electrical circuit. But unlike ordinary incandescent bulbs, they don't have a filament that will burn out, and they don't get especially hot. They are illuminated solely by the movement of electrons in a semiconductor material, and they last just as long as a standard transistor.

What is a Diode?

A diode is the simplest sort of semiconductor device. Broadly speaking, a semiconductor is a material with a varying ability to conduct electrical current. Most semiconductors are made of a poor conductor that has had impurities (atoms of another material) added to it. The process of adding impurities is called doping.

In the case of LEDs, the conductor material is typically aluminum-gallium-arsenide (AlGaAs). In pure aluminum-gallium-arsenide, all of the atoms bond perfectly to their neighbors, leaving no free electrons (negatively-charged particles) to conduct electric current. In doped material, additional atoms change the balance, either adding free electrons or creating holes where electrons can go. Either of these additions make the material more conductive.

A semiconductor with extra electrons is called N-type material, since it has extra negatively-charged particles. In N-type material, free electrons move from a negatively-charged area to a positively charged area.

A semiconductor with extra holes is called P-type material, since it effectively has extra positively-charged particles. Electrons can jump from hole to hole, moving from a negatively-charged area to a positively-charged area. As a result, the holes themselves appear to move from a positively-charged area to a negatively-charged area.

A diode comprises a section of N-type material bonded to a section of P-type material, with electrodes on each end. This arrangement conducts electricity in only one direction. When no voltage is applied to the diode, electrons from the N-type material fill holes from the P-type material along the junction between the layers, forming a depletion zone. In a depletion zone, the semiconductor material is returned to its original insulating state -- all of the holes are filled, so there are no free electrons or empty spaces for electrons, and charge can't flow.

At the junction, free electrons from the N-type material fill holes from the P-type material. This creates an insulating layer in the middle of the diode called the depletion zone.

To get rid of the depletion zone, you have to get electrons moving from the N-type area to the P-type area and holes moving in the reverse direction. To do this, you connect the N-type side of the diode to the negative end of a circuit and the P-type side to the positive end. The free electrons in the N-type material are repelled by the negative electrode and drawn to the positive electrode. The holes in the P-type material move the other way. When the voltage difference between the electrodes is high enough, the electrons in the depletion zone are boosted out of their holes and begin moving freely again. The depletion zone disappears, and charge moves across the diode.

When the negative end of the circuit is hooked up to the N-type

layer and the positive end is hooked up to P-type layer, electrons and holes start moving and the depletion zone disappears.

If you try to run current the other way, with the P-type side connected to the negative end of the circuit and the N-type side connected to the positive end, current will not flow. The negative electrons in the N-type material are attracted to the positive electrode. The positive holes in the P-type material are attracted to the negative electrode. No current flows across the junction because the holes and the electrons are each moving in the wrong direction. The depletion zone increases.

When the positive end of the circuit is hooked up to the N-type layer and the negative end is hooked up to the P-type

layer, free electrons collect on one end of the diode and holes collect on the other. The depletion zone gets bigger.

The interaction between electrons and holes in this setup has an interesting side effect -- it generates light.

Light is a form of energy that can be released by an atom. It is made up of many small particle-like packets that have energy and momentum but no mass. These particles, called photons, are the most basic units of light.

Photons are released as a result of moving electrons. In an atom, electrons move in orbitals around the nucleus. Electrons in different orbitals have different amounts of energy. Generally speaking, electrons with greater energy move in orbitals farther away from the nucleus.

For an electron to jump from a lower orbital to a higher orbital, something has to boost its energy level. Conversely, an electron releases energy when it drops from a higher orbital to a lower one. This energy is released in the form of a photon. A greater energy drop releases a higher-energy photon, which is characterized by a higher frequency. As we saw in the last section, free electrons moving across a diode can fall into empty holes from the P-type layer. This involves a drop from the conduction band to a lower orbital, so the electrons release energy in the form of photons. This happens in any diode, but you can only see the photons when the diode is composed of certain material. The atoms in a standard silicon diode, for example, are arranged in such a way that the electron drops a relatively short distance. As a result, the photon's frequency is so low that it is invisible to the human eye -- it is in the infrared portion of the light spectrum. This isn't necessarily a bad thing, of course: Infrared LEDs are ideal for remote controls, among other things.

Visible light-emitting diodes (VLEDs), such as the ones that light up numbers in a digital clock, are made of materials characterized by a wider gap between the conduction band and the lower orbitals. The size of the gap determines the frequency of the photon -- in other words, it determines the color of the light.

LED Advantages

While all diodes release light, most don't do it very effectively. In an ordinary diode, the semiconductor material itself ends up absorbing a lot of the light energy. LEDs are specially constructed to release a large number of photons outward. Additionally, they are housed in a plastic bulb that concentrates the light in a particular direction. As you can see in the diagram, most of the light from the diode bounces off the sides of the bulb, traveling on through the rounded end.

LEDs have several advantages over conventional incandescent lamps. For one thing, they don't have a filament that will burn out, so they last much longer. Additionally, their small plastic bulb makes them a lot more durable. They also fit more easily into modern electronic circuits.

But the main advantage is efficiency. In conventional incandescent bulbs the light-production process involves generating a lot of heat (the filament must be warmed). This is completely wasted energy, unless you're using the lamp as a heater, because a huge portion of the available electricity isn't going toward producing visible light. LEDs generate very little heat, relatively speaking. A much higher percentage of the electrical power is going directly to generating light, which cuts down on the electricity demands considerably.

Up until recently, LEDs were too expensive to use for most lighting applications because they're built around advanced semiconductor material. The price of semiconductor devices has plummeted over the past decade, however, making LEDs a more cost-effective lighting option for a wide range of situations. While they may be more expensive than incandescent lights up front, their lower cost in the long run can make them a better buy. In the future, they will play an even bigger role in the world of technology.

Light Basics

Light is a form of energy that can be released by an atom. It is made up of many small particle-like packets that have energy and momentum but no mass. These particles, called light photons, are the most basic units of light. Atoms release light photons when their electrons become excited. then you know that electrons are the negatively charged particles that move around an atom's

nucleus (which has a net positive charge). An atom's electrons have different levels of energy, depending on several factors, including their speed and distance from the nucleus. Electrons of different energy levels occupy different orbitals. Generally speaking, electrons with greater energy move in orbitals farther away from the nucleus. When an atom gains or loses energy, the change is expressed by the movement of electrons. When something passes energy on to an atom, an electron may be temporarily boosted to a higher orbital (farther away from the nucleus). The electron only holds this position for a tiny fraction of a second; almost immediately, it is drawn back toward the nucleus, to its original orbital. As it returns to its original orbital, the electron releases the extra energy in the form of a photon, in some cases a light photon.

The wavelength of the emitted light (which determines its color) depends on how much energy is released, which depends on the particular position of the electron. Consequently, different sorts of atoms will release different sorts of light photons. In other words, the color of the light is determined by what kind of atom is excited.

This is the basic mechanism at work in nearly all light sources. The main difference between these sources is the process of exciting the atoms.

When the bulb is hooked up to a power supply, an electric current flows from one contact to the other, through the wires and the filament. Electric current in a solid conductor is the mass movement of free electrons (electrons that are not tightly bound to an atom) from a negatively charged area to a positively charged area.

As the electrons zip along through the filament, they are constantly bumping into the atoms that make up the filament. The energy of each impact vibrates an atom -- in other words, the current heats the atoms up. A thinner conductor heats up more easily than a thicker conductor because it is more resistant to the movement of electrons.

Bound electrons in the vibrating atoms may be boosted temporarily to a higher energy level. When they fall back to their normal levels, the electrons release the extra energy in the form of photons. Metal atoms release mostly infrared light photons, which are invisible to the human eye. But if they are heated to a high enough level -- around 4,000 degrees Fahrenheit (2,200 degrees C) in the case of a light bulb -- they will emit a good deal of visible light.

The filament in a light bulb is made of a long, incredibly thin length of tungsten metal. In a typical 60-watt bulb, the tungsten filament is about 6.5 feet (2 meters) long but only one-hundredth of an inch thick. The tungsten is arranged in a double coil in order to fit it all in a small space. That is, the filament is wound up to make one coil, and then this coil is wound to make a larger coil. In a 60-watt bulb, the coil is less than an inch long.

Tungsten is used in nearly all incandescent light bulbs because it is an ideal filament material.

The Filament

As we saw in the last section, a metal must be heated to extreme temperatures before it will emit a useful amount of visible light. Most metals will actually melt before reaching such extreme

temperatures -- the vibration will break apart the rigid structural bonds between the atoms so that the material becomes a liquid. Light bulbs are manufactured with tungsten filaments because tungsten has an abnormally high melting temperature.

But tungsten will catch on fire at such high temperatures, if the conditions are right. Combustion is caused by a reaction between two chemicals, which is set off when one of the chemicals has reached its ignition temperature. On Earth, combustion is usually a reaction between oxygen in the atmosphere and some heated material, but other combinations of chemicals will combust as well.

The filament in a light bulb is housed in a sealed, oxygen-free chamber to prevent combustion. In the first light bulbs, all the air was sucked out of the bulb to create a near vacuum -- an area with no matter in it. Since there wasn't any gaseous matter present (or hardly any), the material could not combust.

The problem with this approach was the evaporation of the tungsten atoms. At such extreme temperatures, the occasional tungsten atom vibrates enough to detach from the atoms around it and flies into the air. In a vacuum bulb, free tungsten atoms shoot out in a straight line and collect on the inside of the glass. As more and more atoms evaporate, the filament starts to disintegrate, and the glass starts to get darker. This reduces the life of the bulb considerably.

In a modern light bulb, inert gases, typically argon, greatly reduce this loss of tungsten. When a tungsten atom evaporates, chances are it will collide with an argon atom and bounce right back toward the filament, where it will rejoin the solid structure. Since inert gases normally don't react with other elements, there is no chance of the elements combining in a combustion reaction.

Cheap, effective and easy-to-use, the light bulb has proved a monstrous success. It is still the most popular method of bringing light indoors and extending the day after sundown. But by all indications, it will eventually give way to more advanced technologies, because it isn't very efficient.

Incandescent light bulbs give off most of their energy in the form of heat-carrying infrared light photons -- only about 10 percent of the light produced is in the visible spectrum. This wastes a lot of electricity. Cool light sources, such as fluorescent lamps and LEDs, don't waste a lot of energy generating heat -- they give off mostly visible light. For this reason, they are slowly edging out the old reliable light bulb.

Bright, Brighter, Brightest

Light bulbs are ranked by their power -- the amount of light they put out in a certain period of time (measured in watts). Higher-watt bulbs have a bigger filament, so they produce more light.

A three-way bulb has two filaments of different wattage -- typically a 50-watt filament and a 100-watt filament. The filaments are wired to separate circuits, which can be closed initially using a special three-way socket.

The switch in the three-way socket lets you choose from three different light levels. On the lowest level, the switch closes only the circuit for the 50-watt filament. For the medium light level, the switch closes the circuit for the 100-watt filament. For the brightest level, the switch closes the circuits for both filaments, so the bulb operates at 150 watts.

Resistors

From

Woodford, Chris. (2008/2018) Resistors. Retrieved from https://www.explainthatstuff.com/resistors.html. [July 24, 2018]

When you first learn about electricity, you discover that materials fall into two basic categories called conductors and insulators. Conductors (such as metals) let electricity flow through them; insulators (such as plastics and wood) generally do not. But nothing's quite so simple, is it? Any substance will conduct electricity if you put a big enough voltage across it: even air, which is normally an insulator, suddenly becomes a conductor when a powerful voltage builds up in the clouds—and that's what makes lightning. Rather than talking about conductors and insulators, it's often clearer to talk about resistance: the ease with which something will let electricity flow through it. A conductor has low resistance, while an insulator has much higher resistance. Devices called resistors let us introduce precisely controlled amounts of resistance into electrical circuits. Let's take a closer look at what they are and how they work!

Electricity flows through a material carried by electrons, tiny charged particles inside atoms. Broadly speaking, materials that conduct electricity well are ones that allow electrons to flow freely through them. In metals, for example, the atoms are locked into a solid, crystalline structure (a bit like a metal climbing frame in a playground). Although most of the electrons inside these atoms are fixed in place, some can swarm through the structure carrying electricity with them. That's why metals are good conductors: a metal puts up relatively little resistance to electrons flowing through it. Plastics are entirely different. Although often solid, they don't have the same crystalline structure. Their molecules (which are typically very long, repetitive chains called polymers) are bonded together in such a way that the electrons inside the atoms are fully occupied. There are, in short, no free electrons that can move about in plastics to carry an electric current. Plastics are good insulators: they put up a high resistance to electrons flowing through them.

This is all a little vague for a subject like electronics, which requires precise control of electric currents. That's why we define resistance more precisely as the voltage in volts required to make a current of 1 amp flow through a circuit. If it takes 500 volts to make 1 amp flow, the resistance is 500 ohms (written 500 Ω). You might see this relationship written out as a mathematical equation:

V = I × R

This is known as Ohm's Law for German physicist Georg Simon Ohm (1789–1854).

How many times have you heard bad guys say that in movies? It's often true in science as well. If a material has a high resistance, it means electricity will struggle to get through it. The more the electricity has to struggle, the more energy is wasted. That sounds like a bad idea, but sometimes resistance is far from "useless" and actually very helpful.

In an old-style light bulb, for example, electricity is made to flow through an extremely thin piece of wire called a filament. The wire is so thin that the electricity really has to fight to get through it. That makes the wire extremely hot—so much so, in fact, that it gives off light. Without resistance, light bulbs like this wouldn't function. Of course the drawback is that we have to waste a huge amount of energy heating up the filament. Old-style light bulbs like this make light by making heat and that's why they're called incandescent lamps; newer energy-efficient light bulbs make light without making much heat through the entirely different process of fluorescence.

The heat that filaments make isn't always wasted energy. In appliances like electric kettles, electric radiators, electric showers, coffee makers, and toasters, there are bigger and more durable versions of filaments called heating elements. When an electric current flows through them, they get hot enough to boil your water or cook your bread. In heating elements, at least, resistance is far from useless.

Resistance is also useful in things like transistor radios and TV sets. Suppose you want to lower the volume on your TV. You turn the volume knob and the sound gets quieter—but how does that happen? The volume knob is actually part of an electronic component called a variable resistor. If you turn the volume down, you're actually turning up the resistance in an electrical circuit that drives the TV's loudspeaker. When you turn up the resistance, the electric current flowing through the circuit is reduced. With less current, there's less energy to power the loudspeaker—so it sounds much quieter.

Photo: "Variable resistor" is the very general name for a component whose resistance can be varied by moving a dial, lever, or control of some sort. More specific kinds of variable resistors include potentiometers (small electronic components with three terminals) and rheostats (usually much larger and made from multiple turns of coiled wire with a sliding contact that moves across the coils to "tap off" some fraction of the resistance). Photos: 1) A small variable resistor acting as the volume control in a transistor radio. 2) Two large rheostats from a power plant. You can see the dial controls that "tap off" more or less resistance. Photo by Jack Boucher from Historic American Engineering Record courtesy of US Library of Congress.

How resistors work

People who make electric or electronic circuits to do particular jobs often need to introduce precise amounts of resistance. They can do that by adding tiny components called resistors. A resistor is a little package of resistance: wire it into a circuit and you reduce the current by a precise amount. From the outside, all resistors look more or less the same. As you can see in the top photo on this page, a resistor is a short, worm-like component with colored stripes on the side. It has two connections, one on either side, so you can hook it into a circuit.

What's going on inside a resistor? If you break one open, and scratch off the outer coating of insulating paint, you might see an insulating ceramic rod running through the middle with copper wire wrapped around the outside. A resistor like this is described as wire-wound. The number of copper turns controls the resistance very precisely: the more copper turns, and the thinner the copper, the higher the resistance. In smaller-value resistors, designed for lower-power circuits, the copper winding is replaced by a spiral pattern of carbon. Resistors like this are much cheaper to make and are called carbon-film. Generally, wire-wound resistors are more precise and more stable at higher operating temperatures.

Photo: Inside a wire-wound resistor. Break one in half, scratch away the paint, and you can clearly see the insulating ceramic core and the conducting copper wire wrapped around it.

How does the size of a resistor affect its resistance?

Suppose you're trying to force water through a pipe. Different sorts of pipes will be more or less obliging, so a fatter pipe will resist the water less than a thinner one and a shorter pipe will offer less resistance than a longer one. If you fill the pipe with, say, pebbles or sponge, water will still trickle through it but much more slowly. In other words, the length, cross-sectional area (the area you see looking into the pipe to see what's inside), and stuff inside the pipe all affect its resistance to water.

Electrical resistors are very similar—affected by the same three factors. If you make a wire thinner or longer, it's harder for electrons to wiggle through it. And, as we've already seen, it's

harder for electricity to flow through some materials (insulators) than others (conductors). Although Georg Ohm is best known for relating voltage, current, and resistance, he also researched the relationship between resistance and the size and type of material from which a resistor is made. That led him to another important equation:

R = ρ × L / A

In simple words, the resistance (R) of a material increases as its length increases (so longer wires offer more resistance) and increases as its area decreases (thinner wires offer more resistance). The resistance is also related to the type of material from which a resistor is made, and that's indicated in this equation by the symbol ρ, which is called the resistivity, and measured in units of Ωm (ohm meters). Different materials have very different resistivities: conductors have much lower resistivity than insulators. At room temperature, aluminum comes in at about 2.8 x 10−8 Ωm, while copper (a better conductor) is significantly lower at 1.7 −8 Ωm. Silicon (a semiconductor) has a resistivity of about 1000 Ωm and glass (a good insulator) measures about 1012 Ωm. You can see from these figures how vastly different conductors and insulators are in their ability to carry electricity: silicon is about 100 billion times worse than copper and glass is about a billion times worse again!

Resistance and temperature

Chart: The resistance of a material increases with temperature. This chart shows how resistivity (basic resistance of a material, independent of its length or area) increases almost linearly as the temperature increases from absolute zero up to about 600K (327°C) for four common metals. Drawn using original data from "Electrical Resistivity of Selected Elements" by P. Desai et al, J. Phys. Chem. Ref. Data, Vol 13, No 4, 1984 and "Electrical Resistivity of Copper, Gold, Palladium, and Silver" by R. Matula, J. Phys. Chem. Ref. Data, Vol 8, No 4, 1979, courtesy of US National Institute of Standards and Technology Open Data.

The resistance of a resistor isn't constant, even if it's a certain material of a fixed length and area: it steadily increases as the temperature increases. Why? The hotter a material, the more its atoms or ions jiggle about and the harder it is for electrons to wriggle through, which translates into higher electrical resistance. Broadly speaking, the resistivity of most materials increases linearly with temperature (so if you increase the temperature by 10 degrees, the resistivity increases by a certain amount, and if you increase it by another 10 degrees, the resistivity rises by the same amount again). If you cool a material, you lower its resistivity—and if you cool it to an extremely low temperature, you can sometimes make the resistivity disappear altogether, in a phenomenon known as superconductivity.

Resistor color codes

You can figure out the resistance of a resistor from the pattern of colored bands.

1. On most resistors, you'll see there are three rainbow-colored bands, then a space, then a fourth band colored brown, red, gold, or silver.

2. Turn the resistor so the three rainbow bands are on the left. 3. The first two of the rainbow bands tell you the first two digits of the resistance. Suppose

you have a resistor like the one shown here, with colored bands that are brown, black, and red and a fourth golden band. You can see from the color chart below that brown means 1 and black means 0, so the resistance is going to start with "10". The third band is a decimal multiplier: it tells you how many powers of ten to multiply the first two numbers by (or how many zeros to add on the end, if you prefer to think of it that way). Red means 2, so we multiply the 10 we've got already by 10 × 10 = 100 and get 1000. Our resistor is 1000 ohms.

4. The final band is called the tolerance and it tells you how accurate the resistance value you've just figured out is likely to be. If you have a final band colored gold, it means the resistance is accurate to within plus or minus 5 percent. So while the officially stated resistance is 1000 ohms, in practice, the real resistance is likely to be anywhere between 950 and 1050 ohms.

5. If there are five bands instead of four, the first three bands give the value of the resistance, the fourth band is the decimal multiplier, and the final band is the tolerance. Five-band

resistors quoted with three digits and a multiplier, like this, are necessarily more accurate than four-band resistors, so they have a lower tolerance value.

Circuits

From

Soclof, S., Science, E. and Production, E. (2018). How Circuits Work. [online] HowStuffWorks. Available at: https://science.howstuffworks.com/environmental/energy/circuit1.htm [Accessed 24 Jul. 2018].

You've probably heard these terms before. You knew they had something to do with electricity, but maybe you weren't sure quite sure how.

Just as your heart produces the pressure to make blood circulate, a battery or generator produces the pressure or force to push electrons around a circuit. Voltage is the force and is measured in volts (V). A typical flashlight battery produces 1.5V, and the standard household electrical voltage is 110V or 220V.

Electrical current, or flow of electrons, is measured in amperes (A). The product of electric force (in volts) and current (in amperes) is electrical power, measured in watts (W). A battery generating 1.5V and producing a current flow of 1A through a flashlight bulb delivers 1.5V x 1A = 1.5W of electrical power.

The blood flowing through your body doesn't get a free ride. The walls of the blood vessels impede the flow, and the smaller the blood vessel, the more the resistance to flow. Some of the pressure produced by your heart is just for pushing blood through blood vessels. As electrons move through wires, they bump into atoms. This impedes the flow of the electrons. The wire offers resistance to the flow of the current. The amount of resistance depends on the material, diameter and length of the wire. The resistance increases as the diameter of the wire decreases. Resistance is in units of ohms (Ω).

Ohm's Law relates voltage, current and resistance:

Resistance (Ω) = Voltage (V)/ Current (I)

Ohm's Law can be written as R = V/I.

Electric circuits are composed of wires and other components -- like light bulbs, transistors, computer chips and motors. Wires, made of metals called conductors that have a low resistance to current, connect the components. Copper and aluminum are the most common conductors. Gold, because of its resistance to corrosion, is often used for attaching wires to tiny electronic chips.

In an incandescent bulb, the current flows through a thin tungsten wire or a metallic filament that offers high resistance to current flow. When the electrons bump into the atoms, the friction, or loss of kinetic energy, produces heat. If the temperature of the filament is high enough, it starts to glow and give off light. This is incandescence. Typical filament temperatures for light bulbs are around 4,600 degrees F (2,550 degrees C). Unfortunately, 90 to 95 percent of the energy supplied to a light bulb is lost in the form of heat rather than light, so incandescent bulbs are very inefficient.

Fluorescent lights produce light by having electrons pass through a tube filled with mercury vapor and neon or argon gas. As the electrons bump into the mercury atoms, they cause electrons in the atoms to absorb some of their energy. As these electrons return to their normal state, they radiate bundles of light energy called photons. Fluorescent lights are four to five times more efficient than incandescent bulbs.

A closed circuit has a complete path for current to flow. An open circuit doesn't, which means that it's not functional. If this is your first exposure to circuits, you might think that when a circuit is open, it's like an open door or gate that current can flow through. And when it's closed, it's like a shut door that current can't flow through. Actually, it's just the opposite, so it might take awhile to get used to this concept.

A short circuit is a low-resistance path, usually made unintentionally, that bypasses part of a circuit. This can happen when two bare wires in a circuit touch each other. The part of the circuit bypassed by the short circuit ceases to function, and a large amount of current could start to flow. This can generate a lot of heat in the wires and cause a fire. As a safety measure, fuses and circuit breakers automatically open the circuit when there is an excessive current.

In a series circuit, the same current flows through all the components. The total voltage across the circuit is the sum of the voltages across each component, and the total resistance is the sum of the resistances of each component. In this circuit, V = V1 + V2 + V3 and R = R1 + R2 + R3. An example of a series circuit is a string of Christmas lights. If any one of the bulbs is missing or burned out, no current will flow and none of the lights will go on.

Parallel circuits are like the smaller blood vessels that branch off from an artery and then connect to a vein to return blood to the heart. Now think of two wires, each representing an artery and a vein, with some smaller wires connected between them. These smaller wires will have the same voltage applied to them, but different amounts of current flowing through them depending on the resistance of the individual wires.

An example of a parallel circuit is the wiring system of a house. A single electric power source supplies all the lights and appliances with the same voltage. If one of the lights burns out, current can still flow through the rest of the lights and appliances. However, if there is a short circuit, the voltage drops to almost zero, and the entire system goes down.

Circuits are generally very complex combinations of series and parallel circuits. The first circuits were very simple DC circuits.

Early investigations of static electricity go back hundreds of years. Static electricity is a transfer of electrons produced by friction, like when you rub a balloon across a sweater. A spark or very brief

flow of current can occur when charged objects come into contact, but there is no continuous flow of current. In the absence of a continuous current, there is no useful application of electricity.

The invention of the battery -- which could produce a continuous flow of current -- made possible the development of the first electric circuits. Alessandro Volta invented the first battery, the voltaic pile, in 1800. The very first circuits used a battery and electrodes immersed in a container of water. The flow of current through the water produced hydrogen and oxygen.

The first widespread application of electric circuits for practical use was for electric lighting. Shortly after Thomas Edison invented his incandescent light bulb, he sought practical applications for it by developing an entire power generation and distribution system. The first such system in the United States was the Pearl Street Station in downtown Manhattan. It provided a few square blocks of the city with electric power, primarily for illumination.

One classification of circuits has to do with the nature of the current flow. The earliest circuits were battery-powered, which made in a steady, constant current that always flowed in the same direction. This is direct current, or DC. The use of DC continued through the time of the first electric power systems. A major problem with the DC system was that power stations could serve an area of only about a square mile because of power loss in the wires.

In 1883, engineers proposed harnessing the tremendous hydroelectric power potential of Niagara Falls to supply the needs of Buffalo, N.Y. Although this power would ultimately go beyond Buffalo to New York City and even farther, there was an initial problem with distance. Buffalo was only 16 miles from Niagara Falls, but the idea was unworkable -- until Nikola Tesla made it possible.

Engineer Nikola Tesla, aided by theoretical work by Charles Proteus Steinmetz, came up with the idea of using alternating current, or AC. Unlike direct current, AC is always changing and repeatedly reverses direction.

So why was AC the answer to the problem of long-distance power transmission? With AC, it's possible to use transformers to change voltage levels in a circuit. Transformers work on a principle of magnetic induction, which requires a changing magnetic field produced by the alternating current. With transformers, voltages can be increased for long-distance transmission. At the receiving end, the voltage level can decrease to a safer 220V or 110V for business and residential use.

We need high voltages for long distances because wire resistance causes power loss. The electrons bumping into atoms lose energy in the form of heat as they travel. This power loss is proportional to the square of the amount of current moving through the wire.

To measure the amount of power the line transmits, you can multiply the voltage by the current. You can express these two ideas using an equation in which I represents current, V represents voltage and P equals power:

P = V x I

Let's consider the example of transmitting 1 megawatt. If we increase the voltage from 100V to 10,000V, we can then decrease the current from 10,000A to 100A. This will reduce the power loss by (100)2, or 10,000. This was Tesla's concept, and from that idea power transmission from Niagara Falls to Buffalo, and ultimately to New York City and beyond, became a reality.

In the United States and many other countries, the standard frequency for AC power is 60 cycles per second, or 60 hertz. This means that 60 times a second, a complete cycle of the current flows in one direction and then in the other. The current flows in one direction for 1/120th of a second and in the other direction for another 1/120th of a second. The time it takes for one cycle to be completed is called a period, which in this case is 1/60th of a second. In Europe and other areas, the standard frequency for AC power is 50 hertz.

Materials and Supplies: Investigating Circuits: Materials needed for each person/pair/lab group

a) 3 miniature bulbs and holders b) 6 leads with alligator clips c) Flexible copper wire (~ 12 – 15 cm) d) Low current LED bulb e) Standard #2 pencil f) Artist quality graphite pencil g) Charcoal/Charcoal pencil

Lab Practicum: Materials to create seven lab kits with each kit containing (white vinegar, lemon, aluminum foil, text books and distilled water provided in class)

h) plastic storage container i) multi-meter (with leads) j) 5 film-canisters k) caliper l) ruler m) 2 D batteries with holder n) 2 low current LED bulbs o) 2 miniature bulbs and holders p) 7 leads with alligator clips q) Squeeze bottle (containing lemon juice) r) 6 galvanized nails s) 7 pieces of copper wire (10-14 gauge) cut to ~ 8 mm t) 1 pre-1982 penny u) 1 post-1965 nickel, dime, or quarter

Summary/Plan: The purpose of my summer research was to investigate possible hands-on applications for teaching concepts of electrical energy, current and circuits to my students. These are important concepts of physics but also have real-world connections and importance to students. There are many Electricity lessons that can be found on the internet and teacher resources, but I wanted to create units that were created for different academic levels and that could be taken as a whole or individually. I wanted to take the time to go through internet experiments and try them out and find out what works well and what does not. For example, I had researched citrus and film canister electrochemical cells but I had found most of the resources were inaccurate, the labs did not work, or were too basic for an advanced physics classes. My goal was to complete an in-depth and meaningful set of lessons and activities that had already been tested by a teacher. The lessons cover the following topics.

Chemical Reactions: First a demonstration that introduced the students to chemical reactions to lead into a discussion of chemical energy. It is important to relate chemistry and physics and show the students the relationship that ties the two disciplines together. The Cleaning Pennies lab is a simple way to open up dialogue about chemical energy and reactions.

Electrochemical Cells (Citrus and Film Canister batteries): This is a very important topic and great hook for students. The students are all familiar with batteries but are not always sure of the basic components and how they produce electrical energy. Various fruits were tested (lemons, limes, oranges, and grapefruits) along with different types of electrodes (galvanized nails, pre and post 1982 pennies (cleaned and un-cleaned, sanded and un-sanded), copper wire, nickels, dimes, and quarters). For the greatest voltage, best wow factor, and the most cost-efficient my findings were to use a lemon with galvanized nail and pre-1982 penny (un-cleaned and un-sanded). I also determined that two lemon cells in series could power a calculator and four would power a low-current LED.

For cost and re-usability purposes I also wanted to explore creating electrochemical cells using a solution within a film canister. The voltage was measured for several different solutions: white vinegar, cola, diet cola, lemon-lime soda, salt water (iodized and non-iodized), bottled water, Gatorade, distilled water, purified water, lime juice from concentrate, lemon juice, and water and baking soda. For the electrodes I tested various gauges of copper wire. For the greatest voltage, best wow factor, and the most cost-efficient my findings were to use white vinegar, a galvanized nail, and 10-14 gauge copper wire. I also determined that two film canister cells in series could power a calculator and four would power a low-current LED.

Series and Parallel Circuits/Resistors: An investigation lab introduces the concepts of circuits (parallel and series), components of circuits (including materials for conductors) and diodes. The Lab Practicum explores further into this concept using various resistors in series and parallel circuits that would give the closest measurement to the label and calculated total resistance. I found that using too small of a resistance would give too much error especially when calculating equivalent resistance in parallel circuits.

Included in this section are problems on Ohm’s Law and Circuits and two different projects relating to Electrical Circuits (Pioneers in Electricity/Electrical Circuits and Integrating Technology and Design).

Capacitors: The Lab Practicum includes and activity for the students to create a parallel capacitor and obtain results that demonstrate an inverse relationship between capacitance and plate distance.

The main goal of any program like STEP-UP is to provide experience for teachers that can be brought back into the classroom. If the action plan does not meet the standards or is too extensive and/or expensive it will not be done in the classroom and the students will not benefit from the teacher’s experience. I understand this and have created a one lesson or a whole unit on two very crucial concepts: Electrical Energy and Current and Circuits and Circuit Elements. I am using my research on Energy and Utilities to produce meaningful activities, demonstrations, Power Points, and labs for my students. From these items and my experience I hope for the students to:

1) Use critical thinking skills to analyze data and make relevant real-world connections.

2) Complete activities that require inquiry and high order thinking skills.

3) Encourage thought on the environmental impact of the concepts.

4) Increase knowledge of scientific concepts.

5) Encourage further exploration of concepts.

OHM’S LAW AND CIRCUITS

Name: __________________________ Date: __________________ Period: _____

Ohm’s Law: the law relating the voltage applied to a wire to the current produced and the wire’s resistance in a circuit.

Ohm’s Law Problems:

1) A circuit is connected to a 6 V battery and contains 2 A of current. What is the resistance in the circuit?

2) How much current will flow through a circuit with a 4 Ω resistor and a 12 V voltage source?

3) What is the voltage across a resistor with a value of 50 Ω and draws a current of 3 A?

http://www.m0zpk.co.uk/ohms-law-equations-and-triangle/

4) How much resistance allows for a 9 V battery to produce 0.5 A of current?

5) A resistance of 3 Ω is placed in a circuit with a 12 V battery. What amount of current flows within the circuit?

6) A motor with a resistance of 14 Ω is connected to a voltage source with 2.5 A of current flowing through the circuit. What is the voltage of the source?

Electric Circuit: a completely closed, conducting pathway (loop) through which electrical current flows driven by a voltage source. It also may contain various electrical components.

ΔV: Voltage Source for Circuit

I: Current

R: Resistor

1) What is the current in the circuit above if the ΔV is 12 V and R is 6 Ω? Types of Electrical Circuits: There are two types of electrical circuits: Series and Parallel circuits. Series circuits only allows the current to flow through a single pathway (think a one way street). A Parallel circuit provides several pathways for current to flow through (think of a multiple lane highway).

Series Circuit:

- Current is the same through each part of the circuit. o IT = I1 = I2 = I3 …

- The total Voltage applied (from battery, outlet, etc.) is equal to the sum of the individual voltage drops across the resistors.

o VT = V1 + V2 + V3 … - The total Resistance is equal to the sum of the individual resistors.

o RT = R1 + R2 + R3 … Problems:

1) Two resistors are in series and R2 has two times greater the resistance of R1 (See Diagram S-1). Relate the current in R2 compared to R1.

2) Referring to Diagram S-1, if the voltage across the battery is 24 V and R1 = 5 Ω and R2 = 7 Ω, what is the total current and total resistance in the circuit?

A) Two times the value B) The same value C) ½ the value D) ¼ the value Diagram: S-1

3) If a third resistor was added to the circuit in Diagram S-1 would the total resistance increase, decrease or stay the same?

4) Referring to Diagram S-2, if the voltage across the battery is 36 V

and R1 = 2 Ω and R2 = 4 Ω and R3 = 6 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______

5) Find the current going through and the voltage across R1, R2, and R3 from problem 4.

I1 = _______ V1 = _______ I2 = _______ V2 = _______ I3 = _______ V3 = _______

6) Referring to Diagram S-2, if the voltage across the battery is 9 V and R1 = 2 Ω and R2 = 1 Ω and R3 = 3 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______

Diagram: S-2

7) Find the current going through and the voltage across R1, R2, and R3 from problem 6.

I1 = _______ V1 = _______ I2 = _______ V2 = _______ I3 = _______ V3 = _______

8) Referring to Diagram S-2, if the current through Resistor two is 3 A and R1 = 3 Ω and R2 = 1 Ω and R3 = 5 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______

9) Find the voltage across each of the Resistors in problem 8. V1 = _______ V2 = _______ V3 = _______

10) Referring to Diagram S-2, if the voltage across the battery is 12 V, the current through Resistor three is 2 A and R1 = 1 Ω and R3 = 4 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______

Determine the following values for this circuit:

I1 = _______ V1 = _______ R1 = _______ I2 = _______ V2 = _______ R2 = _______ I3 = _______ V3 = _______ R3 = _______

Parallel Circuit:

- The total Current is equal to the sum of each the currents in the individual branches.

o IT = I1 + I2 + I3 … - The total Voltage applied (from battery, outlet, etc.) is the same

for each individual voltage drop across each resistor. o VT = V1 = V2 = V3 …

- The total Resistance is less than any of the individual resistors. o RT = 1/(1/R1 + 1/R2 + 1/R3 …) OR RT = (R1

-1+ R2-1 + R3

-1 …)-1

Problems:

1) Two resistors are in parallel and R2 has two times greater the resistance of R1 (See Diagram P-1). Relate the current in R2 compared to R1.

A) Two times the value B) The same value C) ½ the value D) ¼ the value

2) Referring to Diagram P-1, if the voltage across the battery is 24 V and R1 = 5 Ω and R2 = 7 Ω, what is the total current and total resistance in the circuit?

3) If a third resistor was added to the circuit in Diagram P-1 would the total resistance increase, decrease or stay the same?

4) Referring to Diagram P-2, if the voltage across the battery is 36 V and R1 = 2 Ω and R2 = 4 Ω and R3 = 6 Ω, what is the total current and total resistance in the circuit?

Diagram: P-1

Diagram: P-2

It = _______ Rt = _______

5) Find the current going through and the voltage across R1, R2, and R3 from problem 4.

I1 = _______ V1 = _______ I2 = _______ V2 = _______ I3 = _______ V3 = _______

6) Referring to Diagram P-2, if the voltage across the battery is 9 V and R1 = 2 Ω and R2 = 1 Ω and R3 = 3 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______

7) Find the current going through and the voltage across R1, R2, and R3 from problem 6.

I1 = _______ V1 = _______ I2 = _______ V2 = _______ I3 = _______ V3 = _______

8) Referring to Diagram P-2, if the current through Resistor two is 3 A and R1 = 3 Ω and R2 = 1 Ω and R3 = 5 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______

9) Find the voltage across each of the Resistors in problem 8. V1 = _______ V2 = _______ V3 = _______

10) Referring to Diagram P-2, if the voltage across the battery is 12 V, the current through Resistor two is 2 A and R1 = 1 Ω and R3 = 4 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______

Determine the following values for this circuit:

I1 = _______ V1 = _______ R1 = _______ I2 = _______ V2 = _______ R2 = _______ I3 = _______ V3 = _______ R3 = _______

OHM’S LAW AND CIRCUITS

ANSWER KEY

Name: _______________________________ Date: __________________ Period: _____

Ohm’s Law: the law relating the voltage applied to a wire to the current produced and the wire’s resistance in a circuit.

Ohm’s Law Problems:

1) A circuit is connected to a 6 V battery and contains 2 A of current. What is the resistance in the circuit?

V = 6 V I = 2 A R = ?

R = V/I (6 V)/(2 A) = 3 Ω

2) How much current will flow through a circuit with a 4 Ω resistor and a 12 V voltage source?

V = 12 V I = ? R = 4 Ω I = V/R (12 V)/(4 Ω) = 3 A

3) What is the voltage across a resistor with a value of 50 Ω and draws

a current of 3 A? V = ? I = 3 A R = 50 Ω

V = IR (3 A)(50 Ω) = 150 V

http://www.m0zpk.co.uk/ohms-law-equations-and-triangle/

4) How much resistance allows for a 9 V battery to produce 0.5 A of current?

V = 9 V I = 0.5 A R = ? R = V/I (9 V)/(0.5 A) = 18 Ω

5) A resistance of 3 Ω is placed in a circuit with a 12 V battery. What amount of current flows within the circuit?

V = 12 V I = ? R = 3 Ω I = V/R (12 V)/(3 Ω) = 4 A

6) A motor with a resistance of 14 Ω is connected to a voltage source

with 2.5 A of current flowing through the circuit. What is the voltage of the source?

V = ? I = 2.5 A R = 14 Ω V = IR (2.5 A)(14 Ω) = 35 V

Electric Circuit: a completely closed, conducting pathway (loop) through which electrical current flows driven by a voltage source. It also may contain various electrical components.

What is the current in the circuit above if the ΔV is 12 V and R is 6 Ω?

V = 12 V I = ? R = ?

ΔV: Voltage Source for Circuit

I: Current

R: Resistor

R = V/I (6 V)/(2 A) = 3 Ω

Types of Electrical Circuits: There are two types of electrical circuits: Series and Parallel circuits. Series circuits only allows the current to flow through a single pathway (think a one way street). A Parallel circuit provides several pathways for current to flow through (think of a multiple lane highway).

Series Circuit:

- Current is the same through each part of the circuit. o IT = I1 = I2 = I3 …

- The total Voltage applied (from battery, outlet, etc.) is equal to the sum of the individual voltage drops across the resistors.

o VT = V1 + V2 + V3 … - The total Resistance is equal to the sum of the individual resistors.

o RT = R1 + R2 + R3 … Problems:

1) Two resistors are in series and R2 has two times greater the resistance of R1 (See Diagram S-1). Relate the current in R2 compared to R1.

B) The same value: Series circuits have the same current through each resistor regardless of the resistor’s value.

A) Two times the value B) The same value C) ½ the value D) ¼ the value Diagram: S-1

2) Referring to Diagram S-1, if the voltage across the battery is 24 V and R1 = 5 Ω and R2 = 7 Ω, what is the total current and total resistance in the circuit?

RT = 5 Ω + 7 Ω = 12 Ω VT = 24 V IT = V/R (24 V)/(12 Ω) = 2 A

3) If a third resistor was added to the circuit in Diagram S-1 would the total resistance increase, decrease or stay the same?

It would increase.

4) Referring to Diagram S-2, if the voltage across the battery is 36 V and R1 = 2 Ω and R2 = 4 Ω and R3 = 6 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______ VT = 36 V Rt = 2 Ω + 4 Ω + 6 Ω = 12 Ω It = V/R = (36 V)/(12 Ω) = 3 A

5) Find the current going through and the voltage across R1, R2, and R3 from problem 4.

I1 = _______ V1 = _______ I1 = 3 A; V1 = IR (3 A)(2 Ω) = 6 V I2 = _______ V2 = _______ I1 = 3 A; V1 = IR (3 A)(4 Ω) = 12 V I3 = _______ V3 = _______ I1 = 3 A; V1 = IR (3 A)(6 Ω) = 18 V

Diagram: S-2

6) Referring to Diagram S-2, if the voltage across the battery is 9 V and R1 = 2 Ω and R2 = 1 Ω and R3 = 3 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______ VT = 9 V Rt = 2 Ω + 1 Ω + 3 Ω = 6 Ω It = V/R = (9 V)/(6 Ω) = 1.5 A

7) Find the current going through and the voltage across R1, R2, and R3 from problem 6.

I1 = _______ V1 = _______ I1 = 1.5 A; V1 = IR (1.5 A)(2 Ω) = 3 V I2 = _______ V2 = _______ I2 = 1.5 A; V2 = IR (1.5 A)(1 Ω) = 1.5 V I3 = _______ V3 = _______ I3 = 1.5 A; V3 = IR (1.5 A)(3 Ω) = 4.5 V

8) Referring to Diagram S-2, if the current through Resistor two is 3 A and R1 = 3 Ω and R2 = 1 Ω and R3 = 5 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______ If I2 = 3 A then It = 3 A Rt = 3 Ω + 1 Ω + 5 Ω = 9 Ω Vt = IR (3 A)(9 Ω) = 27 V

9) Find the voltage across each of the Resistors in problem 8.

V1 = _______ V1 = IR (3 A)(3 Ω) = 9 V V2 = _______ V2 = IR (3 A)(1 Ω) = 3 V V3 = _______ V3 = IR (3 A)(5 Ω) = 15 V

10) Referring to Diagram S-2, if the voltage across the battery is 12 V, the current through Resistor three is 2 A and R1 = 1 Ω and R3 = 4 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______ If I3 = 2 A then It = 2 A Rt = V/I (12 V)/(2 A) = 6 Ω Vt = 12 V

Determine the following values for this circuit:

I1 = _______ V1 = _______ R1 = _______ I1 = 2 A R1 = 1 Ω V1 = IR (2 A)(1 Ω) = 2 V I2 = _______ V2 = _______ R2 = _______ I2 = 2 A R2 = Rt – (R1 + R2) 6 Ω - (1 Ω + 4 Ω) = 1 Ω V2 = IR (2 A)(1 Ω) = 2 V I3 = _______ V3 = _______ R3 = _______ I3= 2 A R3 = 4 Ω V3 = IR (2 A)(4 Ω) = 8 V

Parallel Circuit:

- The total Current is equal to the sum of each the currents in the individual branches.

o IT = I1 + I2 + I3 … - The total Voltage applied (from battery, outlet, etc.) is the same

for each individual voltage drop across each resistor. o VT = V1 = V2 = V3 …

- The total Resistance is less than any of the individual resistors. o RT = 1/(1/R1 + 1/R2 + 1/R3 …) OR RT = (R1

-1+ R2-1 + R3

-1 …)-1

Problems:

1) Two resistors are in parallel and R2 has two times greater the resistance of R1 (See Diagram P-1). Relate the current in R2 compared to R1.

2) Referring to Diagram P-1, if the voltage across the battery is 24 V and R1 = 5 Ω and R2 = 7 Ω, what is the total current and total resistance in the circuit?

Vt = 24 V RT = 1/(1/5 Ω + 1/7 Ω ) = 2.9 Ω It = V/R (24 V)/(2.9 Ω) = 8.3 A

Diagram: P-1

A) Two times the value

B) The same value C) ½ the value D) ¼ the value

C) ½ the value

3) If a third resistor was added to the circuit in Diagram P-1 would

the total resistance increase, decrease or stay the same? Decrease

4) Referring to Diagram P-2, if the voltage across the battery is 36 V and R1 = 2 Ω and R2 = 4 Ω and R3 = 6 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______ Vt = 36 V RT = 1/(1/2 Ω + 1/4 Ω + 1/6 Ω) = 1.1 Ω It = V/R (36 V)/(1.1 Ω) = 33 A

5) Find the current going through and the voltage across R1, R2, and R3 from problem 4.

I1 = _______ V1 = _______ V1 = 36 V I1 = V/R (36 V)/(2 Ω) = 18 A I2 = _______ V2 = _______ V2 = 36 V I2 = V/R (36 V)/(4 Ω) = 9 A

Diagram: P-2

I3 = _______ V3 = _______ V3 = 36 V I3 = V/R (36 V)/(6 Ω) = 6 A

6) Referring to Diagram P-2, if the voltage across the battery is 9 V and R1 = 2 Ω and R2 = 1 Ω and R3 = 3 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______ Vt = 9 V RT = 1/(1/2 Ω + 1/1 Ω + 1/3 Ω) = 0.545 Ω It = V/R (9 V)/(0.545 Ω) = 16.5 A

7) Find the current going through and the voltage across R1, R2, and R3 from problem 6.

I1 = _______ V1 = _______ V1 = 9 V I1 = V/R (9 V)/(2 Ω) = 4.5 A I2 = _______ V2 = _______ V2 = 9 V I2 = V/R (9 V)/(1 Ω) = 9 A I3 = _______ V3 = _______ V3 = 9 V I3 = V/R (9 V)/(3 Ω) = 3 A

8) Referring to Diagram P-2, if the current through Resistor two is 3 A and R1 = 3 Ω and R2 = 1 Ω and R3 = 5 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______ I2 = 3 A RT = 1/(1/3 Ω + 1/1 Ω + 1/5 Ω) = 0.65 Ω V2 = I2R2 = (3 A)(1 Ω) = 3 V = Vt = 3 V

It = V/R = (3 V)/(0.65 Ω) = 4.6 A

9) Find the current going through and the voltage across R1, R2, and R3 from problem 8.

I1 = _______ V1 = _______ V1 = 3 V I1 = V/R (3 V)/(3 Ω) = 1 A

I2 = _______ V2 = _______ V2 = 3 V I2 = V/R (3 V)/(1 Ω) = 3 A I3 = _______ V3 = _______ V3 = 3 V I3 = V/R (3 V)/(5 Ω) = 0.6 A

10) Referring to Diagram P-2, if the voltage across the battery is 12 V, the current through Resistor two is 2 A and R1 = 1 Ω and R3 = 4 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______ Vt = 12 V I2 = 2 A R2 = V2/I2 = (12 V)/(2 A) = 6 Ω RT = 1/(1/1 Ω + 1/6 Ω + 1/4 Ω) = 0.706 Ω It = V/R (12 V)/(0.706 Ω) = 17 A

Determine the following values for this circuit:

I1 = _______ V1 = _______ R1 = _______ V1 = 12 V R1 = 1 Ω I1 = V/R (12 V)/(1 Ω) = 12 A I2 = _______ V2 = _______ R2 = _______ V2 = 12 V R2 = 6 Ω (see above) I2 = V/R (12 V)/(6 Ω) = 2 A I3 = _______ V3 = _______ R3 = _______ V1 = 12 V R1 = 4 Ω I1 = V/R (12 V)/(4 Ω) = 3 A

OHM’S LAW AND CIRCUITS Name: _______________________________ Date: __________________ Period: _____

Ohm’s Law: the law relating the voltage applied to a wire to the current produced and the wire’s resistance in a circuit.

Ohm’s Law Problems:

7) A circuit is connected to a 6 V battery and contains 2 mA of current. What is the resistance in the circuit?

8) How much current will flow through a circuit with a 4 kΩ resistor and a 120 V voltage source?

9) What is the voltage across a resistor with a value of 50 Ω and draws a current of 3 A?

10) How much resistance allows for a 9 V battery to produce 0.25 A of current?

11) A resistance of 3 Ω is placed in a circuit with a 12 V battery. What amount of current flows within the circuit?

12) A motor with a resistance of 14 Ω is connected to a voltage source with 2.5 A of current flowing through the circuit. What is the voltage of the source?

Electric Circuit: a completely closed, conducting pathway (loop) through which electrical current flows driven by a voltage source. It also may contain various electrical components.

2) What is the current in the circuit above if the ΔV is 120 V and R is 37 Ω?

3) What direction is the current moving? Why?

ΔV: Voltage Source for Circuit

I: Current

R: Resistor

Types of Electrical Circuits: There are two types of electrical circuits: Series and Parallel circuits. Series circuits only allows the current to flow through a single pathway (think a one way street). A Parallel circuit provides several pathways for current to flow through (think of a multiple lane highway).

Series Circuit:

- Current is the same through each part of the circuit. o IT = I1 = I2 = I3 …

- The total Voltage applied (from battery, outlet, etc.) is equal to the sum of the individual voltage drops across the resistors.

o VT = V1 + V2 + V3 … - The total Resistance is equal to the sum of the individual resistors.

o RT = R1 + R2 + R3 … Problems:

11) Two resistors are in series and R2 has two times greater the resistance of R1 (See Diagram S-1). Relate the current in R2 compared to R1.

Explain your answer.

E) Two times the value F) The same value G) ½ the value H) ¼ the value Diagram: S-1

12) Referring to Diagram S-1, if the voltage across the battery is 24 V and R1 = 5 Ω and R2 = 7 Ω, what is the total current and total resistance in the circuit?

13) If a third resistor was added to the circuit in Diagram S-1 would the total resistance increase, decrease or stay the same? Explain your answer.

14) Referring to Diagram S-2, if the voltage across the battery is 36 V and R1 = 2 Ω and R2 = 4 Ω and R3 = 6 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______

15) Find the current going through and the voltage across R1, R2, and R3 from problem 4.

I1 = _______ V1 = _______ I2 = _______ V2 = _______ I3 = _______ V3 = _______

Diagram: S-2

16) Referring to Diagram S-2, if the voltage across the battery is 9 V and R1 = 2 Ω and R2 = 1 Ω and R3 = 3 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______

17) Find the current going through and the voltage across R1, R2, and R3 from problem 6.

I1 = _______ V1 = _______ I2 = _______ V2 = _______ I3 = _______ V3 = _______

18) Referring to Diagram S-2, if the current through Resistor two is 3 A and R1 = 3 Ω and R2 = 1 Ω and R3 = 5 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______

19) Find the voltage across each of the Resistors in problem 8. V1 = _______ V2 = _______ V3 = _______

20) Referring to Diagram S-2, if the voltage across the battery is 12 V, the current through Resistor three is 2 A and R1 = 1 Ω and R3 = 4

Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______

Determine the following values for this circuit:

I1 = _______ V1 = _______ R1 = _______ I2 = _______ V2 = _______ R2 = _______ I3 = _______ V3 = _______ R3 = _______

Parallel Circuit:

- The total Current is equal to the sum of each the currents in the individual branches.

o IT = I1 + I2 + I3 … - The total Voltage applied (from battery, outlet, etc.) is the same

for each individual voltage drop across each resistor. o VT = V1 = V2 = V3 …

- The total Resistance is less than any of the individual resistors. o RT = 1/(1/R1 + 1/R2 + 1/R3 …) OR RT = (R1

-1+ R2-1 + R3

-1 …)-1

Problems:

11) Two resistors are in parallel and R2 has two times greater the resistance of R1 (See Diagram P-1). Relate the current in R2 compared to R1. Explain.

12) Referring to Diagram P-1, if the voltage across the battery is 24 V and R1 = 5 Ω and R2 = 7 Ω, what is the total current and total resistance in the circuit?

13) If a third resistor was added to the circuit in Diagram P-1 would the total resistance increase, decrease or stay the same? Explain.

14) Referring to Diagram P-2, if the voltage across the battery is 36 V and R1 = 2 Ω and R2 = 4 Ω and R3 = 6 Ω, what is the total current and total resistance in the circuit?

Diagram: P-1

E) Two times the value F) The same value G) ½ the value H) ¼ the value

It = _______ Rt = _______

15) Find the current going through and the voltage across R1, R2, and R3 from problem 4.

I1 = _______ V1 = _______ I2 = _______ V2 = _______ I3 = _______ V3 = _______

16) Referring to Diagram P-2, if the voltage across the battery is 9 V and R1 = 2 Ω and R2 = 1 Ω and R3 = 3 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______

Diagram: P-2

17) Find the current going through and the voltage across R1, R2, and R3 from problem 6.

I1 = _______ V1 = _______ I2 = _______ V2 = _______ I3 = _______ V3 = _______

18) Referring to Diagram P-2, if the current through Resistor two is 3 A and R1 = 3 Ω and R2 = 1 Ω and R3 = 5 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______

19) Find the voltage across each of the Resistors in problem 8. V1 = _______ V2 = _______ V3 = _______

20) Referring to Diagram P-2, if the voltage across the battery is 12 V, the current through Resistor two is 2 A and R1 = 1 Ω and R3 = 4 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______

Determine the following values for this circuit:

I1 = _______ V1 = _______ R1 = _______ I2 = _______ V2 = _______ R2 = _______ I3 = _______ V3 = _______ R3 = _______

Combination (Complex) Circuits:

Most circuits are a combination of both series and parallel circuits. Using the previous rules the circuits will be reduced to an equivalent series or parallel circuit. See handout: “Combination (Complex) Circuits: Solving for Equivalent Resistance”

Solve for the Req and Itotal for the following Circuit. The Voltage across the battery is 120 V and these are the following values for the Resistors:

R1 = 2 Ω, R2 = 4 Ω, R3 = 6 Ω, R4 = 5 Ω, R5 = 3 Ω, R6 = 12 Ω

BONUS: Find the Voltage across and the Current through each Resistor.

V1 = _________ I1 = _________ V2 = _________ I2 = _________

V3 = _________ I3 = _________ V4 = _________ I4 = _________

V5 = _________ I5 = _________ V6 = _________ I6 = _________

ANSWER KEY OHM’S LAW AND CIRCUITS

Name: _______________________________ Date: __________________ Period: _____

Ohm’s Law: the law relating the voltage applied to a wire to the current produced and the wire’s resistance in a circuit.

Ohm’s Law Problems:

1) A circuit is connected to a 6 V battery and contains 2 mA of current. What is the resistance in the circuit?

V = 6 V I = 2 E -3 A R = ? R = V/I (6 V)/(2 E -3 A) = 3000 Ω

2) How much current will flow through a circuit with a 4 kΩ resistor and a 120 V voltage source?

V = 120 V I = ? R = 4,000 Ω R = V/I I = V/R (120 V)/(4,000 Ω) = 0.03 A

4) What is the voltage across a resistor with a value of 50 Ω and draws a current of 3 A?

V = ? I = 3 A R = 50 Ω R = V/I V = IR (3 A)(50 Ω) = 150 V

5) How much resistance allows for a 9 V battery to produce 0.25 A of current?

V = 6 V I = 2 A R = ? R = V/I (9 V)/(0.25 A) = 36 Ω

6) A resistance of 3 Ω is placed in a circuit with a 12 V battery. What amount of current flows within the circuit?

V = 12 V I = ? R = 3 Ω R = V/I I = V/R (12 V)/(3 Ω) = 4 A

7) A motor with a resistance of 14 Ω is connected to a voltage source with 2.5 A of current flowing through the circuit. What is the voltage of the source?

V = ? I = 2.5 A R = 14 Ω R = V/I V = IR (2.5 A)(14 Ω) = 35 V Electric Circuit: a completely closed, conducting pathway (loop) through which electrical current flows driven by a voltage source. It also may contain various electrical components.

8) What is the current in the circuit above if the ΔV is 120 V and R is 37 Ω?

V = 120 V I = ? R = 37 Ω R = V/I I = V/R (120 V)/(37 Ω) = 3.24 A

9) What direction is the current moving? Why? From + (high potential) to – (low potential); conventional current To provide a definition of current independent of the type of charge carriers, conventional current is defined as moving in the same direction as the positive charge flow.

ΔV: Voltage Source for Circuit

I: Current

R: Resistor

Types of Electrical Circuits: There are two types of electrical circuits: Series and Parallel circuits. Series circuits only allows the current to flow through a single pathway (think a one way street). A Parallel circuit provides several pathways for current to flow through (think of a multiple lane highway).

Series Circuit:

- Current is the same through each part of the circuit. o IT = I1 = I2 = I3 …

- The total Voltage applied (from battery, outlet, etc.) is equal to the

sum of the individual voltage drops across the resistors. o VT = V1 + V2 + V3 …

- The total Resistance is equal to the sum of the individual resistors. o RT = R1 + R2 + R3 …

Problems:

1) Two resistors are in series and R2 has two times greater the resistance of R1 (See Diagram S-1). Relate the current in R2 compared to R1.

Explain your answer. B) The same value; In a series circuit the current is the same through each resistor regardless of its resistance

A) Two times the value B) The same value C) ½ the value D) ¼ the value Diagram: S-1

2) Referring to Diagram S-1, if the voltage across the battery is 24 V and R1 = 5 Ω and R2 = 7 Ω, what is the total current and total resistance in the circuit?

RT = 5 Ω + 7 Ω = 12 Ω VT = 24 V IT = V/R (24 V)/(12 Ω) = 2 A

3) If a third resistor was added to the circuit in Diagram S-1 would the total resistance increase, decrease or stay the same? Explain your answer.

It would increase; In a series circuit each resistor adds towards the total resistance; the more resistors leads to a greater overall resistance

4) Referring to Diagram S-2, if the voltage across the battery is 36 V

and R1 = 2 Ω and R2 = 4 Ω and R3 = 6 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______ VT = 36 V Rt = 2 Ω + 4 Ω + 6 Ω = 12 Ω It = V/R = (36 V)/(12 Ω) = 3 A

5) Find the current going through and the voltage across R1, R2, and R3 from problem 4.

I1 = _______ V1 = _______ I1 = 3 A; V1 = IR (3 A)(2 Ω) = 6 V I2 = _______ V2 = _______ I1 = 3 A; V1 = IR (3 A)(4 Ω) = 12 V

Diagram: S-2

I3 = _______ V3 = _______ I1 = 3 A; V1 = IR (3 A)(6 Ω) = 18 V

6) Referring to Diagram S-2, if the voltage across the battery is 9 V and R1 = 2 Ω and R2 = 1 Ω and R3 = 3 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______ VT = 9 V Rt = 2 Ω + 1 Ω + 3 Ω = 6 Ω It = V/R = (9 V)/(6 Ω) = 1.5 A

7) Find the current going through and the voltage across R1, R2, and R3 from problem 6.

I1 = _______ V1 = _______ I1 = 1.5 A; V1 = IR (1.5 A)(2 Ω) = 3 V I2 = _______ V2 = _______ I2 = 1.5 A; V2 = IR (1.5 A)(1 Ω) = 1.5 V I3 = _______ V3 = _______ I3 = 1.5 A; V3 = IR (1.5 A)(3 Ω) = 4.5 V

8) Referring to Diagram S-2, if the current through Resistor two is 3 A and R1 = 3 Ω and R2 = 1 Ω and R3 = 5 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______ If I2 = 3 A then It = 3 A Rt = 3 Ω + 1 Ω + 5 Ω = 9 Ω Vt = IR (3 A)(9 Ω) = 27 V

9) Find the voltage across each of the Resistors in problem 8.

V1 = _______ V1 = IR (3 A)(3 Ω) = 9 V V2 = _______ V2 = IR (3 A)(1 Ω) = 3 V V3 = _______ V3 = IR (3 A)(5 Ω) = 15 V

10) Referring to Diagram S-2, if the voltage across the battery is 12 V, the current through Resistor three is 2 A and R1 = 1 Ω and R3 = 4 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______ If I3 = 2 A then It = 2 A Rt = V/I (12 V)/(2 A) = 6 Ω Vt = 12 V

Determine the following values for this circuit:

I1 = _______ V1 = _______ R1 = _______ I1 = 2 A R1 = 1 Ω V1 V= IR (2 A)(1 Ω) = 2 V I2 = _______ V2 = _______ R2 = _______ I2 = 2 A R2 = Rt – (R1 + R2) 6 Ω - (1 Ω + 4 Ω) = 1 Ω V2 = IR (2 A)(1 Ω) = 2 V I3 = _______ V3 = _______ R3 = _______ I3= 2 A R3 = 4 Ω V3 = IR (2 A)(4 Ω) = 8 V Parallel Circuit:

- The total Current is equal to the sum of each the currents in the individual branches.

o IT = I1 + I2 + I3 … - The total Voltage applied (from battery, outlet, etc.) is the same

for each individual voltage drop across each resistor. o VT = V1 = V2 = V3 …

- The total Resistance is less than any of the individual resistors. o RT = 1/(1/R1 + 1/R2 + 1/R3 …) OR RT = (R1

-1+ R2-1 + R3

-1 …)-1

Problems:

1) Two resistors are in parallel and R2 has two times greater the resistance of R1 (See Diagram P-1). Relate the current in R2 compared to R1. Explain.

C) ½ the value; if the voltage is the same then doubling the resistance will lead to ½ the current I = V/R compared to V/2R ½ I

2) Referring to Diagram P-1, if the voltage across the battery is 24 V and R1 = 5 Ω and R2 = 7 Ω, what is the total current and total resistance in the circuit?

Vt = 24 V RT = 1/(1/5 Ω + 1/7 Ω ) = 2.9 Ω It = V/R (24 V)/(2.9 Ω) = 8.3 A

Diagram: P-1

A) Two times the value B) The same value C) ½ the value D) ¼ the value

3) If a third resistor was added to the circuit in Diagram P-1 would the total resistance increase, decrease or stay the same? Explain.

Decrease; When more resistors are added to a parallel circuit it provides more pathways for the charge to flow through. This increases the overall current but the voltage must remain the same. The only way this is accomplished is if the total resistance is decreased.

4) Referring to Diagram P-2, if the voltage across the battery is 36 V and R1 = 2 Ω and R2 = 4 Ω and R3 = 6 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______ Vt = 36 V RT = 1/(1/2 Ω + 1/4 Ω + 1/6 Ω) = 1.1 Ω It = V/R (36 V)/(1.1 Ω) = 33 A

5) Find the current going through and the voltage across R1, R2, and R3 from problem 4.

Diagram: P-2

I1 = _______ V1 = _______ V1 = 36 V I1 = V/R (36 V)/(2 Ω) = 18 A I2 = _______ V2 = _______ V2 = 36 V I2 = V/R (36 V)/(4 Ω) = 9 A I3 = _______ V3 = _______ V3 = 36 V I3 = V/R (36 V)/(6 Ω) = 6 A

6) Referring to Diagram P-2, if the voltage across the battery is 9 V and R1 = 2 Ω and R2 = 1 Ω and R3 = 3 Ω, what is the total current and total resistance in the circuit?

It = _______ Rt = _______ Vt = 9 V RT = 1/(1/2 Ω + 1/1 Ω + 1/3 Ω) = 0.545 Ω It = V/R (9 V)/(0.545 Ω) = 16.5 A

7) Find the current going through and the voltage across R1, R2, and R3 from problem 6.

I1 = _______ V1 = _______ V1 = 9 V I1 = V/R (9 V)/(2 Ω) = 4.5 A I2 = _______ V2 = _______ V2 = 9 V I2 = V/R (9 V)/(1 Ω) = 9 A I3 = _______ V3 = _______ V3 = 9 V I3 = V/R (9 V)/(3 Ω) = 3 A

8) Referring to Diagram P-2, if the current through Resistor two is 3 A and R1 = 3 Ω and R2 = 1 Ω and R3 = 5 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______ I2 = 3 A RT = 1/(1/3 Ω + 1/1 Ω + 1/5 Ω) = 0.65 Ω V2 = I2R2 = (3 A)(1 Ω) = 3 V = Vt = 3 V

It = V/R = (3 V)/(0.65 Ω) = 4.6 A

9) Find the current going through and the voltage across R1, R2, and R3 from problem 8.

I1 = _______ V1 = _______ V1 = 3 V I1 = V/R (3 V)/(3 Ω) = 1 A I2 = _______ V2 = _______ V2 = 3 V I2 = V/R (3 V)/(1 Ω) = 3 A I3 = _______ V3 = _______ V3 = 3 V I3 = V/R (3 V)/(5 Ω) = 0.6 A

10) Referring to Diagram P-2, if the voltage across the battery is 12 V, the current through Resistor two is 2 A and R1 = 1 Ω and R3 = 4 Ω, what is the total current, total voltage and total resistance in the circuit?

It = _______ Vt = _______ Rt = _______ Vt = 12 V I2 = 2 A R2 = V2/I2 = (12 V)/(2 A) = 6 Ω RT = 1/(1/1 Ω + 1/6 Ω + 1/4 Ω) = 0.706 Ω It = V/R (12 V)/(0.706 Ω) = 17 A

Determine the following values for this circuit:

I1 = _______ V1 = _______ R1 = _______ V1 = 12 V R1 = 1 Ω I1 = V/R (12 V)/(1 Ω) = 12 A I2 = _______ V2 = _______ R2 = _______ V2 = 12 V R2 = 6 Ω (see above) I2 = V/R (12 V)/(6 Ω) = 2 A I3 = _______ V3 = _______ R3 = _______

V1 = 12 V R1 = 4 Ω I1 = V/R (12 V)/(4 Ω) = 3 A

Combination (Complex) Circuits:

Most circuits are a combination of both series and parallel circuits. Using the previous rules the circuits will be reduced to an equivalent series or parallel circuit. See handout: “Combination (Complex) Circuits: Solving for Equivalent Resistance”

Solve for the Req and Itotal for the following Circuit. The Voltage across the battery is 120 V and these are the following values for the Resistors:

R1 = 2 Ω, R2 = 4 Ω, R3 = 6 Ω, R4 = 5 Ω, R5 = 3 Ω, R6 = 12 Ω

Req = 6.5 Ω IT = VT/Req (120 V)/(6.5 Ω) = 18.5 A

BONUS: Find the Voltage across and the Current through each Resistor.

V1 = 30.9 V I1 = 15.45 A V2 = 12.4 V I2 = 3.1 A

V3 = 18.6 V I3 = 3.1 A V4 = 55.5 V I4 = 11.1 A

V5 = 33.3 V I5 = 11.1 A V6 = 88.8 V I6 = 7.4 A

COMBINATION (COMPLEX) CIRCUITS SOLVING FOR EQUIVALENT RESISTANCE

• A "COMBINATION CIRCUIT" is a circuit that is a blend of series paths and parallel paths. See Figure for a visual explanation. Most circuits are of this variety. Don't be afraid to tackle these circuits as far as the math goes. You merely have to break each part of the circuit down into either a series circuit or parallel circuit.

• In the figure to the side there is an 8 Ω resistor in series with two 4 Ω resistors in parallel (This can be written as R2||R3). You must first figure out the equivalent resistance of each individual parallel path circuit.

• 1/Req// = 1/R2 + 1/R3 1/Req// = ¼ + ¼

Req// = 2 Ω ∴ R2||R3 = 2 Ω

• Now that you have combined the resistors in parallel they act as one resistor in series with R1. You can now calculate total resistance for the entire circuit by using the rules for series circuits.

• Req = R1 + (R2||R3) Req = 8 Ω + 2 Ω

Req = 10 Ω

COMBINATION (COMPLEX) CIRCUITS SOLVING FOR VOLTAGE AND CURRENT

• Now that you know how to solve for equivalent resistance in a combination circuit you can work backwards to find the voltage and current across any resistor in the circuit. Just remember the rules for series and parallel circuits.

VARIABLE SERIES PARALLEL

CURRENT (I) Same as total Add to find total

VOLTAGE (V) Add to find total Same as total

• Step one: Identify which parts of your circuit are in parallel and which ones are in series. • Step two: Solve for equivalent resistance for the complete circuit. If you are keeping

your variables in a table make a separate column for each of your equivalent resistors. • Step three: Calculate total current from the total voltage and equivalent resistance. • Step four: Taking the total voltage and total current values, go back to the last step in

the circuit reduction process (determining equivalent resistance) and insert those values where applicable.

• Step five: From the known resistance, voltage, and current use Ohm’s Law to solve for the unknown values (V = IR; R = V/I; I = V/R).

• Step six: Repeat steps 4 and 5 until all values for voltage and current are know in the original circuit configuration. Essentially, you will proceed step-by-step from the simplified version of the circuit back into its original, complex form, plugging in values of voltage and current where appropriate until all values of voltage and current are known.

• Example: Finding equivalent resistance.

RESISTANCE OF INDIVIDUAL RESISTORS AND COMBINATIONS

R1 R2 R3 R4 R5 R6 R7 R2||R3 R4||R5||R6

6Ω 8Ω 3Ω 5Ω 6Ω 4Ω 4Ω 2.18Ω 1.62Ω

R1 + R2||R3 + R4||R5||R6 + R7 Req = 13.80 Ω

• Determine total current.

I = Vtotal/Req (98 V)/(13.80 Ω) 7.10 A

• Rebuild the circuit and solve for the unknown variables. REMEMBER the rules for series and parallel circuits and Ohm’s Law.

I = V/R R = V/I V = IR

R/V/I FOR EACH RESISTOR

R1 R2 R3 R4 R5 R6 R7 R2||R3 R4||R5||R6 TOTAL

6 Ω

8 Ω

3 Ω

5 Ω

6 Ω

4 Ω

4 Ω

2.18 Ω

1.62 Ω

13.80Ω

42.60V

15.48V

15.48V

11.50V

11.50V

11.50V

28.40V

15.48V

11.50V

98V

7.10 A

1.93 A

5.16 A

2.30 A

1.92 A

2.88 A

7.10 A

7.10 A

7.10 A

7.10 A

INVESTIGATING CIRCUITS Name(s): ______________________________________ Date: __________ Period: ____

Part 1: With a copper wire, a bare bulb and a D battery light up the bulb.

Completed correctly: _________ Did not complete: _________

Part 2: With a battery (with holder), two bulbs (with holders) and up to four alligator clips light up the bulbs in a SERIES circuit.

Completed correctly: _________ Did not complete: _________

Unscrew one of the light bulbs and describe what happens.

Draw and Label a schematic diagram of your circuit.

Part 3: With a battery (with holder), two bulbs (with holders) and up to four alligator clips light up the bulbs in a PARALLEL circuit.

Completed correctly: _________ Did not complete: _________

Unscrew one of the light bulbs and describe what happens. Is your answer the same as in Part 2? Explain.

Draw and Label schematic diagram of your circuit.

Part 4: With a LED, a two battery holder (with connecting wires) and two D batteries light up the LED.

Completed correctly: _________ Did not complete: _________

Describe a LED. Which end is the positive of the LED?

Does the LED light up both orientations (positive to positive and negative to negative AND/OR positive to negative and negative to positive)? Explain.

Part 5: With a LED, a two battery holder (with connecting wires), tape and two D batteries create a current detector.

Completed correctly: _________ Did not complete: _________

Color in the following boxes with the indicated medium.

#2 Pencil Charcoal Graphite Pencil

Using your Current Detector (or Multi-meter) test the conductivity of the above materials and rank them from 1-3 (3 being the most conductive; the brighter the light = the more conductive).

1: ______________________ 2: ______________________ 3: ______________________

What is the main ingredient in all of these materials?

Research Question: Why is there a difference in conductivity if all of the materials are made of the same ingredient? Please cite your sources.

TEACHER COPY INVESTIGATING CIRCUITS Name(s): ______________________________________ Date: __________ Period: ____

Part 1: With a copper wire, a bare bulb and a D battery light up the bulb.

(copper tape can also be used)

Completed correctly: _________ Did not complete: _________

This is an easy way to grade the students while moving around the room. You can have the students keep trying until they succeed or just give them one chance to show you their completion.

Part 2: With a battery (with holder), two bulbs (with holders) and up to four alligator clips light up the bulbs in a SERIES circuit.

Copper wire or tape can be used instead of the alligator clips; also a switch and other circuit parts can be included.

Completed correctly: _________ Did not complete: _________

Unscrew one of the light bulbs and describe what happens.

The circuit is broken and none of the lights came on.

Draw and Label a schematic diagram of your circuit.

It should look something like this:

The diagram may look different depending on parts used.

Part 3: With a battery (with holder), two bulbs (with holders) and up to four alligator clips light up the bulbs in a PARALLEL circuit.

Copper wire or tape can be used instead of the alligator clips; also a switch and other circuit parts can be included.

Completed correctly: _________ Did not complete: _________

Unscrew one of the light bulbs and describe what happens. Is your answer the same as in Part 2? Explain.

The circuit should not be broken and the other bulb should still work. This is because there is more than one pathway for the current to move through.

Draw and Label a schematic diagram of your circuit.

The diagram may look different depending on parts used.

Part 4: With a LED, a two battery holder (with connecting wires) and two D batteries light up the LED.

Completed correctly: _________ Did not complete: _________

Describe a LED. Which end is the positive of the LED?

A semiconductor diode (a semiconductor device with two terminals, typically allowing the flow of current in one direction only) that emits light when a voltage is applied to it and that is used especially in electronic devices (as for an indicator light)

Does the LED light up both orientations (positive to positive and negative to negative AND/OR positive to negative and negative to positive)? Explain.

No, An LED has a positive lead, known as the anode and a negative lead known as the cathode. An LED must be connected in a circuit the right way around – LEDs are diodes which means that current can only flow through an LED from the anode to the cathode and not the other way around.

If an LED is connected the wrong way around in a circuit (anode to negative and cathode to positive) it is said to be "reverse biased" and will not emit light. When connected the right way around the LED is said to be "forward biased".

Part 5: With a LED, a two battery holder (with connecting wires), tape and two D batteries create a current detector.

Completed correctly: _________ Did not complete: _________

Color in the following boxes with the indicated medium.

Make sure that the students darken in the whole box.

#2 Pencil Charcoal Graphite Pencil

https://www.build-electronic-circuits.com/what-is-an-led/

Using your Current Detector (or Multi-meter) test the conductivity of the above materials and rank them from 1-3 (3 being the most conductive; the brighter the light = the more conductive).

1: __Charcoal____ 2: ____ #2 Pencil____ 3: ___Graphite Pencil___

What is the main ingredient in all of these materials?

Carbon

Research Question: Why is there a difference in conductivity if all of the materials are made of the same ingredient? Please cite your sources.

Even though these are mainly composed of carbon there are differences in their purity and structure that determine their conductive properties.

“Graphite is one of only three naturally occurring allotropes of carbon (the others being amorphous carbon and diamond). The difference between the three naturally occurring allotropes is the structure and bonding of the atoms within the allotropes; diamond enjoying a diamond lattice crystalline structure, graphite having a honeycomb lattice structure, and amorphous carbon (such as coal or soot) does not have a crystalline structure.

While there are many different forms of carbon, graphite is of an extremely high grade and is the most stable under standard conditions. Therefore, it is commonly used in thermochemistry as the standard state for defining the heat formation of compounds made from carbon. It is found naturally in three different forms: crystalline flake, amorphous and lump or vein graphite, and depending on its form, is used for a number of different applications.

As previously touched upon, graphite has a planar, layered structure; each layer being made up of carbon atoms linked together in a hexagonal lattice. These links, or covalent bonds as they are more technically known, are extremely strong, and the carbon atoms are separated by only 0.142 nm. The carbon atoms are linked together by very sturdy sp2 hybridized bonds in a single layer of atoms, two dimensionally. Each individual, two dimensional, one atom thick layer of sp2 bonded carbon atoms in

graphite is separated by 0.335nm. Essentially, the crystalline flake form of graphite, as mentioned earlier, is simply hundreds of thousands of individual layers of linked carbon atoms stacked together.”

(Fuente, J. (2018). Graphene & Graphite - How Do They Compare?. [online] Graphenea. Available at: https://www.graphenea.com/pages/graphene-graphite#.W0yOPdVKiUk [Accessed 16 Jul. 2018].)

“Allotropy (also referred to as ‘allotropism’) of an element is that element’s ability to exist in multiple forms in the same physical state with a different arrangement of its atoms. The different forms are called allotropes of the given chemical element.”

“Allotropes of the same element have different bonding arrangements, which give rise to different chemical and physical properties for the substance. Furthermore, different allotropes can also differ in the occurrence of molecules in the number of atoms.”

(Scienceabc.com. (2016). "Why is Graphite Soft, But Diamond So Hard?". [online] Available at: https://www.scienceabc.com/pure-sciences/graphite-soft-diamond-structure-properties-hard-carbon-allotrope-tetrahedral-layers.html [Accessed 10 Jul. 2018].)

“Pencil ‘lead’ is made of graphite mixed with clay. Graphite is a form of carbon, and is a conductor of electricity: the carbon atoms’ electrons can move within the material. The pencil line on the paper is a continuous trail of graphite. This trail is a route along which electrons flow – from the negative to the positive terminal of the battery – forming an electric current that lights the LED. Graphite is only a fairly good conductor, and longer circuits have more resistance to current flow. That’s why touching the crocodile clip at different points along the line can make the LED brighter or dimmer.”

Learning-resources.sciencemuseum.org.uk. (2018). [online] Available at: https://learning-resources.sciencemuseum.org.uk/wp-content/uploads/2017/07/Graphite-Circuits.pdf [Accessed 16 Jul. 2018].

A. Weiner, “When Is Carbon an Electrical Conductor?,” Popular Science, 30-Sep-2008. [Online]. Available: https://www.popsci.com/adam-weiner/article/2008-09/when-carbon-electrical-conductor. [Accessed: 16-Jul-2018].A. Weiner, “When Is Carbon an Electrical Conductor?,” Popular Science, 30-Sep-2008. [Online]. Available: https://www.popsci.com/adam-weiner/article/2008-09/when-carbon-electrical-conductor. [Accessed: 16-Jul-2018].

(Courses.lumenlearning.com. (2018). Allotropes of Carbon | Introduction to Chemistry. [online] Available at: https://courses.lumenlearning.com/introchem/chapter/allotropes-of-carbon/ [Accessed 16 Jul. 2018].)

CHEMICAL REACTIONS AND ELECTRICAL ENERGY

CLASS ACTIVITY

PART A: Demonstrating a Chemical Reaction: “Cleaning” Pennies

Objective: To demonstrate a chemical reaction and observe chemical changes.

1) Describe the appearance of the pennies before placed in the solution.

2) When pennies appear “dirty” they are actually covered with copper oxide (CuO). How does the copper oxide form on the penny?

3) What is the penny’s material composition?

4) What type of solution is the vinegar? What type of compound is table salt?

5) Describe the appearance of the pennies after placed in the solution.

6) Predict any change in appearance of the pennies as they dry. (Rinsed vs. Un-rinsed)

7) Describe the appearance of the pennies once they had a chance to dry.

8) Explain why the pennies may have changed in appearance.

9) Describe the paperclip before placed in the solution. Predict changes in appearance (if any) after it has been in the solution.

10) Describe the appearance of the paperclip after it has been removed from the solution.

PART B: Changing Chemical Energy into Electrical Energy

Objective: To create an electrochemical battery from a lemon that produces enough voltage and current to light a LED.

1) Describe the components of the lemon cell?

2) What is occurring for each of the lemon cells to produce a voltage?

3) Record the individual voltages of each of the lemons in Data Table 1.

4) When each lemon cell is placed in series the voltages add up. Record in Data Table 2 the calculated values of the following combinations: A + B, A + B + C, and A + B + C + D (you may have to do more combinations if this does not produce enough electrical energy for the LED).

5) Record in Data Table 2 the measured voltages for the combinations. Is there any difference? If so, explain why.

6) Based on your calculated voltages predict how many lemons will be needed for the LED bulb to work.

7) How many lemon cells were needed to light the LED? How did this compare to your prediction?

DATA TABLE 1: Individual Voltage Readings

LEMON VOLTAGE READING (V) A B C D

DATA TABLE 2: Combination Voltage Readings

COMBINATION PREDICTED VOLTAGE (V)

MEASURED VOLTAGE(V)

A + B A + B + C

A + B + C + D Other Combinations

CHEMICAL REACTIONS AND ELECTRICAL ENERGY

CLASS ACTIVITY TEACHER GUIDE

PART A: Demonstrating a Chemical Reaction: “Cleaning” Pennies

Objective: To demonstrate a chemical reaction and observe chemical changes.

Materials: -Glass/Plastic Container (do not use a metal container because it will react with the vinegar)

-White Vinegar (4-5% acetic acid)

-Table Salt

-Plastic spoon (glass stirrers would work – just not metal)

-20 “dirty” pennies (“dirty” meaning tarnished; a month prior to this experiment you may want to start a collection within your classroom or school)

-3 steel paperclips

-Paper Towels

-Marker/Pen

-Water Source

Procedure and Observations:

1) Carefully pour the white vinegar into the container (usually you need about a ¼ -½ cup but depending on your containers you need enough to cover the pennies when they are added).

2) Add about a teaspoon of salt to the vinegar and stir until the salt dissolves.

3) Pass the dirty pennies around so the students can observe them.

4) Place the pennies in the solution and leave them in for about 10 minutes.

5) During this time mark two paper towels as A and B. Also, unbend one of the paperclips.

6) Remove ten pennies from the vinegar and salt solution and place on towel B. Then remove the other pennies, rinse them thoroughly under water and place them on towel A. DO NOT discard the solution.

7) Set the towels aside to let the pennies dry till the next day.

8) Place the unbent paperclip in the container so half of it is in the solution and the other half is out. Also place one of the bent paperclips into the solution. Leave overnight and remove the paperclips and place them on a towel. Pass them around so the students may observe them.

11) Describe the appearance of the pennies before placed in the solution.

Dull and dingy looking

12) When pennies appear “dirty” they are actually covered with copper oxide (CuO). How does the copper oxide form on the penny?

The dullness of pennies occurs when the copper in the pennies reacts with the oxygen in the air and forms copper oxide (CuO)

13) What is the penny’s material composition? Pennies prior to 1982 were made almost completely out of copper. However, post 1982 pennies are copper plated around a zinc core (97.5% Zn/2.5% Cu).

14) What type of solution is the vinegar? What type of compound is table salt?

Acidic solution; ionic compound

15) Describe the appearance of the pennies after placed in the solution.

The students should notice that the pennies are shiny and most of the tarnish was removed; placing the pennies in a salt and vinegar solution allows for the acetic acid from the vinegar to dissolve the CuO. The copper from the CuO remains in the solution.

16) Predict any change in appearance of the pennies as they dry. (Rinsed vs. Un-rinsed)

Predictions may vary.

17) Describe the appearance of the pennies once they had a chance to dry.

The students should notice that the pennies on A (rinsed) remained shiny and the pennies on B (un-rinsed) turned a blue-green color

18) Explain why the pennies may have changed in appearance.

The vinegar and salt dissolve the CuO layer making it easier for the copper atoms to join oxygen from the air and chlorine from the salt to make blue-green compounds called copper carbonate, and copper chloride (verdigris). The vinegar can also cause copper acetate to form. The students may not know these particular compounds but they should observe that leaving the vinegar and salt solution on the pennies causes a reaction in air).

Statue of Liberty, copper pots, other copper (brass, bronze) materials; The Statue of Liberty was not always green but the copper clad would have oxidized quickly being surrounded by water. Unfortunately, there was not color photography in the mid 1880’s to see the difference. There is a same-size replica of the face shown as part of the exhibit in one of the corridors of the Statue's pedestal. The patina (oxidized layer) on the statue acts as a barrier protecting the underlying copper from any further oxidation.

19) Describe the paperclip before placed in the solution. Predict changes in appearance (if any) after it has been in the solution.

They are stainless steel so they appear shiny and silvery.

20) Describe the appearance of the paperclip after it has been removed from the solution.

The students should note a color change to the paperclip. When the pennies are removed the copper from the CuO remains in the solution. The copper in the solution are positively charged copper ions. The paperclip is made of steel (an alloy of iron and carbon) and the salt and vinegar solution dissolves some of the iron and its oxides on the surface of the paperclip leaving its surface negatively charged. The positive copper ions are attracted to the negatively charged surface of the paperclip which will coat the paperclip.

PART B: Changing Chemical Energy into Electrical Energy

Materials:

- At least 4 lemons (limes may be used instead; other citrus fruit can be used but usually are less acidic and will produce a smaller voltage)

-4 pre-1982 pennies (these pennies are almost 100% Copper and work better than the post-1982 pennies; if post-1982 pennies must be used more lemon cells may be needed; copper wire (10-14 gauge is another alternative)

-4 galvanized nails (this means they are zinc coated; they are easily found at building supply or hardware stores)

-7 connecting wires with alligator clips (pieces of thin copper or insulated wire can be used instead)

-multi-meter (or voltmeter)

- 1 LED light (clear bulbs with red or pink glow are usually the easiest to detect; Recommended: 5mm Red LED (clear-color lens; typical voltage 1.7 V, with a max voltage of 2.4 V; 20 mA max); these can be purchased through an electronics store or on-line; To view LED better in a classroom setting the room should be darken)

-marker

-knife/razor blade

Procedure and Observations:

1. Using the marker label each lemon as A, B, C, and D.

2. Puncture on one side of each lemon with the galvanized nails. (See Figure 1.)

3. With the knife/razor blade make a small incision (large enough to fit a penny) on the other side of each lemon and insert the pennies. (See Figure 1.)

Figure 1.

4. Using the multi-meter and the connecting wires measure the voltage of each one of the lemon batteries cells. Have the students record their data in Data Table 1.

If the multi-meter displays a negative voltage then the positive and negative connections are switched.

Objective: To create an electrochemical battery from a lemon that produces enough voltage and current to light a LED.

8) Describe the components of the lemon cell? galvanized nail = negative and copper = positive; With copper and zinc electrodes, the zinc more readily loses electrons (e-) than the copper. Therefore, e- flow from the zinc to the copper electrode.

9) What is occurring for each of the lemon cells to produce a voltage?

Electrolyte: lemon juice citric acid; 2 Electrodes – nail (-) and copper coin (+); The nail (zinc) is the anode since this where oxidation occurs (loses e-). The copper coin is the cathode since this where reduction occurs (gains e-).

A chemical reaction occurs when two or more molecules interact and a chemical change occurs. All chemical reactions involve a change in energy. (Law of Conservation: matter or energy cannot be created or destroyed only changed).

This energy change allows for the conversion of chemical energy to electrical energy.

Metals differ in their tendency to give up electrons. This difference allows for an imbalance of charges and since opposite charges attract the electrons released from one metal will be attracted to the other metal. These metals are called electrodes. For the electron exchange between the two metals to begin and/or continue they must be placed in an electrolyte (a solution of ions). With copper and zinc electrodes, the zinc more readily loses electrons (e-) than the copper. Therefore, e- flow from the zinc to the copper electrode. For this example citric acid (C6H8O7) is used as the electrolyte. An acid has an easily detached hydrogen ion. Hydrogen ions are positive, and the remaining part of the

acid (citrate) becomes negative when the hydrogen is removed. As the zinc electrode loses e-, positive zinc ions are produced. The acetate ions are more attracted to the zinc ions than the H+ and will pair up leaving the H+ to bond with the extra electrons on the copper electrode. Since the copper electrode has lost its extra e- to the H+ it can accept more from the zinc electrode. This continues the process until the zinc is used up and the battery is dead.

10) Record the individual voltages of each of the lemons in Data Table 1. Each one should read 0.85 V to 1 V; if the measurements are producing very low or zero volt readings the electrodes may be touching one another or there is a problem with the wire connections.

11) When each lemon cell is placed in series the voltages add up. Record in Data Table 2 the calculated values of the following combinations: A + B, A + B + C, and A + B + C + D (you may have to do more combinations if this does not produce enough electrical energy for the LED).

Indicate the correct way to connect the lemons in series: positive terminal Cu penny; Zn nail (on same cell) Cu penny (2nd cell); Zn nail (on 2nd cell) negative terminal

12) Record in Data Table 2 the measured voltages for the combinations. Is there any difference? If so, explain why.

There usually is a slight difference due to internal resistance of the cells and in the connecting wires.

13) Based on your calculated voltages predict how many lemons will be needed for the LED bulb to work.

14) How many lemon cells were needed to light the LED? How did this compare to your prediction? Usually about 3 or 4 lemons will be needed

(REMEMBER: since the LED is a diode it will only work in one direction)

DATA TABLE 1: Individual Voltage Readings

LEMON VOLTAGE READING (V) A B C D

DATA TABLE 2: Combination Voltage Readings

COMBINATION PREDICTED VOLTAGE (V)

MEASURED VOLTAGE(V)

A + B A + B + C

A + B + C + D Other Combinations

Changing Chemical Energy into Electrical Energy: Film Canister Battery

Name: ___________________________ Date: ___________ Period: _______

Objective:

To create an electrochemical battery that produces enough voltage and current to light a LED.

Materials:

-4 film canisters

-white vinegar

-4 pieces of copper wire (~8 cm)

-4 galvanized nails

-7 connecting wires with alligator clips

-multi-meter (or voltmeter)

- 2 incandescent lamps in holders

- 1 LED light

-2 D batteries with holder

-masking tape

-marker/pen

-Optional: a calculator or another LCD display, soldering iron and solder or electrical tape, screwdriver

Procedure and Observations:

5. Using the marker/pen and masking tape label each canister as A, B, C, and D. Remove the caps and carefully pour the vinegar into each of the film canisters (3/4 of the way full) and then replace the caps.

6. Place the galvanized nails in one of the holes of the caps of each canister.

7. Place the copper wire in the other hole of each canister.

8. Using the multi-meter (set to the voltage reading specified by your teacher) and the connecting wires measure the voltage of each one of your film canister batteries. Place your measurements in Data Table 1.

If the multi-meter displays a negative voltage then the positive and negative connections are switched. Noting this, which part of your film canister battery (the galvanized nail or the copper wire) is the positive electrode and which is the negative electrode? What does this mean in terms of electron movement?

DATA TABLE 1: Individual Voltage Readings

Film Canister Voltage Reading (V)

A

B

C

D

9. Choose two of the canisters and using the rules of series circuits predict the total voltage. Connect the two canisters in series with one another and measure their total voltage. Repeat this procedure with three and then four in series. Place all information in Data Table 2.

DATA TABLE 2: Combination Voltage Readings

Combination Predicted Voltage Measured Voltage

Two in Series ( and )

Three in Series ( , and )

Four in Series

10. Place two D batteries in series. Knowing that each is about 1.5 V what would be the total voltage for two in series? _____________. Using the multi-meter determine the total voltage: _____________.

11. Connect the two D batteries to the screws on the base of the lamp holder. Does the light come on?

Switch the leads on the lamp. Does the light come on?

Using the combination of Film Canister Cells that produces the voltage closest to the two D batteries connect to the base of the other lamp holder. Does the light come on?

Switch the leads on the lamp. Does the light come on?

12. Using the combination of Film Canister Cells that produces a voltage between 1.7 V – 2.4 V (or the combination specified by the teacher) connect to the leads of an LED lamp. Does the light come on?

Switch the leads on the LED. Does the light come on?

Optional:

13. Unscrew and remove the back plate of an inexpensive calculator.

14. Remove battery from the calculator.

15. Solder, tape or clip wire leads to the positive and negative terminals of the battery holder. (Calculators that require an AAA or AA battery are easier to attach, but button battery powered calculators will also work – it just takes a little more time to make the connections and keep them in place. Remind students who are soldering (not recommended) the solder safety precautions:

• DO NOT touch the tip of the soldering iron. • ALWAYS return the soldering iron to its stand when not in use. • Work in a well-ventilated area. • Wash your hands after soldering.

16. Connect your leads to two Film Canister Cells.

17. Turn on the calculator and perform the following calculations:

476 x 219 = ________________ 698/42 = _________________

Questions:

1. Name the three main parts of a battery. Assign each of these parts to the components of the Film Canister Cell.

2. Why must the Film Canister Cells be placed in a series arrangement rather than a parallel arrangement?

3. Why will the batteries light up the incandescent lamp but not the Film Canister Battery even though their voltages are similar?

4. What is an LED and why is it able to be lit by the Film Canister Battery?

5. Why will the LED only light in one direction?

TEACHER RESOURCE GUIDE

B. Changing Chemical Energy into Electrical Energy: Film Canister Battery

Objective:

To create an electrochemical battery that produces enough voltage and current to light a LED.

Materials:

-4 film canisters (many photo labs will give you their empty film canister – the problem now is that most people have a digital camera and do not use film; film canisters can be ordered from Educational Innovations (www.teachersource.com) for $12.95 (~40 count) or $49.95 (~200 count))

-white vinegar (4-5% acetic acid; lemon-lime soda or lemon juice works as well)

-4 pieces of copper wire (~8 cm) (10-14 gauge wire works best since it sturdy but still malleable enough to work with. Copper can be ordered on-line but it usually comes in large and expensive rolls. A building supply store allows for a portion of the wire to be cut from the roll (7½ ft is enough for 28 pieces (~8 cm)). It is suggested that the teacher cut the wire prior to the lab)

-4 galvanized nails (this means they are zinc coated; they are easily found at building supply or hardware stores)

-7 connecting wires with alligator clips (pieces of thin copper or insulated wire can be used instead)

-multi-meter (or voltmeter)

- 2 incandescent lamps in holders (2.5 V – 3.5 V are recommended)

- 1 LED light (clear bulbs with red or pink glow are usually the easiest to detect; Recommended: 5mm Red LED (clear-color lens; typical voltage 1.7 V, with a max voltage of 2.4 V; 20 mA max); these can be purchased through an electronics store or on-line)

-2 D batteries with holder (2 series battery holders are recommended or the batteries will need to be connected in series with wire; battery holders can be purchased at a electronics store or on-line)

-masking tape

-marker/pen

-Optional: a calculator or another LCD display, soldering iron and solder or electrical tape, screwdriver

The teacher should punch two holes in each of the film canister caps using the galvanized nail (see pictures below) prior to lab.

Procedure and Observations:

18. Using the marker/pen and masking tape label each canister as A, B, C, and D. Remove the caps and carefully pour the vinegar into each of the film canisters (3/4 of the way full) and then replace the caps.

19. Place the galvanized nails in one of the holes of the caps of each canister.

20. Place the copper wire in the other hole of each canister.

21. Using the multi-meter (set to the voltage reading specified by your teacher) and the connecting wires measure the voltage of each one of your film canister batteries. Place your measurements in Data Table 1.

If the multi-meter displays a negative voltage then the positive and negative connections are switched. Noting this, which part of your film canister battery (the galvanized nail or the copper wire) is the positive electrode and which is the negative electrode? What does this mean in terms of electron movement?

(galvanized nail = negative and copper = positive; With copper and zinc electrodes, the zinc more readily loses electrons (e-) than the copper. Therefore, e- flow from the zinc to the copper electrode.)

DATA TABLE 1: Individual Voltage Readings

Film Canister Voltage Reading (V)

A

B

C

D

(Each one should read 0.85 V to 1 V; if the students are getting very low or zero volt readings their electrodes may be touching one another or there is a problem with the wire connections)

22. Choose two of the canisters and using the rules of series circuits predict the total voltage. Connect the two canisters in series with one another (indicate the correct way to connect the canisters: positive terminal Cu wire; Zn nail (on same cell) Cu wire (2nd cell); Zn nail (on 2nd cell) negative terminal) and measure their total voltage. Repeat this procedure with three and then four in series. Place all information in Data Table 2.

DATA TABLE 2: Combination Voltage Readings

Combination Predicted Voltage Measured Voltage

Two in Series ( and )

Three in Series ( , and

)

Four in Series

23. Place two D batteries in series. Knowing that each is about 1.5 V what would be the total voltage for two in series? _____________. Using the multi-meter determine the total voltage: _____________. (this should give a reading of about 3.1 V)

24. Connect the two D batteries to the screws on the base of the lamp holder. Does the light come on? (Yes)

Switch the leads on the lamp. Does the light come on? (Yes)

Using the combination of Film Canister Cells that produces the voltage closest to the two D batteries connect to the base of the other lamp holder. Does the light come on? (No)

Switch the leads on the lamp. Does the light come on? (No)

25. Using the combination of Film Canister Cells that produces a voltage between 1.7 V – 2.4 V (or the combination specified by the teacher) connect to the leads of an LED lamp. Does the light come on?

Switch the leads on the LED. Does the light come on?

(since the LED is a diode it will only work in one direction)

Optional: (If the students have their own calculators out have them put them away)

26. Unscrew and remove the back plate of an inexpensive calculator.

27. Remove battery from the calculator.

28. Solder, tape or clip wire leads to the positive and negative terminals of the battery holder. (Calculators that require an AAA or AA battery are easier to attach, but button battery powered calculators will also work – it just takes a little more time to make the connections and keep them in place. Remind students who are soldering (not recommended) the solder safety precautions:

• DO NOT touch the tip of the soldering iron. • ALWAYS return the soldering iron to its stand when not in use. • Work in a well-ventilated area. • Wash your hands after soldering.

29. Connect your leads to two Film Canister Cells.

30. Turn on the calculator and perform the following calculations:

476 x 219 = (104,244) 698/42 = (16.62)

Questions:

6. Name the three main parts of a battery. Assign each of these parts to the components of the Film Canister Cell. (Electrolyte: vinegar; 2 Electrodes – nail (-) and copper wire (+); The nail (zinc) is the anode since this where oxidation occurs (loses e-). The copper wire is the cathode since this where reduction occurs (gains e-).

A chemical reaction occurs when two or more molecules interact and a chemical change occurs. All chemical reactions involve a change in energy. (Law of Conservation: matter or energy cannot be created or destroyed only changed).

This energy change allows for the conversion of chemical energy to electrical energy.

Metals differ in their tendency to give up electrons. This difference allows for an imbalance of charges and since opposite charges attract the electrons released from one metal will be attracted to the other metal. These metals are called electrodes. For the electron exchange between the two metals to begin and/or continue they must be placed in an electrolyte (a solution of ions). With copper and zinc electrodes, the zinc more readily loses electrons (e-) than the copper. Therefore, e- flow from the zinc to the copper electrode. For this example acetic acid (CH3COOH) is used as the electrolyte. An acid has an easily detached hydrogen ion. Hydrogen ions are positive, and the remaining part of the acid (acetate) becomes negative when the hydrogen is removed (H+ and CH3COO- respectively). As the zinc electrode loses e-, positive zinc ions are produced. The acetate ions are more attracted to the zinc ions than the H+ and will pair up leaving the H+ to bond with the extra electrons on the copper electrode. Since the copper electrode has lost its extra e- to the H+ it can accept more from the zinc electrode. This continues the process until the zinc is used up and the battery is dead.

7. Why must the Film Canister Cells be placed in a series arrangement rather than a parallel arrangement? (In series circuits the voltage of each individual cell is added together for the total voltage of the circuit. In parallel the voltage remains the same regardless how many cells you connect together. Since the voltage needs to be increased to power the lights a series circuit is the only option.)

8. Why will the batteries light up the incandescent lamp but not the Film Canister Battery even though their voltages are similar?

(The Film Canister Battery only produces a very small amount of current due its internal resistance. The D batteries have less internal resistance and will allow more current to the bulb. Even with multiple canister cells, the amount of current flowing through the wire is not enough. Though the voltage is high enough (~2.5 volts with three cells), the current is too weak.

Connecting a bulb to a voltage source allows charges to flow from one contact to the other. The charges will also pass through the wires and filament of the bulb. As the charges move through the filament they are continuously bumping into the filament’s atoms (usually tungsten). These collisions cause the atoms to vibrate which releases energy in the form of heat. The bound electrons in the vibrating atoms are raised to a higher energy level temporarily. When they fall back to their original levels, the electrons release extra energy in the form of photons. Unfortunately, this process is not very efficient. Most incandescent bulbs allow only 10% of its energy to go to light the rest is lost as heat. This is why there is such a push for alternative light bulbs.)

9. What is an LED and why is it able to be lit by the Film Canister Battery? (Light Emitting Diode; a light source created by current moving through a semiconducting diode.

A diode is a very simple semiconducting device. A semiconductor is a material with a varying ability to conduct current. Usually semiconductors are created from a poor conductor that has had atoms of another material added to it. This process is called doping. As electrons move through the semiconductor they will drop from a higher orbital to a lower orbital. When this occurs energy is released as a photon (light particle). A larger drop in energy releases a higher-energy photon (higher frequency – can vary color of LED). All diodes release light, however most are not efficient since the semiconducting material will absorb a lot of the light energy. LEDs are made to release a large number of photons outward and in a particular direction. For more information about how LEDs and semiconductors work HowStuffWorks has good diagrams about these subjects.)

10. Why will the LED only light in one direction?

(This is due to the semiconducting nature of the diode.

Since extra atoms are used in doped materials (see question 4) free electrons (N-type; negative e- are attracted to + area) or holes (P-type; basically positively charged particles; since e- can jump from hole to hole the holes appear to move towards the negatively-charged area) where electrons can go are created. Either of these additions makes the material more electrically conductive.

A diode is composed of a N-type material attached to a portion of a P-type material, with electrodes on each end. This set up allows current in only one direction. When no potential difference is applied across the diode, e- from the N-type material fill holes from the P-type

material along the junction between the layers, forming a depletion zone (becomes an insulator and charges cannot flow).

To remove the depletion zone, e- from the N-type area must move to the P-type area and the holes would then appear to move in the opposite direction. This is accomplished by connecting the N-type side to the negative end of a voltage source and the P-type side to the positive end. The free e- in the N-type material are attracted to the + electrode and repelled by the negative electrode. The holes in the P-type material move in the opposite direction. When the potential difference between the electrodes is large enough, the e- in the depletion zone move out of their holes and begin to move freely. The depletion zone disappears, and charge moves across the diode.

If the electrodes are switched so the P-type side is connected to the negative end and the N-type side is connected to the positive end, current will not flow. This is due to the e- in the N-type material being attracted to the positive electrode and the positive holes in the P-type material being attracted to the negative electrode. No current moves across the diode because the holes and the e- are moving in the wrong direction and the depletion zone increases.

For more information about how LEDs and semiconductors work HowStuffWorks has good diagrams about these subjects.)

Changing Chemical Energy into Electrical Energy: The Lemon Battery

Name: ____________________________ Date: _______________ Period: ____

Objective:

To create an electrochemical battery that produces enough voltage and current to light a LED.

Materials:

-4 lemons

-4 pre-1982 pennies

-4 galvanized nails

-7 connecting wires with alligator clips

-multi-meter

- 2 incandescent lamps in holders

- 1 LED light

-2 D batteries with holder

-marker

-knife/razor blade

-Optional: nickel, dime, and/or quarter (post-1965; before this time they were mainly composed of silver now they are made of copper alloys)

Procedure and Observations:

1. Using the marker label each lemon as A, B, C, and D.

2. Puncture on one side of each lemon with the galvanized nails.

3. With the knife/razor blade make a small incision (large enough to fit a penny) on the other side of each lemon and insert the pennies.

4. Using the multi-meter (set to the voltage reading specified by your teacher) and the connecting wires measure the voltage of each one of your lemon batteries. Place your measurements in Data Table 1.

If the multi-meter displays a negative voltage then the positive and negative connections are switched. Noting this, which part of your Lemon Battery Cell (the galvanized nail or the copper coin) is the positive electrode and which is the negative electrode? What does this mean in terms of electron movement?

DATA TABLE 1: Individual Voltage Readings

LEMON VOLTAGE READING (V)

A

B

C

D

5. Choose two of the lemons and using the rules of series circuits predict the total voltage. Connect the two lemons in series with one another and measure their total voltage. Repeat this procedure with three and then four in series. Place all information in Data Table 2.

DATA TABLE 2: Combination Voltage Readings

COMBINATION PREDICTED VOLTAGE (V)

MEASURED VOLTAGE(V)

Two in Series ( and )

Three in Series ( , and )

Four in Series

6. Place two D batteries in series. Knowing that each is about 1.5 V what would be the total voltage for two in series? _____________. Using the multi-meter determine the total voltage: _____________.

7. Connect the two D batteries to the screws on the base of the lamp holder. Does the light come on?

Switch the leads on the lamp. Does the light come on?

Using the combination of Lemon Battery Cells that produces the voltage closest to the two D batteries connect to the base of the other lamp holder. Does the light come on?

Switch the leads on the lamp. Does the light come on?

8. Using the combination of Lemon Battery Cells that produces a voltage between 1.7 V – 2.4 V (or the combination specified by the teacher) connect to the leads of an LED lamp. Does the light come on? Switch the leads on the LED. Does the light come on?

Extension:

1. Predict if the nail or the penny can be replaced by a nickel, dime, or quarter. Explain your prediction.

2. Remove the nail from one of your lemons. Using the knife/razor blade make an incision right above or below the puncture mark. The incision should be large enough to fit a quarter.

3. Place one of the coins (nickel, dime or quarter) in this incision. Connect this coin to the negative terminal and the penny to the positive terminal of the multi-meter. Record your data in Data Table 3.

4. Repeat procedure 2 with the other two coins and record your data in Data Table 3.

5. Remove the coins from the lemon. Replace the nail into the lemon. In the penny’s location place one of the other coins (nickel, dime or quarter). Connect the nail to the negative terminal and the coin to the positive terminal of the multi-meter. Record your data in Data Table 4.

6. Repeat procedure 2 with the other two coins and record your data in Data Table 3.

DATA TABLE 3: Voltage Readings with Replacement of Galvanized Nail with Coins

COIN TYPE VOLTAGE (V)

Nickel

Dime

Quarter

DATA TABLE 4: Voltage Readings with Replacement of Penny with other Coins

COIN TYPE VOLTAGE (V)

Nickel

Dime

Quarter

7. Based on your results what can you conclude about the composition of the nickel, dime and quarter? Support your answer.

Questions:

1. Name the three main parts of a battery. Assign each of these parts to the components of the Lemon Battery Cell.

2. Why must the Lemon Battery Cells be placed in a series arrangement rather than a parallel arrangement?

3. Why will the batteries light up the incandescent lamp but not the Lemon Battery even though their voltages are similar?

4. What is an LED and why is it able to be lit by the Lemon Battery?

5. Why will the LED only light in one direction?

TEACHER RESOURCE GUIDE

C. Changing Chemical Energy into Electrical Energy: The Lemon Battery

Objective:

To create an electrochemical battery that produces enough voltage and current to light a LED.

Materials:

-4 lemons (limes may be used instead; other citrus fruit can be used but usually are less acidic and will produce a smaller voltage)

-4 pre-1982 pennies (these pennies are almost 100% Copper and work better than the post-1982 pennies; if post-1982 pennies must be used more lemon cells may be needed; copper wire (10-14 gauge is another alternative)

-4 galvanized nails (this means they are zinc coated; they are easily found at building supply or hardware stores)

-7 connecting wires with alligator clips (pieces of thin copper or insulated wire can be used instead)

-multi-meter (or voltmeter)

- 2 incandescent lamps in holders (2.5 V – 3.5 V are recommended)

- 1 LED light (clear bulbs with red or pink glow are usually the easiest to detect; Recommended: 5mm Red LED (clear-color lens; typical voltage 1.7 V, with a max voltage of 2.4 V; 20 mA max); these can be purchased through an electronics store or on-line)

-2 D batteries with holder (2 series battery holders are recommended or the batteries will need to be connected in series with wire; battery holders can be purchased at a electronics store or on-line)

-marker

-knife/razor blade

-Optional: nickel, dime, and/or quarter (post-1965; before this time they were mainly composed of silver now they are made of copper alloys)

Procedure and Observations:

31. Using the marker label each lemon as A, B, C, and D.

32. Puncture on one side of each lemon with the galvanized nails. (See pictures below.)

33. With the knife/razor blade make a small incision (large enough to fit a penny) on the other side of each lemon and insert the pennies. (The teacher may want to assist the students with this task or prior to the lab already have made the incisions; See pictures below.)

34. Using the multi-meter (set to the voltage reading specified by your teacher) and the connecting wires measure the voltage of each one of your lemon batteries. Place your measurements in Data Table 1.

If the multi-meter displays a negative voltage then the positive and negative connections are switched. Noting this, which part of your Lemon Battery Cell (the galvanized nail or the copper coin) is the positive electrode and which is the negative electrode? What does this mean in terms of electron movement?

(galvanized nail = negative and copper = positive; With copper and zinc electrodes, the zinc more readily loses electrons (e-) than the copper. Therefore, e- flow from the zinc to the copper electrode.)

DATA TABLE 1: Individual Voltage Readings

Lemon Voltage Reading (V)

A

B

C

D

(Each one should read 0.85 V to 1 V; if the students are getting very low or zero volt readings their electrodes may be touching one another or there is a problem with the wire connections)

35. Choose two of the lemons and using the rules of series circuits predict the total voltage. Connect the two lemons in series with one another (indicate the correct way to connect the lemons: positive terminal Cu penny; Zn nail (on same cell) Cu penny (2nd cell); Zn nail (on 2nd cell) negative terminal) and measure their total voltage. Repeat this procedure with three and then four in series. Place all information in Data Table 2.

DATA TABLE 2: Combination Voltage Readings

Combination Predicted Voltage Measured Voltage

Two in Series ( and )

Three in Series ( , and

)

Four in Series

36. Place two D batteries in series. Knowing that each is about 1.5 V what would be the total voltage for two in series? _____________. Using the multi-meter determine the total voltage: _____________. (this should give a reading of about 3.1 V)

37. Connect the two D batteries to the screws on the base of the lamp holder. Does the light come on? (Yes)

Switch the leads on the lamp. Does the light come on? (Yes)

Using the combination of Lemon Battery Cells that produces the voltage closest to the two D batteries connect to the base of the other lamp holder. Does the light come on? (No)

Switch the leads on the lamp. Does the light come on? (No)

38. Using the combination of Lemon Battery Cells that produces a voltage between 1.7 V – 2.4 V (or the combination specified by the teacher) connect to the leads of an LED lamp. Does the light come on?

Switch the leads on the LED. Does the light come on?

(since the LED is a diode it will only work in one direction)

Extension:

8. Predict if the nail or the penny can be replaced by a nickel, dime, or quarter. Explain your prediction. (any appropriate hypothesis is acceptable – many students will probably answer that the coins could replace the nail due to color)

9. Remove the nail from one of your lemons. Using the knife/razor blade make an incision right above or below the puncture mark. The incision should be large enough to fit a quarter. (The incision making may need to be done by the teacher)

10. Place one of the coins (nickel, dime or quarter) in this incision. Connect this coin to the negative terminal and the penny to the positive terminal of the multi-meter. Record your data in Data Table 3. (the voltage reading should be very close to zero since the coins are mainly made from copper – just like the penny)

11. Repeat procedure 2 with the other two coins and record your data in Data Table 3. (similar results should occur)

12. Remove the coins from the lemon. Replace the nail into the lemon. In the penny’s location place one of the other coins (nickel, dime or quarter). Connect the nail to the negative terminal and the coin to the positive terminal of the multi-meter. Record your data in Data Table 4. (the voltage reading should be very similar to that of the penny: 0.9 V – 1 V)

13. Repeat procedure 2 with the other two coins and record your data in Data Table 3. (similar results should occur)

DATA TABLE 3: Voltage Readings with Replacement of Galvanized Nail with Coins

COIN TYPE VOLTAGE (V)

Nickel

Dime

Quarter

DATA TABLE 4: Voltage Readings with Replacement of Penny with other Coins

COIN TYPE VOLTAGE (V)

Nickel

Dime

Quarter

14. Based on your results what can you conclude about the composition of the nickel, dime and quarter? Support your answer.

The composition of the nickel, dime and quarter is similar to the penny’s make-up. If two electrodes are made of the same material then they would have the same reaction in the electrolyte and no potential difference would be detected.

The coins are made of a Cupro-nickel alloy. The nickel is 25% Ni and 75% Cu. Both the dime and quarter are 8.33% Ni and 91.77% Cu. So even though the coins appear to be more like the nail they are actually more like the penny.

Questions:

11. Name the three main parts of a battery. Assign each of these parts to the components of the Lemon Battery Cell. (Electrolyte: lemon juice citric acid; 2 Electrodes – nail (-) and copper coin (+); The nail (zinc) is the anode since this where oxidation occurs (loses e-). The copper coin is the cathode since this where reduction occurs (gains e-).

A chemical reaction occurs when two or more molecules interact and a chemical change occurs. All chemical reactions involve a change in energy. (Law of Conservation: matter or energy cannot be created or destroyed only changed).

This energy change allows for the conversion of chemical energy to electrical energy.

Metals differ in their tendency to give up electrons. This difference allows for an imbalance of charges and since opposite charges attract the electrons released from one metal will be attracted to the other metal. These metals are called electrodes. For the electron exchange between the two metals to begin and/or continue they must be placed in an electrolyte (a solution of ions). With copper and zinc electrodes, the zinc more readily loses electrons (e-) than the copper. Therefore, e- flow from the zinc to the copper electrode. For this example citric acid (C6H8O7) is used as the electrolyte. An acid has an easily detached hydrogen ion. Hydrogen ions are positive, and the remaining part of the acid (citrate) becomes negative when the hydrogen is removed. As the zinc electrode loses e-, positive zinc ions are produced. The acetate ions are more attracted to the zinc ions than the H+ and will pair up leaving the H+ to bond with the extra electrons on the copper electrode. Since the copper electrode has lost its extra e- to the H+ it can accept more from the zinc electrode. This continues the process until the zinc is used up and the battery is dead.

12. Why must the Lemon Battery Cells be placed in a series arrangement rather than a parallel arrangement? (In series circuits the voltage of each individual cell is added together for the total voltage of the circuit. In parallel the voltage remains the same regardless how many cells you connect together. Since the voltage needs to be increased to power the lights a series circuit is the only option.)

13. Why will the batteries light up the incandescent lamp but not the Lemon Battery even though their voltages are similar?

(The Lemon Battery only produces a very small amount of current due its internal resistance. The D batteries have less internal resistance and will allow more current to the bulb. Even with multiple canister cells, the amount of current flowing through the wire is not enough. Though the voltage is high enough (~2.5 volts with three cells), the current is too weak.

Connecting a bulb to a voltage source allows charges to flow from one contact to the other. The charges will also pass through the wires and filament of the bulb. As the charges move through the filament they are continuously bumping into the filament’s atoms (usually tungsten). These collisions cause the atoms to vibrate which releases energy in the form of heat. The bound electrons in the vibrating atoms are raised to a higher energy level temporarily. When they fall back to their original levels, the electrons release extra energy in the form of photons. Unfortunately, this process is not very efficient. Most incandescent bulbs allow only 10% of its energy to go to light the rest is lost as heat. This is why there is such a push for alternative light bulbs.)

14. What is an LED and why is it able to be lit by the Lemon Battery? (Light Emitting Diode; a light source created by current moving through a semiconducting diode.

A diode is a very simple semiconducting device. A semiconductor is a material with a varying ability to conduct current. Usually semiconductors are created from a poor

conductor that has had atoms of another material added to it. This process is called doping. As electrons move through the semiconductor they will drop from a higher orbital to a lower orbital. When this occurs energy is released as a photon (light particle). A larger drop in energy releases a higher-energy photon (higher frequency – can vary color of LED). All diodes release light, however most are not efficient since the semiconducting material will absorb a lot of the light energy. LEDs are made to release a large number of photons outward and in a particular direction. For more information about how LEDs and semiconductors work HowStuffWorks has good diagrams about these subjects.)

15. Why will the LED only light in one direction?

(This is due to the semiconducting nature of the diode.

Since extra atoms are used in doped materials (see question 4) free electrons (N-type; negative e- are attracted to + area) or holes (P-type; basically positively charged particles; since e- can jump from hole to hole the holes appear to move towards the negatively-charged area) where electrons can go are created. Either of these additions makes the material more electrically conductive.

A diode is composed of an N-type material attached to a portion of a P-type material, with electrodes on each end. This set up allows current in only one direction. When no potential difference is applied across the diode, e- from the N-type material fill holes from the P-type material along the junction between the layers, forming a depletion zone (becomes an insulator and charges cannot flow).

To remove the depletion zone, e- from the N-type area must move to the P-type area and the holes would then appear to move in the opposite direction. This is accomplished by connecting the N-type side to the negative end of a voltage source and the P-type side to the positive end. The free e- in the N-type material are attracted to the + electrode and repelled by the negative electrode. The holes in the P-type material move in the opposite direction. When the potential difference between the electrodes is large enough, the e- in the depletion zone move out of their holes and begin to move freely. The depletion zone disappears, and charge moves across the diode.

If the electrodes are switched so the P-type side is connected to the negative end and the N-type side is connected to the positive end, current will not flow. This is due to the e- in the N-type material being attracted to the positive electrode and the positive holes in the P-type material being attracted to the negative electrode. No current moves across the diode because the holes and the e- are moving in the wrong direction and the depletion zone increases.

For more information about how LEDs and semiconductors work HowStuffWorks has good diagrams about these subjects.)

LAB PRATICUM: ELECTRICAL ENERGY, CURRENT AND CIRCUITS

Name: _________________________ Group Number: _____ Date: ___________ Period: ______

SAFETY:

• Never eat or drink anything in the laboratory. • Always follow proper laboratory dress code. • Always read all instructions before beginning the experiments. • Never rewire or take apart any element within the circuit. • Never short circuit the batteries. • Unhook all circuit components when finish (or at the end of each period) • Clean all appropriate materials • Allow all equipment to cool before storing.

PART A: Creating an Electrochemical Cell

OBJECTIVE: To identify the main components of an electrochemical cell and the factors that affects its voltage.

MATERIALS:

• Multi-meter (with probes) • Film canister (with top) • An 8mm piece of copper wire (12-14 gauge) • Two galvanized nails (steel nail with zinc coating) • Distilled or purified water • Lemon juice (from concentrate) or table salt • A lemon (or other citrus fruit) • Knife (suggested that the instructor be in charge of the knife or razor blade) • A pre-1982 penny • A post-1965 nickel, dime, or quarter • Two alligator clip wires

Parts of a Film Canister Electrochemical Cell

Procedure:

1. Take the film canister lid and one of the nails and poke two holes in the top as shown in Figure 1 (this may already have been done for you).

2. Fill the canister with the distilled or purified water about ¾ the way full.

3. Place the lid on the canister and insert one of the nails in one of the holes and the copper wire in the other.

4. DESCRIBE the three main parts of an electrochemical cell. Assign each of these parts to the components of the Film Canister Cell.

5. PREDICT the voltage of your Film Canister Electrochemical Cell. ________ V

6. Set the multi-meter to your teacher’s instructions and attach the probes.

7. Attach one alligator clip end to the galvanized nail and the other to one of the multi-meter probes.

8. Attach the other alligator clip end to the copper wire and the other to the second multi-meter probe.

ANALYZE: If the multi-meter displays a negative voltage then the positive and negative connections are switched. Noting this, which part of your Film Canister Cell (the

Figure 1

galvanized nail or the copper wire) is the positive electrode and which is the negative electrode? What does this mean in terms of electron movement?

9. COMPARE the multi-meter voltage reading (________ V) to your predicted reading. Explain any differences or similarities between the two voltages.

10. Unhook the alligator clips from the galvanized nail and copper wire. Remove lid and add 20 drops of lemon juice (or ~ 6 g salt). Replace top and gently shake container.

11. PREDICT if the voltage will change with the addition of the lemon juice. If so will it increase or decrease. Explain your prediction.

12. Connect the alligator clips back to the galvanized nail and copper wire.

13. COMPARE the multi-meter voltage reading (________ V) to your predicted reading. Explain any differences or similarities between the two voltages.

Parts of a Lemon Electrochemical Cell

14. Take the lemon and insert a galvanized nail in one top side of the lemon and with the knife make a small slit on the opposite side large enough for a penny. Insert penny.

Figure 2

15. Connect the alligator clips to the nail and penny and attach to the multi-meter. Record the voltage across the lemon in Data Table 1. REMEMBER if the multi-meter displays a negative voltage then the positive and negative connections are switched. Noting this, which part of the Lemon Cell (the galvanized nail or the penny) is the positive electrode and which is the negative electrode?

16. PREDICT if the nail or the penny can be replaced by a nickel, dime, or quarter. Explain your prediction.

17. Remove the nail from one of your lemons. Using the knife/razor blade make an incision right above or below the puncture mark. The incision should be large enough to fit a quarter.

18. Place one of the coins (nickel, dime or quarter) in this incision. Connect this coin to the negative terminal and the penny to the positive terminal of the multi-meter. RECORD your data in Data Table 1.

19. Remove the coin from the lemon. Replace the nail into the lemon. In the penny’s location place one of the other coins (nickel, dime or quarter). Connect the nail to the negative terminal and the coin to the positive terminal of the multi-meter. RECORD your data in Data Table 1.

DATA TABLE 1: Voltage Readings with Replacement of Galvanized Nail with Coins

ELECTRODES VOLTAGE (V)

Galvanized Nail and Penny

Nickel, Dime or Quarter and Penny

Galvanized Nail and Nickel, Dime or Quarter

20. Based on your results what can you conclude about the composition of the nickel, dime and quarter? Support your answer.

PART B: Changing Chemical Energy into Electrical Energy

OBJECTIVE: To create an electrochemical battery that produces enough voltage and current to light a LED.

MATERIALS:

• 4 film canisters • white vinegar • 4 pieces of copper wire (~8 cm) • 4 galvanized nails • 7 connecting wires with alligator clips • multi-meter (or voltmeter) • 2 incandescent lamps in holders • 1 LED light • 2 D batteries with holder • masking tape • marker/pen

PROCEDURE:

39. Using the galvanized nail poke holes in each of the film canister lids like in Part A: Figure 1. With the marker/pen and masking tape label each canister as A, B, C, and D. Carefully pour the vinegar into each of the film canisters (3/4 of the way full) and replace the caps.

40. Place the galvanized nails in one of the holes of the caps of each canister.

41. Place the copper wire in the other hole of each canister.

42. Using the multi-meter (set to the voltage reading specified by your teacher) and the connecting wires measure the voltage of each one of your film canister batteries. RECORD your measurements in Data Table 2.

DATA TABLE 2: Individual Voltage Readings

Film Canister Voltage Reading (V)

A

B

C

D

43. Choose two of the canisters and using the rules of series circuits predict the total voltage. Connect the two canisters in series with one another and measure their total voltage. Repeat this procedure with three and then four in series. RECORD your information in Data Table 3.

DATA TABLE 3: Combination Voltage Readings

Combination Predicted Voltage Measured Voltage

Two in Series ( and )

Three in Series ( , and )

Four in Series

44. Place two D batteries in series. Knowing that each is about 1.5 V PREDICT the total voltage for two in series? _____________. Using the multi-meter MEASURE the total voltage: _________V.

45. ANALYZE: Connect the two D batteries to the screws on the base of the lamp holder with the incandescent bulb. Does the light come on?

Switch the leads on the lamp. Does the light come on?

Using the combination of Film Canister Cells that produces the voltage closest to the two D batteries connect to the base of the other lamp holder. Does the light come on?

Switch the leads on the lamp. Does the light come on?

46. ANALYZE: Using the combination of Film Canister Cells that produces a voltage between 1.7 V – 2.4 V (or the combination specified by the teacher) connect to the leads of an LED lamp. Does the light come on?

Switch the leads on the LED. Does the light come on?

QUESTIONS:

16. Why must the Film Canister Cells be placed in a series arrangement rather than a parallel arrangement?

17. Why will the batteries light up the incandescent lamp but not the Film Canister Battery even though their voltages are similar?

18. What is an LED and why is it able to be lit by the Film Canister Battery? Why will the LED only light in one direction?

PART C: Resistors in Series and Parallel Circuits

OBJECTIVE: To demonstrate how resistance changes with series and parallel circuits.

MATERIALS:

• three resistors • a multi-meter with probes • six alligator clip wires

PROCEDURE:

1. RECORD the amount of resistance across each resistor (indicated by the label) and include the tolerance for each.

DATA TABLE 4: Resistance from Labels RESISTOR RESISTANCE (Ω) TOLERANCE (%)

R1 R2 R3

2. MEASURE the amount of resistance across each resistor using the multi-meter.

DATA TABLE 5: Resistance from the Multi-meter

RESISTOR RESISTANCE (Ω) R1 R2 R3

3. DRAW a schematic diagram for a series circuit including the three resistors. You can label

your resistors as R1, R2 and R3.

4. What is the equation for calculating equivalent resistance in a series circuit? What

happens to the equivalent resistance of a series circuit when additional resistors are added?

5. CALCULATE the total resistance for a series combination of your three resistors.

a) Equivalent resistance (series – label readings) = _________________ Ω b) Equivalent resistance (series – multi-meter) = _________________ Ω

6. MEASURE the equivalent resistance for your three resistors in a series combination using the multi-meter.

Equivalent resistance (series) = _________________ Ω

7. CALCULATE percent error. % Error = [(O - A)/A] x 100

O (observed) = values from 5 (a) and 5 (b) A (accepted) = value from 6

a) Percent error (series – label readings 5(a)) = ___________________% b) Percent error (series- multi-meter 5 (b)) = ___________________%

8. Which reading gave you a smaller percent error? Why?

9. Draw a schematic diagram for a parallel circuit including the three resistors. You can label your resistors as R1, R2 and R3.

10. What is the equation for calculating equivalent resistance in a parallel circuit? What happens to the equivalent resistance of a parallel circuit when additional resistors are added?

11. CALCULATE the total resistance for a parallel combination of your three resistors.

a) Equivalent resistance (parallel – label readings) = _________________ Ω b) Equivalent resistance (parallel – multimeter) = _________________ Ω

12. MEASURE the equivalent resistance for your three resistors in a parallel combination using the multi-meter.

Equivalent resistance (parallel) = _________________ Ω

13. Calculate percent error. % Error = [(O - A)/A] x 100

O (observed) = values from 11 a) and 11 b) A (accepted) = value from 12

a) Percent error (parallel – label readings) = ___________________% b) Percent error (parallel- multi-meter) = ___________________%

14. Which reading gave you a smaller percent error? Why?

15. What are some possible error sources that you may have encountered?

PART D: Electrical Energy and Capacitance

OBJECTIVE: To construct a capacitor and to calculate its capacitance. To determine the relationship between the capacitance and the distance between the parallel plates.

MATERIALS:

• Aluminum foil • Three Hardcover textbooks • 2 leads with alligator clips • Multi-meter • Calipers • Ruler

PROCEDURE:

1. Cut two pieces of aluminum foil so that they are slightly larger than the book you are using (this will be the plates of your capacitor). MEASURE the length and width of these foil sheets (they should be the same size. CALCULATE the area of your capacitor and record it in Data Table 6.

2. Make sure both pieces of foil are smooth and flat. Place one piece on top of page #201 in your book. Place the second piece on top of page #199. You now have one sheet of paper serving as a dielectric between the foil pieces.

3. Connect one lead to each foil piece. Connect the other end of the leads to the multi-meter. See Figure 3.

4. Place the two other books on top of the book containing the aluminum foil. Make sure that the foil is not touching the desk, the other piece of foil or any other object.

5. MEASURE and record the capacitance reading from the multi-meter in Data Table 7.

6. Repeat steps 2-5 for ten more trials with 2, 3, 4, 5, 10, 15, 20, 30, 40, and 50 sheets of paper of your textbook acting as the dielectric between the foil pieces.

7. Using the calipers MEASURE the total thickness of several hundred sheets of paper in your textbook (Remember each page has 2 page numbers). Record your answer in Data Table 6.

8. RECORD the total number of sheets you measured and calculate the thickness of one sheet of paper. Record your answers in Data Table 6.

9. CALCULATE the Capacitance using your information in Data Table 1 and your dielectric constant as κ = 3.5. Record your answers in Data Table 2. Calculate percent error.

DATA TABLE 6: Capacitance Components

MEASUREMENT AMOUNT

Total thickness of all sheets

Total number of sheets measured

Thickness per sheet

Area of Capacitor

DATA TABLE 7: Capacitance Measurements

TRIAL

NUMBER OF SHEETS

CAPACITANCE

MEASURED

(F)

(OBSERVED)

CAPACITANCE

CALCULATED

(F)

(ACTUAL)

PERCENT ERROR

1

10

2

20

3

30

4

40

5

50

6

60

7

70

8

80

9

90

10

100

• Plot capacitance (y-axis) versus the number of sheets of paper (x-axis) on the graph on the following page. Draw the best-fit line through your data points.

PROBLEMS:

1. Using your graph and data describe the relationship between the capacitance and plate separation. Is the relationship a 1/d, 1/d², or 1/d³ relationship? How are you able to tell from your graph or data?

2. Calculate ε (permittivity constant) if C = 1.51 x 10-5 F, κ = 3.5, A = 0.061 m² and d = 1.01 x 10 -4m.

3. Why is a 1 F capacitor potentially dangerous?

4. Why do people say not to open a TV set even though you just turned it off?

TEACHER’S GUIDE TO LAB PRACTICUM: ELECTRICAL ENERGY, CURRENT AND CIRCUITS

Name: _____________________________ Group Number: _____ Date: _____________ Period: ______

SAFETY:

• Never eat or drink anything in the laboratory. • Always follow proper laboratory dress code. • Always read all instructions before beginning the experiments. • Never rewire or take apart any element within the circuit. • Never short circuit the batteries. • Unhook all circuit components when finish (or at the end of each period) • Clean all appropriate materials • Allow all equipment to cool before storing.

PART A: Creating an Electrochemical Cell

OBJECTIVE: To identify the main components of an electrochemical cell and the factors that affects its voltage.

MATERIALS:

• Multi-meter (with probes) (or Voltmeter) • Film canister (with top) (many photo labs will give you their empty film canister – the

problem now is that most people have a digital camera and do not use film; film canisters can be ordered from Educational Innovations (www.teachersource.com) for $12.95 (~40 count) or $49.95 (~200 count))

• An 8mm piece of copper wire (12-14 gauge) (12-14 gauge wire works best since it sturdy but still malleable enough to work with. Copper can be ordered on-line but it usually comes in large and expensive rolls. A building supply store allows for a portion of the wire to be cut from the roll (7½ ft is enough for 28 pieces (~8 cm)). It is suggested that the teacher cut the wire prior to the lab)

• Two galvanized nails (this means they are zinc coated; they are easily found at building supply or hardware stores)

• Distilled or purified water • Lemon juice (from concentrate) or table salt (salt is cheaper but lemon juice is easier to use

and dissolve and gives a more dramatic result)

• A lemon (or other citrus fruit) • Knife (suggested that the instructor be in charge of the knife or razor blade)

A pre-1982 penny (these pennies are almost 100% Copper and work better than the post-1982 pennies; if post-1982 pennies must be used more lemon cells may be needed; copper wire (10-14 gauge is another alternative)

• A post-1965 nickel, dime, or quarter Two alligator clip wires (pieces of thin copper or insulated wire can be used instead)

Parts of a Film Canister Electrochemical Cell

Procedure:

21. Take the film canister lid and one of the nails and poke two holes in the top as shown in Figure 1 (this may already have been done for you).

(It is suggested that the teacher pre-punch all the lids for

Safety and Time Concerns)

22. Fill the canister with the distilled or purified water about ¾ the way full.

23. Place the lid on the canister and insert one of the nails in one of the holes and the copper wire in the other.

24. DESCRIBE the three main parts of an electrochemical cell. Assign each of these parts to the components of the Film Canister Cell.

The Main Parts of an Electrochemical Cell are: two electrodes (the nail and copper wire) in an electrolyte (vinegar) solution. An electrode is an electric conductor which current will enter or leave the electrolytic cell. The electrodes will need to of different materials and follow redox reactions so one is reduced and the other oxidized. An electrolyte is a dissolved compound that has broken into ions (anions and cations). When an electric field across the electrolyte the anions and cations move in opposite directions and current is conducted.

25. PREDICT the voltage of your Film Canister Electrochemical Cell. ________ V Answers will vary

Figure 1

26. Set the multi-meter to your teacher’s instructions (the voltage should be between 0.25 – 0.6 V and the meter should be set accordingly) and attach the probes.

27. Attach one alligator clip end to the galvanized nail and the other to one of the multi-meter probes.

28. Attach the other alligator clip end to the copper wire and the other to the second multi-meter probe.

ANALYZE: If the multi-meter displays a negative voltage then the positive and negative connections are switched. Noting this, which part of your Film Canister Cell (the galvanized nail or the copper wire) is the positive electrode and which is the negative electrode? What does this mean in terms of electron movement?

(The galvanized nail = negative and copper = positive; With copper and zinc electrodes, the zinc more readily loses electrons (e-) than the copper. Therefore, electrons flow from the zinc to the copper electrode.)

29. COMPARE the multi-meter voltage reading (________ V) to your predicted reading. Explain any differences or similarities between the two voltages.

Answers will vary but the students should reasonably explain any differences. For example: The student may predict a lower voltage than the reading thinking that there would be no reaction.

30. Unhook the alligator clips from the galvanized nail and copper wire. Remove lid and add 20 drops of lemon juice (or ~ 6 g salt). Replace top and gently shake container.

31. PREDICT if the voltage will change with the addition of the lemon juice. If so will it increase or decrease. Explain your prediction. The students should have a basic knowledge of electrolytes and know that lemon juice is acidic and will break easily into ions. Therefore the voltage should rise.

(Electrolyte: water and lemon juice; 2 Electrodes – nail (-) and copper wire (+); The nail (zinc) is the anode since this where oxidation occurs (loses e-). The copper wire is the cathode since this where reduction occurs (gains e-).

A chemical reaction occurs when two or more molecules interact and a chemical change occurs. All chemical reactions involve a change in energy. (Law of Conservation: matter or energy cannot be created or destroyed only changed).

This energy change allows for the conversion of chemical energy to electrical energy.

Metals differ in their tendency to give up electrons. This difference allows for an imbalance of charges and since opposite charges attract the electrons released from one metal will be attracted to the other metal. These metals are called electrodes. For the electron exchange between the two metals to begin and/or continue they must be placed in an electrolyte (a solution of ions). With copper and zinc electrodes, the zinc more readily loses electrons (e-) than the copper. Therefore, electrons flow from the zinc to the copper electrode. For this example acetic acid (CH3COOH) is used as the electrolyte. An acid has an easily detached hydrogen ion. Hydrogen ions are positive, and the remaining part of the acid (acetate) becomes negative when the hydrogen is removed (H+ and CH3COO- respectively). As the zinc electrode loses e-, positive zinc ions are produced. The acetate ions are more attracted to the zinc ions than the H+ and will pair up leaving the H+ to bond with the extra electrons on the copper electrode. Since the copper electrode has lost its extra e- to the H+ it can accept more from the zinc electrode. This continues the process until the zinc is used up and the battery is dead.

32. Connect the alligator clips back to the galvanized nail and copper wire.

33. COMPARE the multi-meter voltage reading (________ V) to your predicted reading. Explain any differences or similarities between the two voltages.

(The voltage should read between 0.85 V to 1 V; if the students are getting very low or zero volt readings their electrodes may be touching one another or there is a problem with the wire connections)

Parts of a Lemon Electrochemical Cell

34. Take the lemon and insert a galvanized nail in one top side of the lemon and with the knife make a small slit on the opposite side large enough for a penny. Insert penny.

(It is suggested that the teacher slice all the lemons for

Safety and Time Concerns)

Figure 2

35. Connect the alligator clips to the nail and penny and attach to the multi-meter. Record the voltage across the lemon in Data Table 1. REMEMBER if the multi-meter displays a negative voltage then the positive and negative connections are switched. Noting this, which part of the Lemon Cell (the galvanized nail or the penny) is the positive electrode and which is the negative electrode?

(The galvanized nail = negative and copper penny = positive. With copper and zinc electrodes, the zinc more readily loses electrons (e-) than the copper. Therefore, electrons flow from the zinc to the copper electrode.)

36. PREDICT if the nail or the penny can be replaced by a nickel, dime, or quarter. Explain your prediction.

(Any appropriate hypothesis is acceptable – many students will probably answer that the coins could replace the nail due to similar color)

37. Remove the nail from one of your lemons. Using the knife/razor blade make an incision right above or below the puncture mark. The incision should be large enough to fit a quarter. (The incision making may need to be done by the teacher)

38. Place one of the coins (nickel, dime or quarter) in this incision. Connect this coin to the negative terminal and the penny to the positive terminal of the multi-meter. RECORD your data in Data Table 1. (The voltage reading should be very close to zero since the coins are mainly made from copper – just like the penny)

39. Remove the coin from the lemon. Replace the nail into the lemon. In the penny’s location place one of the other coins (nickel, dime or quarter). Connect the nail to the negative terminal and the coin to the positive terminal of the multi-meter. RECORD your data in Data Table 1. (The voltage reading should be very similar to that of the penny: 0.85 V – 1 V)

DATA TABLE 1: Voltage Readings with Replacement of Galvanized Nail with Coins

ELECTRODES VOLTAGE (V)

Galvanized Nail and Penny

Nickel, Dime or Quarter and Penny

Galvanized Nail and Nickel, Dime or Quarter

40. Based on your results what can you conclude about the composition of the nickel, dime and quarter? Support your answer.

The composition of the nickel, dime and quarter is similar to the penny’s make-up. If two electrodes are made of the same material then they would have the same reaction in the electrolyte and no potential difference would be detected.

The coins are made of a Cupro-nickel alloy. The nickel is 25% Ni and 75% Cu. Both the dime and quarter are 8.33% Ni and 91.77% Cu. So even though the coins appear to be more like the nail they are actually more like the penny.

PART B: Changing Chemical Energy into Electrical Energy

OBJECTIVE: To create an electrochemical battery that produces enough voltage and current to light a LED.

MATERIALS:

• 4 film canisters (many photo labs will give you their empty film canister – the problem now is that most people have a digital camera and do not use film; film canisters can be ordered from Educational Innovations (www.teachersource.com) for $12.95 (~40 count) or $49.95 (~200 count))

• white vinegar (4-5% acetic acid; lemon-lime soda or lemon juice works as well)

• 4 pieces of copper wire (~8 cm) (10-14 gauge wire works best since it sturdy but still malleable enough to work with. Copper can be ordered on-line but it usually comes in large and expensive rolls. A building supply store allows for a portion of the wire to be cut from the roll (7½ ft is enough for 28 pieces (~8 cm)). It is suggested that the teacher cut the wire prior to the lab)

• 4 galvanized nails (this means they are zinc coated; they are easily found at building supply or hardware stores)

• 7 connecting wires with alligator clips (pieces of thin copper or insulated wire can be used instead)

• multi-meter (or voltmeter) • 2 incandescent lamps in holders (2.5 V – 3.5 V are recommended) • 1 LED light (clear bulbs with red or pink glow are usually the easiest to detect;

Recommended: 5mm Red LED (clear-color lens; typical voltage 1.7 V, with a max voltage of 2.4 V; 20 mA max); these can be purchased through an electronics store or on-line)

• 2 D batteries with holder (2 series battery holders are recommended or the batteries will need to be connected in series with wire; battery holders can be purchased at a electronics store or on-line)

• masking tape • marker/pen

PROCEDURE:

47. Using the galvanized nail poke holes in each of the film canister lids like in Part A: Figure 1. (It is suggested that the teacher pre-punch the lids for safety and time concerns) With the marker/pen and masking tape label each canister as A, B, C, and D. Carefully pour the vinegar into each of the film canisters (3/4 of the way full) and replace the caps.

48. Place the galvanized nails in one of the holes of the caps of each canister.

49. Place the copper wire in the other hole of each canister.

50. Using the multi-meter (set to the voltage reading specified by your teacher) and the connecting wires measure the voltage of each one of your film canister batteries. RECORD your measurements in Data Table 2.

DATA TABLE 2: Individual Voltage Readings

Film Canister Voltage Reading (V)

A

B

C

D

(Each one should read 0.85 V to 1 V; if the students are getting very low or zero volt readings their electrodes may be touching one another or there is a problem with the wire connections)

51. Choose two of the canisters and using the rules of series circuits predict the total voltage. Connect the two canisters in series with one another and measure their total voltage. Repeat this procedure with three and then four in series. RECORD your information in Data Table 3.

(Indicate the correct way to connect the canisters: positive terminal Cu wire; Zn nail (on same cell) Cu wire (2nd cell); Zn nail (on 2nd cell) negative terminal)

DATA TABLE 3: Combination Voltage Readings

Combination Predicted Voltage Measured Voltage

Two in Series ( and )

Three in Series ( , and )

Four in Series

52. Place two D batteries in series. Knowing that each is about 1.5 V PREDICT the total voltage for two in series? _____________. Using the multi-meter MEASURE the total voltage: _________V.

Since voltage adds in series circuits the students should predict a voltage around 3 V. They will probably measure slightly lower/higher due to the difference in emf (open circuit terminal voltage and the multi-meter requires very little current) and terminal voltage (voltage of a battery within a closed circuit with a load).

53. ANALYZE: Connect the two D batteries to the screws on the base of the lamp holder with the incandescent bulb. Does the light come on? (YES)

Switch the leads on the lamp. Does the light come on? (YES)

Using the combination of Film Canister Cells that produces the voltage closest to the two D batteries connect to the base of the other lamp holder. Does the light come on? (NO)

Switch the leads on the lamp. Does the light come on? (NO)

54. ANALYZE: Using the combination of Film Canister Cells that produces a voltage between 1.7 V – 2.4 V (or the combination specified by the teacher) connect to the leads of an LED lamp. Does the light come on? (YES/NO)

Switch the leads on the LED. Does the light come on? (YES/NO)

(Since the LED is a diode it will only work in one direction.)

QUESTIONS:

19. Why must the Film Canister Cells be placed in a series arrangement rather than a parallel arrangement?

(In series circuits the voltage of each individual cell is added together for the total voltage of the circuit. In parallel the voltage remains the same regardless how many cells you connect together. Since the voltage needs to be increased to power the lights a series circuit is the only option.)

20. Why will the batteries light up the incandescent lamp but not the Film Canister Battery even though their voltages are similar?

(The Film Canister Battery only produces a very small amount of current due its internal resistance. The D batteries have less internal resistance and will allow more current to the bulb. Even with multiple canister cells, the amount of current flowing through the wire is not enough. Though the voltage is high enough (~2.5 volts with three cells), the current is too weak.

Connecting a bulb to a voltage source allows charges to flow from one contact to the other. The charges will also pass through the wires and filament of the bulb. As the charges move through the filament they are continuously bumping into the filament’s atoms (usually tungsten). These collisions cause the atoms to vibrate which releases energy in the form of heat. The bound electrons in the vibrating atoms are raised to a higher energy level temporarily. When they fall back to their original levels, the electrons release extra energy in the form of photons. Unfortunately, this process is not very efficient. Most incandescent bulbs allow only 10% of its energy to go to light the rest is lost as heat. This is why there is such a push for alternative light bulbs.)

21. What is an LED and why is it able to be lit by the Film Canister Battery? Why will the LED only light in one direction?

(Light Emitting Diode; a light source created by current moving through a semiconducting diode.

A diode is a very simple semiconducting device. A semiconductor is a material with a varying ability to conduct current. Usually semiconductors are created from a poor conductor that has had atoms of another material added to it. This process is called doping. As electrons move through the semiconductor they will drop from a higher orbital to a lower orbital. When this occurs energy is released as a photon (light particle). A larger drop in energy releases a higher-energy photon (higher frequency – can vary color of LED). All diodes release light, however most are not efficient since the semiconducting material will absorb a lot of the light energy. LEDs are made to release a large number of photons outward and in a particular direction.

(This is due to the semiconducting nature of the diode.

Since extra atoms are used in doped materials (see question 4) free electrons (N-type; negative e- are attracted to + area) or holes (P-type; basically positively charged particles; since e- can jump from hole to hole the holes appear to move towards the negatively-charged area) where electrons can go are created. Either of these additions makes the material more electrically conductive.

A diode is composed of an N-type material attached to a portion of a P-type material, with electrodes on each end. This set up allows current in only one direction. When no potential difference is applied across the diode, e- from the N-type material fill holes from the P-type

material along the junction between the layers, forming a depletion zone (becomes an insulator and charges cannot flow).

To remove the depletion zone, e- from the N-type area must move to the P-type area and the holes would then appear to move in the opposite direction. This is accomplished by connecting the N-type side to the negative end of a voltage source and the P-type side to the positive end. The free e- in the N-type material are attracted to the + electrode and repelled by the negative electrode. The holes in the P-type material move in the opposite direction. When the potential difference between the electrodes is large enough, the e- in the depletion zone move out of their holes and begins to move freely. The depletion zone disappears, and charge moves across the diode.

If the electrodes are switched so the P-type side is connected to the negative end and the N-type side is connected to the positive end, current will not flow. This is due to the e- in the N-type material being attracted to the positive electrode and the positive holes in the P-type material being attracted to the negative electrode. No current moves across the diode because the holes and the e- are moving in the wrong direction and the depletion zone increases.

For more information about how LEDs and semiconductors work HowStuffWorks has good diagrams about these subjects.)

PART C: Resistors in Series and Parallel Circuits

OBJECTIVE: To demonstrate how resistance changes with series and parallel circuits. MATERIALS:

• Three resistors (I prefer the large block capacitors since the resistance and tolerance is printed on them; the banded resistors work fine but the students will have to be instructed how to read them. It is also suggested not to use any with too high or two low of a resistance (there tends to be more percent error). I usually use 3 Ω - 50Ω.

• Multi-meter with probes • 6 alligator clip wires

PROCEDURE:

16. RECORD the amount of resistance across each resistor (indicated by the label) and include the tolerance for each.

DATA TABLE 4: Resistance from Labels RESISTOR RESISTANCE (Ω) TOLERANCE (%)

R1 R2 R3

17. MEASURE the amount of resistance across each resistor using the multi-meter.

DATA TABLE 5: Resistance from the Multi-meter

RESISTOR RESISTANCE (Ω)

R1 R2 R3

18. DRAW a schematic diagram for a series circuit including the three resistors. You can label

your resistors as R1, R2 and R3.

19. What is the equation for calculating equivalent resistance in a series circuit? What

happens to the equivalent resistance of a series circuit when additional resistors are added?

R1 + R2 + R3 = Req

20. CALCULATE the total resistance for a series combination of your three resistors.

a) Equivalent resistance (series – label readings) = _________________ Ω b) Equivalent resistance (series – multi-meter) = _________________ Ω

21. MEASURE the equivalent resistance for your three resistors in a series combination using the multi-meter.

Equivalent resistance (series) = _________________ Ω

22. CALCULATE percent error. % Error = [(O - A)/A] x 100

O (observed) = values from 5 (a) and 5 (b) A (accepted) = value from 6

a) Percent error (series – label readings 5(a)) = ___________________% b) Percent error (series- multi-meter 5 (b)) = ___________________%

23. Which reading gave you a smaller percent error? Why?

24. Draw a schematic diagram for a parallel circuit including the three resistors. You can label your resistors as R1, R2 and R3.

(Diagram can vary)

25. What is the equation for calculating equivalent resistance in a parallel circuit? What happens to the equivalent resistance of a parallel circuit when additional resistors are added?

1/R1 + 1/R2 + 1/R3 = 1/Req OR Req = (R1 + R2 + R3)-1

26. CALCULATE the total resistance for a parallel combination of your three resistors.

a) Equivalent resistance (parallel – label readings) = _________________ Ω b) Equivalent resistance (parallel – multi-meter) = _________________ Ω

27. MEASURE the equivalent resistance for your three resistors in a parallel combination using the multi-meter.

Equivalent resistance (parallel) = _________________ Ω

28. Calculate percent error. % Error = [(O - A)/A] x 100

O (observed) = values from 11 a) and 11 b) A (accepted) = value from 12

a) Percent error (parallel – label readings) = ___________________% b) Percent error (parallel- multi-meter) = ___________________%

29. Which reading gave you a smaller percent error? Why?

30. What are some possible error sources that you may have encountered? Connection error; tolerance; calculation error Due to the equation the parallel readings tend to give a higher percent error.

PART D: Electrical Energy and Capacitance

OBJECTIVE: To construct a capacitor and to calculate its capacitance. Determine the relationship between the capacitance and the distance between the parallel plates.

MATERIALS:

• Aluminum foil (the heavy duty foil is recommended) • Copper wire (this may be optional but most multi-meters require leads from the +/- voltage

source area) • Three Hardcover textbooks (one to act as the capacitor and the other two for weight; the

students can press down on the book but this can tend to vary the readings if pressure is not consistent)

• 2 leads with alligator clips • Multi-meter (must read capacitance) • Calipers (optional- students can measure thickness with ruler) • Ruler

PROCEDURE:

10. Cut two pieces of aluminum foil so that they are slightly larger than the book you are using (this will be the plates of your capacitor). MEASURE the length and width of these foil sheets (they should be the same size. CALCULATE the area of your capacitor and record it in Data Table 6.

11. Make sure both pieces of foil are smooth and flat. Place one piece on top of a page in your book, place ten pages on top of this foil and then place your second sheet of foil. You now have ten sheets of paper serving as a dielectric between the foil pieces.

Figure 3:

12. Connect one lead to each foil piece. Connect the other end of the leads to the multi-meter. See Figure 3.

13. Place the two other books on top of the book containing the aluminum foil. Make sure that the foil is not touching the desk, the other piece of foil or any other object.

14. MEASURE and record the capacitance reading from the multi-meter in Data Table 7. (If the students have a difficult time getting a reading they should check to make sure the foil and leads are not touching anything. Another problem could be the voltage source so they should check that the copper wires are inserted properly. Also, more weigh may need to be added)

15. Repeat steps 2-5 for ten more trials with 20, 30, 40, 50, 60, 70, 80, 90, and 100 sheets of paper of your textbook acting as the dielectric between the foil pieces.

16. Using the calipers MEASURE the total thickness of several hundred sheets of paper in your textbook (Remember each page has 2 page numbers). Record your answer in Data Table 6.

17. RECORD the total number of sheets you measured and calculate the thickness of one sheet of paper. Record your answers in Data Table 6.

DATA TABLE 6: Capacitance Components

MEASUREMENT AMOUNT

Total thickness of all sheets

If the students measure in centimeters they must convert to meters.

Total number of sheets measured

Thickness per sheet

Units should be in meters.

Area of Capacitor

Units should be in square meters.

DATA TABLE 7: Capacitance Measurements

TRIAL NUMBER OF SHEETS BETWEEN PARALLEL

PLATES

MEASURED CAPACITANCE (FARADS)

1

10

2

20

3

30

4

40

5

50

6

60

7

70

8

80

9

90

10

100

18. CALCULATE the dielectric constant using your information in Data Table 6 and one of the measured capacitance trials (choose one trial and repeat three times to obtain an average) in Data Table 7. Show your work in the space provided.

Trial Number of Sheets Measured Capacitance (Farads)

1 2 3

Averaged Capacitance: _________________ F

Calculation:

Capacitance for paper varies from 1.5 to 4. It is recommended that data is collected from all groups and a class average is determined.

• Plot capacitance (y-axis) versus the number of sheets of paper (x-axis) on the graph on the following page. Draw the best-fit line through your data points.

Remind the students to label the x- and y-axis. It is recommended that the students put all their values in the same exponents to help them plot the data. They should plot an inverse relationship graph.

PROBLEMS:

5. Using your graph and data describe the relationship between the capacitance and plate separation. Is the relationship a 1/d, 1/d², or 1/d³ relationship? How are you able to tell from your graph or data?

According to the parallel plate capacitance equation (C = κε0A/d) the students should discover an 1/d relationship.

6. Calculate ε (permittivity constant) if C = 1.51 x 10-8 F, κ = 3.0, A = 0.061 m² and d = 1.01 x 10 -4m.

ε = Cd/κA ([(1.51 x 10-8 F)( 1.01 x 10 -4m)]/[(3.0)( 0.061 m²)] = 8.33 E -12 F/m

(Permittivity of free space is 8.85 E -12 F/m)

7. Why is a 1 F capacitor potentially dangerous?

1 F stores a lot of charge. Most capacitors are in the mF – pF range.

8. Why do people say not to open a TV set even though you just turned it off?

The TV has a large capacitor with a large amount of stored charge. Touching the capacitor could allow it to discharge.

PIONEERS OF ELECTRICITY/ELECTRIC CIRCUITS PROJECT

This project examines the lives of the people integral to development and understanding of electromagnetism and electric circuits. You may work by yourself or with one other person in your class period. You will have to present your project and it can be in the form of a report, PowerPoint, model, poster, brochure, video or cartoon. Most of this project will be done on your own but some time will be given in the classroom/computer lab. You will need to properly document your sources and have at least two of them (not counting Wikipedia).

This project is due ________________. For every day (weekend counts as one day) that it is late (without excuse) 10 points will be deducted. This will count as a project grade.

Choose one of the following scientist (only one scientist can be chosen per class):

George Simon Ohm Nikola Tesla James Watt

James Clerk Maxwell Michael Faraday Luigi Galavini

Charles Coulomb Andre Ampere Lewis Howard Latimer

Thomas Edison Granville Woods James Prescott Joule

Guglielmo Marconi Joseph Henry Gustave Robert Kirchhoff

Heinrich Rudolf Hertz Alessandro Volta Alexander Graham Bell

Joseph Henry Ada Lovelace Charles William Siemens

John Bardeen/William Shockley/Walter Brattain Hans Orsted

Benjamin Franklin Charles P. Steinmetz Leon Charles Thevenin

You need to include:

1) A biography of their life and the contributions they made to the field of Electricity/Electric Circuits.

2) Describe their experiments, inventions and findings.

3) Explain any hardships, obstacles or controversies they had endure and/or overcome.

4) Describe how their research and/or inventions contribute to our modern lives.

INTEGRATING TECHNOLOGY AND DESIGN PROJECT

This project examines how technology and design have been integrated with one another with a focus on electricity and electrical circuits. You may work by yourself or with one other person in your class period. You will have to present your project and it can be in the form of a report, PowerPoint (or other presentation format), model, poster, brochure, video or cartoon. Most of this project will be done on your own but some time will be given in the classroom/computer lab. You will need to properly document your sources and have at least two of them (not counting Wikipedia).

This project is due ________________. For every day (weekend counts as one day) that it is late (without excuse) 10 points will be deducted. This will count as a project grade.

Choose one of the following topics to discuss:

Technology and Fashion

Technology and Architecture

Technology and Art

Technology and Industrial Design

Technology and Music

Technology and Furniture Design

Project should include:

1) History/background on topic.

2) At least three examples that represent topic.

3) Future direction of topic.

4) Your opinion on the importance of technology and design.

Schematics and Circuits Power Point Presentation

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