Physics 132 Textbook Unit 5

Transcription

1 1 Physics 132 Textbook Unit 5 Collection edited by: UMass Physics Content authors: UMass Physics and OpenStax Online: < This selection and arrangement of content as a collection is copyrighted by UMass Physics. Creative Commons Attribution License Collection structure revised: 2018/08/16 PDF Generated: 2018/08/16 12:19:44 For copyright and attribution information for the modules contained in this collection, see the "Attributions" section at the end of the collection.

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3 Table of Contents 1 Introduction Unit 5 on a Page Introduction Current Circuit Elements Ideal Batteries Capacitors and Dielectrics Ohm s Law: Resistance and Simple Circuits Resistance and Resistivity Circuits Electric Power and Energy Kirchhoff's Principles Index

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5 Chapter 1 Introduction 5 1 INTRODUCTION 1.1 Unit 5 on a Page This section is also available as a video on the UMass Physics 13X YouTube page. ( UCUMejMY1La0t8qi-jwFTyTw) The link can be found here. ( Current and charge must be conserved item The current is the amount of charge per second I = Q units: C/s = A t Direction of current is direction of positive charges motion The amount of current going into a junction must equal the amount going out (otherwise it would pile up) The changes in potential around any closed loop is zero The potential drop across a battery is fixed The potential drop across a resistor is related to the current and resistance V = IR Resistors dissipate electrical potential energy into something else (heat, motion, light, etc.) The resistance is fixed by the material, units: V A = Ω The potential drop across a wire is zero ( R = 0 ) The potential drop across a capacitor is related to the charge and the capacitance: C = Q V A capacitor is two pieces of metal that don t touch The capacitance is a property of the geometry of the plates For parallel plates C II = ε A d units: C V = F Capacitors store charges separated, they store energy The power provided/stored/dissipated by a circuit element is P = IV Must do element-by-element

6 6 Chapter 1 Introduction 1.2 Introduction This section is also available as a video on the UMass Physics 13X YouTube page. ( UCUMejMY1La0t8qi-jwFTyTw) The link can be found here. ( How many electrical devices do you have on you at the moment? Electrical circuits are everywhere, I have my tablet, tablet pen, the remote, the microphone, my phone, and my watch. I've got six different electrical circuits on me right now. There are a lot of things that you might not think of as being an electric circuit, but we can analyze using the properties of circuits, which we are going to discuss over the course of this unit. By the end of this unit, we will be able to draw a rough circuit and think about a neuron as a circuit. A cell membrane is essentially a capacitor, so we can think about cell membrane ion transport, so we'll be able to talk a little bit about cell membranes in this context. We will introduce what a current is but other than that there are not a lot of new fundamental physics in this unit, we're mostly going to be dealing with the idea that if I put a charge and a potential, I get a potential energy. That is going to be the vast majority of what we're going to be working with. It's really a lot of application of those ideas and seeing how things work. 1.3 Current By the end of this section you should know: Electric current I is the rate at which charge flows, given by I = ΔQ Δt, where ΔQ is the amount of charge passing through an area in time Δt. The direction of conventional current is taken as the direction in which positive charge moves. The SI unit for current is the ampere (A), where 1 A = 1 C/s. Current is the flow of free charges, such as electrons and ions. Drift velocity v d is the average speed at which these charges move. (1.1) Current I is proportional to drift velocity v d, as expressed in the relationship I = nqav d. Here, I is the current through a wire of cross-sectional area A. The wire s material has a free-charge density n, and each carrier has charge q and a drift velocity v d. Electrical signals travel at speeds about times greater than the drift velocity of free electrons. Electric Current Electric current is defined to be the rate at which charge flows. A large current, such as that used to start a truck engine, moves a large amount of charge in a small time, whereas a small current, such as that used to operate a hand-held calculator, moves a small amount of charge over a long period of time. In equation form, electric current I is defined to be I = ΔQ Δt, (1.2) where ΔQ is the amount of charge passing through a given area in time Δt. (As in previous chapters, initial time is often taken to be zero, in which case Δt = t.) (See Figure 1.1.) The SI unit for current is the ampere (A), named for the French physicist André-Marie Ampère ( ). Since I = ΔQ / Δt, we see that an ampere is one coulomb per second: 1 A = 1 C/s (1.3) This OpenStax book is available for free at

7 Chapter 1 Introduction 7 Not only are fuses and circuit breakers rated in amperes (or amps), so are many electrical appliances. Figure 1.1 The rate of flow of charge is current. An ampere is the flow of one coulomb through an area in one second. Example 1.1 Calculating Currents: Current in a Truck Battery and a Handheld Calculator (a) What is the current involved when a truck battery sets in motion 720 C of charge in 4.00 s while starting an engine? (b) How long does it take 1.00 C of charge to flow through a handheld calculator if a mA current is flowing? Strategy We can use the definition of current in the equation I = ΔQ / Δt to find the current in part (a), since charge and time are given. In part (b), we rearrange the definition of current and use the given values of charge and current to find the time required. for (a) Entering the given values for charge and time into the definition of current gives Discussion for (a) I = ΔQ = Δt C = 180 C/s s = 180 A. This large value for current illustrates the fact that a large charge is moved in a small amount of time. The currents in these starter motors are fairly large because large frictional forces need to be overcome when setting something in motion. for (b) Solving the relationship I = ΔQ / Δt for time Δt, and entering the known values for charge and current gives Discussion for (b) Δt = ΔQ I = s. = 1.00 C C/s This time is slightly less than an hour. The small current used by the hand-held calculator takes a much longer time to move a smaller charge than the large current of the truck starter. So why can we operate our calculators only seconds after turning them on? It s because calculators require very little energy. Such small current and energy demands allow handheld calculators to operate from solar cells or to get many hours of use out of small batteries. Remember, calculators do not have moving parts in the same way that a truck engine has with cylinders and pistons, so the technology requires smaller currents. (1.4) (1.5) Figure 1.2 shows a simple circuit and the standard schematic representation of a battery, conducting path, and load (a resistor). Schematics are very useful in visualizing the main features of a circuit. A single schematic can represent a wide variety of situations. The schematic in Figure 1.2 (b), for example, can represent anything from a truck battery connected to a headlight lighting the street in front of the truck to a small battery connected to a penlight lighting a keyhole in a door. Such schematics are useful because the analysis is the same for a wide variety of situations. We need to understand a few schematics to apply the concepts and analysis to many more situations.

8 8 Chapter 1 Introduction Figure 1.2 (a) A simple electric circuit. A closed path for current to flow through is supplied by conducting wires connecting a load to the terminals of a battery. (b) In this schematic, the battery is represented by the two parallel red lines, conducting wires are shown as straight lines, and the zigzag represents the load. The schematic represents a wide variety of similar circuits. Note that the direction of current flow in Figure 1.2 is from positive to negative. The direction of conventional current is the direction that positive charge would flow. Depending on the situation, positive charges, negative charges, or both may move. In metal wires, for example, current is carried by electrons that is, negative charges move. In ionic solutions, such as salt water, both positive and negative charges move. This is also true in nerve cells. A Van de Graaff generator used for nuclear research can produce a current of pure positive charges, such as protons. Figure 1.3 illustrates the movement of charged particles that compose a current. The fact that conventional current is taken to be in the direction that positive charge would flow can be traced back to American politician and scientist Benjamin Franklin in the 1700s. He named the type of charge associated with electrons negative, long before they were known to carry current in so many situations. Franklin, in fact, was totally unaware of the smallscale structure of electricity. It is important to realize that there is an electric field in conductors responsible for producing the current, as illustrated in Figure 1.3. Unlike static electricity, where a conductor in equilibrium cannot have an electric field in it, conductors carrying a current have an electric field and are not in static equilibrium. An electric field is needed to supply energy to move the charges. Making Connections: Take-Home Investigation Electric Current Illustration Find a straw and little peas that can move freely in the straw. Place the straw flat on a table and fill the straw with peas. When you pop one pea in at one end, a different pea should pop out the other end. This demonstration is an analogy for an electric current. Identify what compares to the electrons and what compares to the supply of energy. What other analogies can you find for an electric current? Note that the flow of peas is based on the peas physically bumping into each other; electrons flow due to mutually repulsive electrostatic forces. This OpenStax book is available for free at

9 Chapter 1 Introduction 9 Figure 1.3 Current I is the rate at which charge moves through an area A, such as the cross-section of a wire. Conventional current is defined to move in the direction of the electric field. (a) Positive charges move in the direction of the electric field and the same direction as conventional current. (b) Negative charges move in the direction opposite to the electric field. Conventional current is in the direction opposite to the movement of negative charge. The flow of electrons is sometimes referred to as electronic flow. Example 1.2 Calculating the Number of Electrons that Move through a Calculator If the mA current through the calculator mentioned in the Example 1.1 example is carried by electrons, how many electrons per second pass through it? Strategy The current calculated in the previous example was defined for the flow of positive charge. For electrons, the magnitude is the same, but the sign is opposite, I electrons = C/s.Since each electron (e ) has a charge of C, we can convert the current in coulombs per second to electrons per second. Starting with the definition of current, we have I electrons = ΔQ electrons Δt We divide this by the charge per electron, so that Discussion = C s. e s = C s 1 e C 15 = e s. There are so many charged particles moving, even in small currents, that individual charges are not noticed, just as individual water molecules are not noticed in water flow. Even more amazing is that they do not always keep moving forward like soldiers in a parade. Rather they are like a crowd of people with movement in different directions but a general trend to move forward. There are lots of collisions with atoms in the metal wire and, of course, with other electrons. (1.6) (1.7) Drift Velocity Electrical signals are known to move very rapidly. Telephone conversations carried by currents in wires cover large distances without noticeable delays. Lights come on as soon as a switch is flicked. Most electrical signals carried by currents travel at speeds on the order of 10 8 m/s, a significant fraction of the speed of light. Interestingly, the individual charges that make up the current move much more slowly on average, typically drifting at speeds on the order of 10 4 m/s. How do we reconcile these two speeds, and what does it tell us about standard conductors? The high speed of electrical signals results from the fact that the force between charges acts rapidly at a distance. Thus, when a free charge is forced into a wire, as in Figure 1.4, the incoming charge pushes other charges ahead of it, which in turn push on charges farther down the line. The density of charge in a system cannot easily be increased, and so the signal is passed on rapidly. The resulting electrical shock wave moves through the system at nearly the speed of light. To be precise, this rapidly

10 10 Chapter 1 Introduction moving signal or shock wave is a rapidly propagating change in electric field. Figure 1.4 When charged particles are forced into this volume of a conductor, an equal number are quickly forced to leave. The repulsion between like charges makes it difficult to increase the number of charges in a volume. Thus, as one charge enters, another leaves almost immediately, carrying the signal rapidly forward. Good conductors have large numbers of free charges in them. In metals, the free charges are free electrons. Figure 1.5 shows how free electrons move through an ordinary conductor. The distance that an individual electron can move between collisions with atoms or other electrons is quite small. The electron paths thus appear nearly random, like the motion of atoms in a gas. But there is an electric field in the conductor that causes the electrons to drift in the direction shown (opposite to the field, since they are negative). The drift velocity v d is the average velocity of the free charges. Drift velocity is quite small, since there are so many free charges. If we have an estimate of the density of free electrons in a conductor, we can calculate the drift velocity for a given current. The larger the density, the lower the velocity required for a given current. Figure 1.5 Free electrons moving in a conductor make many collisions with other electrons and atoms. The path of one electron is shown. The average velocity of the free charges is called the drift velocity, v d, and it is in the direction opposite to the electric field for electrons. The collisions normally transfer energy to the conductor, requiring a constant supply of energy to maintain a steady current. Conduction of Electricity and Heat Good electrical conductors are often good heat conductors, too. This is because large numbers of free electrons can carry electrical current and can transport thermal energy. The free-electron collisions transfer energy to the atoms of the conductor. The electric field does work in moving the electrons through a distance, but that work does not increase the kinetic energy (nor speed, therefore) of the electrons. The work is transferred to the conductor s atoms, possibly increasing temperature. Thus a continuous power input is required to keep a current flowing. An exception, of course, is found in superconductors, for reasons we shall explore in a later chapter. Superconductors can have a steady current without a continual supply of energy a great energy savings. In contrast, the supply of energy can be useful, such as in a lightbulb filament. The supply of energy is necessary to increase the temperature of the tungsten filament, so that the filament glows. Making Connections: Take-Home Investigation Filament Observations Find a lightbulb with a filament. Look carefully at the filament and describe its structure. To what points is the filament connected? We can obtain an expression for the relationship between current and drift velocity by considering the number of free charges in a segment of wire, as illustrated in Figure 1.6. The number of free charges per unit volume is given the symbol n and depends on the material. The shaded segment has a volume Ax, so that the number of free charges in it is nax. The charge ΔQ in this segment is thus qnax, where q is the amount of charge on each carrier. (Recall that for electrons, q is C.) Current is charge moved per unit time; thus, if all the original charges move out of this segment in time Δt, the current is This OpenStax book is available for free at

11 Chapter 1 Introduction 11 I = ΔQ Δt = qnax. Δt Note that x / Δt is the magnitude of the drift velocity, v d, since the charges move an average distance x in a time Δt. Rearranging terms gives I = nqav d, (1.8) (1.9) where I is the current through a wire of cross-sectional area A made of a material with a free charge density n. The carriers of the current each have charge q and move with a drift velocity of magnitude v d. Figure 1.6 All the charges in the shaded volume of this wire move out in a time t, having a drift velocity of magnitude v d = x / t. See text for further discussion. Note that simple drift velocity is not the entire story. The speed of an electron is much greater than its drift velocity. In addition, not all of the electrons in a conductor can move freely, and those that do might move somewhat faster or slower than the drift velocity. So what do we mean by free electrons? Atoms in a metallic conductor are packed in the form of a lattice structure. Some electrons are far enough away from the atomic nuclei that they do not experience the attraction of the nuclei as much as the inner electrons do. These are the free electrons. They are not bound to a single atom but can instead move freely among the atoms in a sea of electrons. These free electrons respond by accelerating when an electric field is applied. Of course as they move they collide with the atoms in the lattice and other electrons, generating thermal energy, and the conductor gets warmer. In an insulator, the organization of the atoms and the structure do not allow for such free electrons. Example 1.3 Calculating Drift Velocity in a Common Wire Calculate the drift velocity of electrons in a 12-gauge copper wire (which has a diameter of mm) carrying a 20.0-A current, given that there is one free electron per copper atom. (Household wiring often contains 12-gauge copper wire, and the maximum current allowed in such wire is usually 20 A.) The density of copper is kg/m 3. Strategy We can calculate the drift velocity using the equation I = nqav d. The current I = 20.0 A is given, and q = C is the charge of an electron. We can calculate the area of a cross-section of the wire using the formula A = πr 2, where r is one-half the given diameter, mm. We are given the density of copper, kg/m 3, and the periodic table shows that the atomic mass of copper is g/mol. We can use these two quantities along with Avogadro s number, atoms/mol, to determine n, the number of free electrons per cubic meter. First, calculate the density of free electrons in copper. There is one free electron per copper atom. Therefore, is the same as the number of copper atoms per m 3. We can now find n as follows: The cross-sectional area of the wire is n = atom 1 e atoms mol mol g 1000 g kg kg 1 m 3 = e /m 3. (1.10)

12 12 Chapter 1 Introduction A = πr 2 Rearranging I = nqav d to isolate drift velocity gives = π m 2 = m 2. 2 (1.11) Discussion v d = I nqa = 20.0 A ( /m 3 )( C)( m 2 ) = m/s. The minus sign indicates that the negative charges are moving in the direction opposite to conventional current. The small value for drift velocity (on the order of 10 4 m/s ) confirms that the signal moves on the order of times faster (about 10 8 m/s ) than the charges that carry it. (1.12) Glossary ampere: (amp) the SI unit for current; 1 A = 1 C/s drift velocity: the average velocity at which free charges flow in response to an electric field electric current: the rate at which charge flows, I = ΔQ/Δt This OpenStax book is available for free at

13 Chapter 2 Circuit Elements 13 2 CIRCUIT ELEMENTS 2.1 Ideal Batteries This section is also available as a video on the UMass Physics 13X YouTube page. ( UCUMejMY1La0t8qi-jwFTyTw) The link can be found here. ( This section is about the power coming out of the wall or a battery, which is probably the first thing you think of when you start thinking about electricity and electric circuits. One of the important things is what is the electron volt as a unit, to go through the math, we're starting with the definition of potential, if we put a charge in a potential we get a potential energy, V, charge, Q, is coulombs which gives me joules or one electron volt. And in fact, this is what an electron volt is, the potential energy gained by a single electron going over a volt. We've mentioned this number before, the energy you need to ionize hydrogen gas is 13.6 electron volts, meaning that you have to give that electron 13.6 electron volts to rip it away from its nucleus. Based upon this you can say that the potential difference, not the potential energy, between the ground state of the electron and very far away is 13.6 volts. Keeping this distinction between electron volts as energy and volts as potential is going to be really important for this section. Our question for this section is: what is a battery? Essentially, it's two pieces of metal in contact. If you think back to Unit 1 we talked about this photoelectric effect, and we explained how much energy you need to remove electrons from a metal. That was described by the work function, which we used to find how much energy we need to remove an electron from the surface of a metal. We connected this with photons and other things but the key thing you need right now is that there is this material dependent number that essentially tells you how much energy you need to rip an electron off the surface and it's called that work function. If we take gold and platinum and touch them together, we can make a simple battery. Gold has a work function of about 5eV and platinum have a work function of about 6 1 ev. This means that an electron in the surface of the gold has a potential energy 3 of 5.1eV because the charge of an electron is negative, so the potential energy might take this positive and we get a negative potential. What's going to happen when we touch these two things together? Well the electron in the gold is going to slide down because now that's a lower energy state in gold it has a potential energy of 5.1eV in platinum it has a potential energy of 6.35eV, it can lower its potential energy by down. electron volt by sliding The energy sliding the electrons sliding over it is losing 1.25eV, if the potential energy difference is 1.25eV then the potential

14 14 Chapter 2 Circuit Elements difference is 1.25Volts, so we have a potential difference and electrons spontaneously moving from two metals in contact. What is a battery? A potential difference that, if I connect the two metals across the terminals the electrons spontaneously move. This is essentially the simplest battery I can think of. Now this isn't a very good battery because the electrons from the gold are going to kind of run across which means we re going to end up with a negatively charged piece of platinum and a positively charged piece of gold. And eventually this repulsion is going to stop the flow, you're going to build up a negatively charged thing and electrons aren't going to flow anymore. So that's what makes this kind of a crummy battery, the electrons flow once and then they stop pretty quickly, you get kind of a spark, you don't get a constant flow. But the potential difference between the two metals states that you keep that potential difference, the electrons just stop flowing because the repulsion. The battery we just talked about, the gold and platinum in contact, well that's not particularly helpful because the electrons flow once and then you're done. We want the battery to keep running, so we're actually going to repeat a famous experiment by a guy named Volta, and so for his setup, shown below, we have a piece of copper and a piece of zinc, and they're not touching each other, and we have no potential difference between them. Then we bring them into contact. Then we've got 1.5 moles per liter sulfuric acid and add this in until there's enough in there to bring the two metals into contact, through the sulfuric acid. So now we're getting a potential difference between these two pieces of metal, because now we've dropped them into contact using the sulfuric acid. Figure 2.1 A voltaic cell. (credit: Wikipedia) This is the first battery voltage, and you can see this potential difference is just sitting there, and these electrons keep flowing. In sulfuric acid, the potential of an electron in zinc is 3.68Volts, the fact that there is sulfuric acid changes these work function numbers. The charge of an electron is negative, so zinc has an electron and zinc has a potential energy of 6.38eV, an electron in the copper has a potential energy of 4.78eV. This number is bigger, and so electrons will flow to the lower energy state, which is towards the copper, because electrons go against the potential, from low potential to high potential. Be careful here we're talking about electrons moving and usually when you're dealing with electronics that's what's going on, but not always in a cell. For example, you can and often do have positive charges moving potassium ions, sodium ions, calcium ions, so those charges would move from high potential to low potential, because they're positive, electrons go low to high, so you got to think about what's actually doing the moving. This OpenStax book is available for free at

15 Chapter 2 Circuit Elements 15 Now let's go through and actually talk through how this thing keeps running. So, the gold-platinum, when we put them in contact the electrons run across and they stopped, because it's a buildup of charge and the repulsion stops the thing from going. Why does the zinc-copper keep going? Well zinc dissolves in sulfuric acid, and you spit off a Zinc 2 + ion, which will eventually precipitate out a zinc sulfate. What does that leave behind in the metal? Electrons. Zinc 2 + popped off, now I've got two electrons, and now my zinc has a slight negative charge to it. Well we just said that copper is the lower energy state, so the electrons would prefer to be in copper because it's a lower energy state. They go over and run through the wire because it's less resistance to run through the wire than to run through the solution. We would have the same problem that we did with our platinum-gold we'd eventually build up charges on this side and the whole thing would stop, but the solution comes in again. Because it's sulfuric acid, there's these positive hydrogen ions sitting in there and these electrons are attracted to these positive hydrogen ions and jump off. Which then form a nice diatomic hydrogen, which bubbles out and now we're back where we started. We have neutral-zinc, neutral-copper and the whole thing can just keep going, and so we can continue moving these electrons across and the reaction doesn't stop because we come back to a neutral state each time and we maintain a fixed potential difference. The reason this reaction keeps going is probably the number one thing that people miss in this unit. Batteries have a fixed potential difference between the terminals, they don't have a fixed current, and they don't have a fixed power output. This is the number one thing people miss in this unit right here. The big takeaway is the battery has a fixed potential difference between its terminals, they do not maintain constant current, they do not maintain constant power output fixed potential difference fixed potential difference. 2.2 Capacitors and Dielectrics By the end of this section you should know: A capacitor is a device used to store charge. The amount of charge Q a capacitor can store depends on two major factors the voltage applied and the capacitor s physical characteristics, such as its size. The capacitance C is the amount of charge stored per volt, or C = Q V. (2.1) The capacitance of a parallel plate capacitor is C = ε 0 A d, when the plates are separated by air or free space. ε 0 is called the permittivity of free space. A parallel plate capacitor with a dielectric between its plates has a capacitance given by C = κε 0 A d, (2.2) where κ is the dielectric constant of the material. The maximum electric field strength above which an insulating material begins to break down and conduct is called dielectric strength. A capacitor is a device used to store electric charge. Capacitors have applications ranging from filtering static out of radio reception to energy storage in heart defibrillators. Typically, commercial capacitors have two conducting parts close to one

16 16 Chapter 2 Circuit Elements another, but not touching, such as those in Figure 2.2. (Most of the time an insulator is used between the two plates to provide separation see the discussion on dielectrics below.) When battery terminals are connected to an initially uncharged capacitor, equal amounts of positive and negative charge, +Q and Q, are separated into its two plates. The capacitor remains neutral overall, but we refer to it as storing a charge Q in this circumstance. Capacitor A capacitor is a device used to store electric charge. Figure 2.2 Both capacitors shown here were initially uncharged before being connected to a battery. They now have separated charges of +Q and Q on their two halves. (a) A parallel plate capacitor. (b) A rolled capacitor with an insulating material between its two conducting sheets. The amount of charge Q a capacitor can store depends on two major factors the voltage applied and the capacitor s physical characteristics, such as its size. The Amount of Charge Q a Capacitor Can Store The amount of charge Q a capacitor can store depends on two major factors the voltage applied and the capacitor s physical characteristics, such as its size. A system composed of two identical, parallel conducting plates separated by a distance, as in Figure 2.3, is called a parallel plate capacitor. It is easy to see the relationship between the voltage and the stored charge for a parallel plate capacitor, as shown in Figure 2.3. Each electric field line starts on an individual positive charge and ends on a negative one, so that there will be more field lines if there is more charge. (Drawing a single field line per charge is a convenience, only. We can draw many field lines for each charge, but the total number is proportional to the number of charges.) The electric field strength is, thus, directly proportional to Q. This OpenStax book is available for free at

17 Chapter 2 Circuit Elements 17 Figure 2.3 Electric field lines in this parallel plate capacitor, as always, start on positive charges and end on negative charges. Since the electric field strength is proportional to the density of field lines, it is also proportional to the amount of charge on the capacitor. The field is proportional to the charge: E Q, (2.3) where the symbol means proportional to. From the discussion in Electric Potential in a Uniform Electric Field ( we know that the voltage across parallel plates is V = Ed. Thus, V E. (2.4) It follows, then, that V Q, and conversely, Q V. (2.5) This is true in general: The greater the voltage applied to any capacitor, the greater the charge stored in it. Different capacitors will store different amounts of charge for the same applied voltage, depending on their physical characteristics. We define their capacitance C to be such that the charge Q stored in a capacitor is proportional to C. The charge stored in a capacitor is given by Q = CV. (2.6) This equation expresses the two major factors affecting the amount of charge stored. Those factors are the physical characteristics of the capacitor, C, and the voltage, V. Rearranging the equation, we see that capacitance C is the amount of charge stored per volt, or Capacitance Capacitance C is the amount of charge stored per volt, or C = Q V. (2.7) C = Q V. (2.8) The unit of capacitance is the farad (F), named for Michael Faraday ( ), an English scientist who contributed to the fields of electromagnetism and electrochemistry. Since capacitance is charge per unit voltage, we see that a farad is a coulomb per volt, or

18 18 Chapter 2 Circuit Elements 1 F = 1 C 1 V. (2.9) A 1-farad capacitor would be able to store 1 coulomb (a very large amount of charge) with the application of only 1 volt. One farad is, thus, a very large capacitance. Typical capacitors range from fractions of a picofarad 1 pf = F to millifarads 1 mf = 10 3 F. Figure 2.4 shows some common capacitors. Capacitors are primarily made of ceramic, glass, or plastic, depending upon purpose and size. Insulating materials, called dielectrics, are commonly used in their construction, as discussed below. Figure 2.4 Some typical capacitors. Size and value of capacitance are not necessarily related. (credit: Windell Oskay) Parallel Plate Capacitor The parallel plate capacitor shown in Figure 2.5 has two identical conducting plates, each having a surface area A, separated by a distance d (with no material between the plates). When a voltage V is applied to the capacitor, it stores a charge Q, as shown. We can see how its capacitance depends on A and d by considering the characteristics of the Coulomb force. We know that like charges repel, unlike charges attract, and the force between charges decreases with distance. So it seems quite reasonable that the bigger the plates are, the more charge they can store because the charges can spread out more. Thus C should be greater for larger A. Similarly, the closer the plates are together, the greater the attraction of the opposite charges on them. So C should be greater for smaller d. Figure 2.5 Parallel plate capacitor with plates separated by a distance d. Each plate has an area A. It can be shown that for a parallel plate capacitor there are only two factors ( A and d ) that affect its capacitance C. The capacitance of a parallel plate capacitor in equation form is given by (2.10) C = ε A 0. d This OpenStax book is available for free at

19 Chapter 2 Circuit Elements 19 Capacitance of a Parallel Plate Capacitor C = ε 0 A d (2.11) A is the area of one plate in square meters, and d is the distance between the plates in meters. The constant ε 0 is the permittivity of free space; its numerical value in SI units is ε 0 = F/m. The units of F/m are equivalent to C 2 /N m 2. The small numerical value of ε 0 is related to the large size of the farad. A parallel plate capacitor must have a large area to have a capacitance approaching a farad. (Note that the above equation is valid when the parallel plates are separated by air or free space. When another material is placed between the plates, the equation is modified, as discussed below.) Example 2.1 Capacitance and Charge Stored in a Parallel Plate Capacitor (a) What is the capacitance of a parallel plate capacitor with metal plates, each of area 1.00 m 2, separated by 1.00 mm? (b) What charge is stored in this capacitor if a voltage of V is applied to it? Strategy Finding the capacitance C is a straightforward application of the equation C = ε 0 A / d. Once C is found, the charge stored can be found using the equation Q = CV. for (a) Entering the given values into the equation for the capacitance of a parallel plate capacitor yields Discussion for (a) C = ε A 0 = d m F 1.00 m m = F = 8.85 nf. This small value for the capacitance indicates how difficult it is to make a device with a large capacitance. Special techniques help, such as using very large area thin foils placed close together. for (b) The charge stored in any capacitor is given by the equation Q = CV. Entering the known values into this equation gives (2.12) Q = CV = F V = 26.6 μc. (2.13) Discussion for (b) This charge is only slightly greater than those found in typical static electricity. Since air breaks down at about V/m, more charge cannot be stored on this capacitor by increasing the voltage. Another interesting biological example dealing with electric potential is found in the cell s plasma membrane. The membrane sets a cell off from its surroundings and also allows ions to selectively pass in and out of the cell. There is a potential difference across the membrane of about 70 mv. This is due to the mainly negatively charged ions in the cell and the predominance of positively charged sodium ( Na + ) ions outside. Things change when a nerve cell is stimulated. Na + ions are allowed to pass through the membrane into the cell, producing a positive membrane potential the nerve signal. The cell membrane is about 7 to 10 nm thick. An approximate value of the electric field across it is given by This electric field is enough to cause a breakdown in air. Dielectric E = V d = V m = V/m. The previous example highlights the difficulty of storing a large amount of charge in capacitors. If d is made smaller to produce a larger capacitance, then the maximum voltage must be reduced proportionally to avoid breakdown (since E = V / d ). An important solution to this difficulty is to put an insulating material, called a dielectric, between the plates of a capacitor and allow d to be as small as possible. Not only does the smaller d make the capacitance greater, but many insulators can withstand (2.14)

20 20 Chapter 2 Circuit Elements greater electric fields than air before breaking down. There is another benefit to using a dielectric in a capacitor. Depending on the material used, the capacitance is greater than that given by the equation C = ε A 0 by a factor κ, called the dielectric constant. A parallel plate capacitor with a dielectric between d its plates has a capacitance given by C = κε 0 A d (parallel plate capacitor with dielectric). (2.15) Values of the dielectric constant κ for various materials are given in Table 2.1. Note that κ for vacuum is exactly 1, and so the above equation is valid in that case, too. If a dielectric is used, perhaps by placing Teflon between the plates of the capacitor in Example 2.1, then the capacitance is greater by the factor κ, which for Teflon is 2.1. Take-Home Experiment: Building a Capacitor How large a capacitor can you make using a chewing gum wrapper? The plates will be the aluminum foil, and the separation (dielectric) in between will be the paper. Table 2.1 Dielectric Constants and Dielectric Strengths for Various Materials at 20ºC Material Dielectric constant κ Dielectric strength (V/m) Vacuum Air Bakelite Fused quartz Neoprene rubber Nylon Paper Polystyrene Pyrex glass Silicon oil Strontium titanate Teflon Water 80 Note also that the dielectric constant for air is very close to 1, so that air-filled capacitors act much like those with vacuum between their plates except that the air can become conductive if the electric field strength becomes too great. (Recall that E = V / d for a parallel plate capacitor.) Also shown in Table 2.1 are maximum electric field strengths in V/m, called dielectric strengths, for several materials. These are the fields above which the material begins to break down and conduct. The dielectric strength imposes a limit on the voltage that can be applied for a given plate separation. For instance, in Example 2.1, the separation is 1.00 mm, and so the voltage limit for air is V = E d = ( V/m)( m) = 3000 V. However, the limit for a 1.00 mm separation filled with Teflon is 60,000 V, since the dielectric strength of Teflon is V/m. So the same capacitor filled with Teflon has a greater capacitance and can be subjected to a much greater voltage. Using the capacitance we calculated in the above example for the air-filled parallel plate capacitor, we find that the Teflon-filled capacitor can store a maximum charge of (2.16) This OpenStax book is available for free at

21 Chapter 2 Circuit Elements 21 Q = CV = κc air V This is 42 times the charge of the same air-filled capacitor. = (2.1)(8.85 nf)( V) = 1.1 mc. (2.17) Dielectric Strength The maximum electric field strength above which an insulating material begins to break down and conduct is called its dielectric strength. Microscopically, how does a dielectric increase capacitance? Polarization of the insulator is responsible. The more easily it is polarized, the greater its dielectric constant κ. Water, for example, is a polar molecule because one end of the molecule has a slight positive charge and the other end has a slight negative charge. The polarity of water causes it to have a relatively large dielectric constant of 80. The effect of polarization can be best explained in terms of the characteristics of the Coulomb force. Figure 2.6 shows the separation of charge schematically in the molecules of a dielectric material placed between the charged plates of a capacitor. The Coulomb force between the closest ends of the molecules and the charge on the plates is attractive and very strong, since they are very close together. This attracts more charge onto the plates than if the space were empty and the opposite charges were a distance d away. Figure 2.6 (a) The molecules in the insulating material between the plates of a capacitor are polarized by the charged plates. This produces a layer of opposite charge on the surface of the dielectric that attracts more charge onto the plate, increasing its capacitance. (b) The dielectric reduces the electric field strength inside the capacitor, resulting in a smaller voltage between the plates for the same charge. The capacitor stores the same charge for a smaller voltage, implying that it has a larger capacitance because of the dielectric. Another way to understand how a dielectric increases capacitance is to consider its effect on the electric field inside the capacitor. Figure 2.6(b) shows the electric field lines with a dielectric in place. Since the field lines end on charges in the dielectric, there are fewer of them going from one side of the capacitor to the other. So the electric field strength is less than if there were a vacuum between the plates, even though the same charge is on the plates. The voltage between the plates is V = Ed, so it too is reduced by the dielectric. Thus there is a smaller voltage V for the same charge Q ; since C = Q / V, the capacitance C is greater. The dielectric constant is generally defined to be κ = E 0 / E, or the ratio of the electric field in a vacuum to that in the dielectric material, and is intimately related to the polarizability of the material.

22 22 Chapter 2 Circuit Elements Things Great and Small The Submicroscopic Origin of Polarization Polarization is a separation of charge within an atom or molecule. As has been noted, the planetary model of the atom pictures it as having a positive nucleus orbited by negative electrons, analogous to the planets orbiting the Sun. Although this model is not completely accurate, it is very helpful in explaining a vast range of phenomena and will be refined elsewhere, such as in Atomic Physics ( The submicroscopic origin of polarization can be modeled as shown in Figure 2.7. Figure 2.7 Artist s conception of a polarized atom. The orbits of electrons around the nucleus are shifted slightly by the external charges (shown exaggerated). The resulting separation of charge within the atom means that it is polarized. Note that the unlike charge is now closer to the external charges, causing the polarization. We will find in Atomic Physics ( that the orbits of electrons are more properly viewed as electron clouds with the density of the cloud related to the probability of finding an electron in that location (as opposed to the definite locations and paths of planets in their orbits around the Sun). This cloud is shifted by the Coulomb force so that the atom on average has a separation of charge. Although the atom remains neutral, it can now be the source of a Coulomb force, since a charge brought near the atom will be closer to one type of charge than the other. Some molecules, such as those of water, have an inherent separation of charge and are thus called polar molecules. Figure 2.8 illustrates the separation of charge in a water molecule, which has two hydrogen atoms and one oxygen atom H 2 O. The water molecule is not symmetric the hydrogen atoms are repelled to one side, giving the molecule a boomerang shape. The electrons in a water molecule are more concentrated around the more highly charged oxygen nucleus than around the hydrogen nuclei. This makes the oxygen end of the molecule slightly negative and leaves the hydrogen ends slightly positive. The inherent separation of charge in polar molecules makes it easier to align them with external fields and charges. Polar molecules therefore exhibit greater polarization effects and have greater dielectric constants. Those who study chemistry will find that the polar nature of water has many effects. For example, water molecules gather ions much more effectively because they have an electric field and a separation of charge to attract charges of both signs. Also, as brought out in the previous chapter, polar water provides a shield or screening of the electric fields in the highly charged molecules of interest in biological systems. Figure 2.8 Artist s conception of a water molecule. There is an inherent separation of charge, and so water is a polar molecule. Electrons in the molecule are attracted to the oxygen nucleus and leave an excess of positive charge near the two hydrogen nuclei. (Note that the schematic on the right is a rough illustration of the distribution of electrons in the water molecule. It does not show the actual numbers of protons and electrons involved in the structure.) PhET Explorations: Capacitor Lab Explore how a capacitor works! Change the size of the plates and add a dielectric to see the effect on capacitance. Change the voltage and see charges built up on the plates. Observe the electric field in the capacitor. Measure the voltage and the electric field. This OpenStax book is available for free at

23 Chapter 2 Circuit Elements 23 Figure 2.9 Capacitor Lab ( 2.3 Ohm s Law: Resistance and Simple Circuits What drives current? We can think of various devices such as batteries, generators, wall outlets, and so on which are necessary to maintain a current. All such devices create a potential difference and are loosely referred to as voltage sources. When a voltage source is connected to a conductor, it applies a potential difference V that creates an electric field. The electric field in turn exerts force on charges, causing current. Ohm s Law The current that flows through most substances is directly proportional to the voltage V applied to it. The German physicist Georg Simon Ohm ( ) was the first to demonstrate experimentally that the current in a metal wire is directly proportional to the voltage applied: I V. This important relationship is known as Ohm s law. It can be viewed as a cause-and-effect relationship, with voltage the cause and current the effect. This is an empirical law like that for friction an experimentally observed phenomenon. Such a linear relationship doesn t always occur. Resistance and Simple Circuits If voltage drives current, what impedes it? The electric property that impedes current (crudely similar to friction and air resistance) is called resistance R. Collisions of moving charges with atoms and molecules in a substance transfer energy to the substance and limit current. Resistance is defined as inversely proportional to current, or I 1 R. Thus, for example, current is cut in half if resistance doubles. Combining the relationships of current to voltage and current to resistance gives I = V R. This relationship is also called Ohm s law. Ohm s law in this form really defines resistance for certain materials. Ohm s law (like Hooke s law) is not universally valid. The many substances for which Ohm s law holds are called ohmic. These include good conductors like copper and aluminum, and some poor conductors under certain circumstances. Ohmic materials have a resistance R that is independent of voltage V and current I. An object that has simple resistance is called a resistor, even if its resistance is small. The unit for resistance is an ohm and is given the symbol Ω (upper case Greek omega). Rearranging I = V/R gives R = V/I, and so the units of resistance are 1 ohm = 1 volt per ampere: 1 Ω = 1 V A. Figure 2.10 shows the schematic for a simple circuit. A simple circuit has a single voltage source and a single resistor. The wires connecting the voltage source to the resistor can be assumed to have negligible resistance, or their resistance can be included in R. (2.18) (2.19) (2.20) (2.21) Figure 2.10 A simple electric circuit in which a closed path for current to flow is supplied by conductors (usually metal wires) connecting a load to the terminals of a battery, represented by the red parallel lines. The zigzag symbol represents the single resistor and includes any resistance in the connections to the voltage source.

24 24 Chapter 2 Circuit Elements Example 2.2 Calculating Resistance: An Automobile Headlight What is the resistance of an automobile headlight through which 2.50 A flows when 12.0 V is applied to it? Strategy We can rearrange Ohm s law as stated by I = V/R and use it to find the resistance. Rearranging I = V/R and substituting known values gives Discussion R = V I = V = 4.80 Ω. A This is a relatively small resistance, but it is larger than the cold resistance of the headlight. As we shall see in Resistance and Resistivity ( resistance usually increases with temperature, and so the bulb has a lower resistance when it is first switched on and will draw considerably more current during its brief warm-up period. (2.22) Resistances range over many orders of magnitude. Some ceramic insulators, such as those used to support power lines, have resistances of Ω or more. A dry person may have a hand-to-foot resistance of 10 5 Ω, whereas the resistance of the human heart is about 10 3 Ω. A meter-long piece of large-diameter copper wire may have a resistance of 10 5 Ω, and superconductors have no resistance at all (they are non-ohmic). Resistance is related to the shape of an object and the material of which it is composed, as will be seen in Resistance and Resistivity ( Additional insight is gained by solving I = V/R for V, yielding V = IR. (2.23) This expression for V can be interpreted as the voltage drop across a resistor produced by the flow of current I. The phrase IR drop is often used for this voltage. For instance, the headlight in Example 2.2 has an IR drop of 12.0 V. If voltage is measured at various points in a circuit, it will be seen to increase at the voltage source and decrease at the resistor. Voltage is similar to fluid pressure. The voltage source is like a pump, creating a pressure difference, causing current the flow of charge. The resistor is like a pipe that reduces pressure and limits flow because of its resistance. Conservation of energy has important consequences here. The voltage source supplies energy (causing an electric field and a current), and the resistor converts it to another form (such as thermal energy). In a simple circuit (one with a single simple resistor), the voltage supplied by the source equals the voltage drop across the resistor, since PE = qδv, and the same q flows through each. Thus the energy supplied by the voltage source and the energy converted by the resistor are equal. (See Figure 2.11.) Figure 2.11 The voltage drop across a resistor in a simple circuit equals the voltage output of the battery. Making Connections: Conservation of Energy In a simple electrical circuit, the sole resistor converts energy supplied by the source into another form. Conservation of energy is evidenced here by the fact that all of the energy supplied by the source is converted to another form by the resistor alone. We will find that conservation of energy has other important applications in circuits and is a powerful tool in circuit analysis. PhET Explorations: Ohm's Law See how the equation form of Ohm's law relates to a simple circuit. Adjust the voltage and resistance, and see the current change according to Ohm's law. The sizes of the symbols in the equation change to match the circuit diagram. (This media type is not supported in this reader. Click to open media in browser.) ( m67937/1.1/#eip-idm ) This OpenStax book is available for free at

25 Chapter 2 Circuit Elements 25 Section Summary A simple circuit is one in which there is a single voltage source and a single resistance. One statement of Ohm s law gives the relationship between current I, voltage V, and resistance R in a simple circuit to be I = V R. Resistance has units of ohms ( Ω ), related to volts and amperes by 1 Ω = 1 V/A. There is a voltage or IR drop across a resistor, caused by the current flowing through it, given by V = IR. 2.4 Resistance and Resistivity For this section I am NOT expecting you to remember the formula exactly. What I want you to know is that resistance increases with length (more atoms to run into), decreases with area, and is dependent upon the resistivity of the material. Material and Shape Dependence of Resistance The resistance of an object depends on its shape and the material of which it is composed. The cylindrical resistor in Figure 2.12 is easy to analyze, and, by so doing, we can gain insight into the resistance of more complicated shapes. As you might expect, the cylinder s electric resistance R is directly proportional to its length L, similar to the resistance of a pipe to fluid flow. The longer the cylinder, the more collisions charges will make with its atoms. The greater the diameter of the cylinder, the more current it can carry (again similar to the flow of fluid through a pipe). In fact, R is inversely proportional to the cylinder s crosssectional area A. Figure 2.12 A uniform cylinder of length L and cross-sectional area A. Its resistance to the flow of current is similar to the resistance posed by a pipe to fluid flow. The longer the cylinder, the greater its resistance. The larger its cross-sectional area A, the smaller its resistance. For a given shape, the resistance depends on the material of which the object is composed. Different materials offer different resistance to the flow of charge. We define the resistivity ρ of a substance so that the resistance R of an object is directly proportional to ρ. Resistivity ρ is an intrinsic property of a material, independent of its shape or size. The resistance R of a uniform cylinder of length L, of cross-sectional area A, and made of a material with resistivity ρ, is R = ρl A. (2.24) Table 2.2 gives representative values of ρ. The materials listed in the table are separated into categories of conductors, semiconductors, and insulators, based on broad groupings of resistivities. Conductors have the smallest resistivities, and insulators have the largest; semiconductors have intermediate resistivities. Conductors have varying but large free charge densities, whereas most charges in insulators are bound to atoms and are not free to move. Semiconductors are intermediate, having far fewer free charges than conductors, but having properties that make the number of free charges depend strongly on the type and amount of impurities in the semiconductor. These unique properties of semiconductors are put to use in modern electronics, as will be explored in later chapters.

26 26 Chapter 2 Circuit Elements Table 2.2 Resistivities ρ of Various materials at 20ºC Material Resistivity ρ ( Ω m ) Conductors Silver Copper Gold Aluminum Tungsten Iron Platinum Steel Lead Manganin (Cu, Mn, Ni alloy) Constantan (Cu, Ni alloy) Mercury Nichrome (Ni, Fe, Cr alloy) Semiconductors [1] Carbon (pure) Carbon (3.5 60) 10 5 Germanium (pure) Germanium (1 600) 10 3 Silicon (pure) 2300 Silicon Insulators Amber Glass Lucite >10 13 Mica Quartz (fused) Rubber (hard) Sulfur Teflon >10 13 Wood This OpenStax book is available for free at

27 Chapter 2 Circuit Elements 27 Example 2.3 Calculating Resistor Diameter: A Headlight Filament A car headlight filament is made of tungsten and has a cold resistance of Ω. If the filament is a cylinder 4.00 cm long (it may be coiled to save space), what is its diameter? Strategy We can rearrange the equation R = ρl A to find the cross-sectional area A of the filament from the given information. Then its diameter can be found by assuming it has a circular cross-section. The cross-sectional area, found by rearranging the expression for the resistance of a cylinder given in R = ρl A, is A = ρl R. (2.25) Substituting the given values, and taking ρ from Table 2.2, yields A = ( Ω m)( m) Ω = m 2. The area of a circle is related to its diameter D by A = πd2 4. Solving for the diameter D, and substituting the value found for A, gives Discussion 1 1 D = 2 A p 2 = m = m. The diameter is just under a tenth of a millimeter. It is quoted to only two digits, because ρ is known to only two digits. (2.26) (2.27) (2.28) Temperature Variation of Resistance The resistivity of all materials depends on temperature. Some even become superconductors (zero resistivity) at very low temperatures. (See Figure 2.13.) Conversely, the resistivity of conductors increases with increasing temperature. Since the atoms vibrate more rapidly and over larger distances at higher temperatures, the electrons moving through a metal make more collisions, effectively making the resistivity higher. Over relatively small temperature changes (about 100ºC or less), resistivity ρ varies with temperature change ΔT as expressed in the following equation ρ = ρ 0 (1 + αδt), where ρ 0 is the original resistivity and α is the temperature coefficient of resistivity. (See the values of α in Table 2.3 below.) For larger temperature changes, α may vary or a nonlinear equation may be needed to find ρ. Note that α is positive for metals, meaning their resistivity increases with temperature. Some alloys have been developed specifically to have a small temperature dependence. Manganin (which is made of copper, manganese and nickel), for example, has α close to zero (to three digits on the scale in Table 2.3), and so its resistivity varies only slightly with temperature. This is useful for making a temperature-independent resistance standard, for example. (2.29) 1. Values depend strongly on amounts and types of impurities

28 28 Chapter 2 Circuit Elements Figure 2.13 The resistance of a sample of mercury is zero at very low temperatures it is a superconductor up to about 4.2 K. Above that critical temperature, its resistance makes a sudden jump and then increases nearly linearly with temperature. Table 2.3 Tempature Coefficients of Resistivity α Material Coefficient α (1/ C) [2] Conductors Silver Copper Gold Aluminum Tungsten Iron Platinum Lead Manganin (Cu, Mn, Ni alloy) Constantan (Cu, Ni alloy) Mercury Nichrome (Ni, Fe, Cr alloy) Semiconductors Carbon (pure) Germanium (pure) Silicon (pure) Note also that α is negative for the semiconductors listed in Table 2.3, meaning that their resistivity decreases with increasing temperature. They become better conductors at higher temperature, because increased thermal agitation increases the number of free charges available to carry current. This property of decreasing ρ with temperature is also related to the type and amount of impurities present in the semiconductors. The resistance of an object also depends on temperature, since R 0 is directly proportional to ρ. For a cylinder we know 2. Values at 20 C. This OpenStax book is available for free at

29 Chapter 2 Circuit Elements 29 R = ρl / A, and so, if L and A do not change greatly with temperature, R will have the same temperature dependence as ρ. (Examination of the coefficients of linear expansion shows them to be about two orders of magnitude less than typical temperature coefficients of resistivity, and so the effect of temperature on L and A is about two orders of magnitude less than on ρ.) Thus, R = R 0 (1 + αδt) (2.30) is the temperature dependence of the resistance of an object, where R 0 is the original resistance and R is the resistance after a temperature change ΔT. Numerous thermometers are based on the effect of temperature on resistance. (See Figure 2.14.) One of the most common is the thermistor, a semiconductor crystal with a strong temperature dependence, the resistance of which is measured to obtain its temperature. The device is small, so that it quickly comes into thermal equilibrium with the part of a person it touches. Figure 2.14 These familiar thermometers are based on the automated measurement of a thermistor s temperature-dependent resistance. (credit: Biol, Wikimedia Commons) Example 2.4 Calculating Resistance: Hot-Filament Resistance Although caution must be used in applying ρ = ρ 0 (1 + αδt) and R = R 0 (1 + αδt) for temperature changes greater than 100ºC, for tungsten the equations work reasonably well for very large temperature changes. What, then, is the resistance of the tungsten filament in the previous example if its temperature is increased from room temperature ( 20ºC ) to a typical operating temperature of 2850ºC? Strategy This is a straightforward application of R = R 0 (1 + αδt), since the original resistance of the filament was given to be R 0 = Ω, and the temperature change is ΔT = 2830ºC. The hot resistance R is obtained by entering known values into the above equation: R = R 0 (1 + αδt) (2.31) = (0.350 Ω)[1 + ( / ºC)(2830ºC)] = 4.8 Ω. Discussion This value is consistent with the headlight resistance example in Ohm s Law: Resistance and Simple Circuits ( PhET Explorations: Resistance in a Wire Learn about the physics of resistance in a wire. Change its resistivity, length, and area to see how they affect the wire's resistance. The sizes of the symbols in the equation change along with the diagram of a wire. (This media type is not supported in this reader. Click to open media in browser.) ( m67938/1.2/#eip-idm ) Section Summary The resistance R of a cylinder of length L and cross-sectional area A is R = ρl A, where ρ is the resistivity of the material. Values of ρ in Table 2.2 show that materials fall into three groups conductors, semiconductors, and insulators. Temperature affects resistivity; for relatively small temperature changes ΔT, resistivity is ρ = ρ 0 (1 + αδt), where ρ 0 is the original resistivity and α is the temperature coefficient of resistivity.

30 30 Chapter 2 Circuit Elements Table 2.3 gives values for α, the temperature coefficient of resistivity. The resistance R of an object also varies with temperature: R = R 0 (1 + αδt), where R 0 is the original resistance, and R is the resistance after the temperature change. Conceptual Questions Exercise 2.1 Does the capacitance of a device depend on the applied voltage? What about the charge stored in it? Exercise 2.2 Use the characteristics of the Coulomb force to explain why capacitance should be proportional to the plate area of a capacitor. Similarly, explain why capacitance should be inversely proportional to the separation between plates. Exercise 2.3 Give the reason why a dielectric material increases capacitance compared with what it would be with air between the plates of a capacitor. What is the independent reason that a dielectric material also allows a greater voltage to be applied to a capacitor? (The dielectric thus increases C and permits a greater V.) Exercise 2.4 How does the polar character of water molecules help to explain water s relatively large dielectric constant? (???) Exercise 2.5 Sparks will occur between the plates of an air-filled capacitor at lower voltage when the air is humid than when dry. Explain why, considering the polar character of water molecules. Exercise 2.6 Water has a large dielectric constant, but it is rarely used in capacitors. Explain why. Exercise 2.7 Membranes in living cells, including those in humans, are characterized by a separation of charge across the membrane. Effectively, the membranes are thus charged capacitors with important functions related to the potential difference across the membrane. Is energy required to separate these charges in living membranes and, if so, is its source the metabolization of food energy or some other source? This OpenStax book is available for free at

31 Chapter 2 Circuit Elements 31 Figure 2.15 The semipermeable membrane of a cell has different concentrations of ions inside and out. Diffusion moves the K + (potassium) and Cl (chloride) ions in the directions shown, until the Coulomb force halts further transfer. This results in a layer of positive charge on the outside, a layer of negative charge on the inside, and thus a voltage across the cell membrane. The membrane is normally impermeable to Na + (sodium ions). Exercise 2.8 The IR drop across a resistor means that there is a change in potential or voltage across the resistor. Is there any change in current as it passes through a resistor? Explain. Exercise 2.9 How is the IR drop in a resistor similar to the pressure drop in a fluid flowing through a pipe? Exercise 2.10 In which of the three semiconducting materials listed in Table 2.2 do impurities supply free charges? (Hint: Examine the range of resistivity for each and determine whether the pure semiconductor has the higher or lower conductivity.) Exercise 2.11 Does the resistance of an object depend on the path current takes through it? Consider, for example, a rectangular bar is its resistance the same along its length as across its width? (See Figure 2.16.) Figure 2.16 Does current taking two different paths through the same object encounter different resistance? Exercise 2.12 If aluminum and copper wires of the same length have the same resistance, which has the larger diameter? Why? Exercise 2.13 Explain why R = R 0 (1 + αδt) for the temperature variation of the resistance R of an object is not as accurate as

32 32 Chapter 2 Circuit Elements ρ = ρ 0 (1 + αδt), which gives the temperature variation of resistivity ρ. title Exercise 2.14 What charge is stored in a 180 µf capacitor when 120 V is applied to it? 21.6 mc Exercise 2.15 Find the charge stored when 5.50 V is applied to an 8.00 pf capacitor. Exercise 2.16 What charge is stored in the capacitor in???? 80.0 mc Exercise 2.17 Calculate the voltage applied to a 2.00 µf capacitor when it holds 3.10 µc of charge. Exercise 2.18 What voltage must be applied to an 8.00 nf capacitor to store mc of charge? 20.0 kv Exercise 2.19 What capacitance is needed to store 3.00 µc of charge at a voltage of 120 V? Exercise 2.20 What is the capacitance of a large Van de Graaff generator s terminal, given that it stores 8.00 mc of charge at a voltage of 12.0 MV? 667 pf Exercise 2.21 Find the capacitance of a parallel plate capacitor having plates of area 5.00 m 2 that are separated by mm of Teflon. Exercise 2.22 (a)what is the capacitance of a parallel plate capacitor having plates of area 1.50 m 2 that are separated by mm of neoprene rubber? (b) What charge does it hold when 9.00 V is applied to it? This OpenStax book is available for free at

33 Chapter 2 Circuit Elements 33 (a) 4.4 µf (b) C Exercise 2.23 Integrated Concepts A prankster applies 450 V to an 80.0 µf capacitor and then tosses it to an unsuspecting victim. The victim s finger is burned by the discharge of the capacitor through g of flesh. What is the temperature increase of the flesh? Is it reasonable to assume no phase change? Exercise 2.24 Unreasonable Results (a) A certain parallel plate capacitor has plates of area 4.00 m 2, separated by mm of nylon, and stores C of charge. What is the applied voltage? (b) What is unreasonable about this result? (c) Which assumptions are responsible or inconsistent? (a) 14.2 kv (b) The voltage is unreasonably large, more than 100 times the breakdown voltage of nylon. (c) The assumed charge is unreasonably large and cannot be stored in a capacitor of these dimensions. Exercise 2.25 What current flows through the bulb of a 3.00-V flashlight when its hot resistance is 3.60 Ω? A Exercise 2.26 Calculate the effective resistance of a pocket calculator that has a 1.35-V battery and through which ma flows. Exercise 2.27 What is the effective resistance of a car s starter motor when 150 A flows through it as the car battery applies 11.0 V to the motor? Ω Exercise 2.28 How many volts are supplied to operate an indicator light on a DVD player that has a resistance of 140 Ω, given that 25.0 ma passes through it? Exercise 2.29 (a) Find the voltage drop in an extension cord having a Ω resistance and through which 5.00 A is flowing. (b) A cheaper cord utilizes thinner wire and has a resistance of Ω. What is the voltage drop in it when 5.00 A flows? (c) Why is the voltage to whatever appliance is being used reduced by this amount? What is the effect on the appliance? (a) V

34 34 Chapter 2 Circuit Elements (b) 1.50 V (c) The voltage supplied to whatever appliance is being used is reduced because the total voltage drop from the wall to the final output of the appliance is fixed. Thus, if the voltage drop across the extension cord is large, the voltage drop across the appliance is significantly decreased, so the power output by the appliance can be significantly decreased, reducing the ability of the appliance to work properly. Exercise 2.30 A power transmission line is hung from metal towers with glass insulators having a resistance of current flows through the insulator if the voltage is 200 kv? (Some high-voltage lines are DC.) Ω. What Exercise 2.31 What is the resistance of a 20.0-m-long piece of 12-gauge copper wire having a mm diameter? Ω Exercise 2.32 The diameter of 0-gauge copper wire is mm. Find the resistance of a 1.00-km length of such wire used for power transmission. Exercise 2.33 If the mm diameter tungsten filament in a light bulb is to have a resistance of Ω at 20.0ºC, how long should it be? m Exercise 2.34 Find the ratio of the diameter of aluminum to copper wire, if they have the same resistance per unit length (as they might in household wiring). Exercise 2.35 What current flows through a 2.54-cm-diameter rod of pure silicon that is 20.0 cm long, when V is applied to it? (Such a rod may be used to make nuclear-particle detectors, for example.) A Exercise 2.36 (a) To what temperature must you raise a copper wire, originally at 20.0ºC, to double its resistance, neglecting any changes in dimensions? (b) Does this happen in household wiring under ordinary circumstances? Exercise 2.37 A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.00% from its value at 20.0ºC. Over what temperature range can it be used? 5ºC to 45ºC This OpenStax book is available for free at

35 Chapter 2 Circuit Elements 35 Exercise 2.38 Of what material is a resistor made if its resistance is 40.0% greater at 100ºC than at 20.0ºC? Exercise 2.39 An electronic device designed to operate at any temperature in the range from 10.0ºC to 55.0ºC contains pure carbon resistors. By what factor does their resistance increase over this range? 1.03 Exercise 2.40 (a) Of what material is a wire made, if it is 25.0 m long with a mm diameter and has a resistance of 77.7 Ω at 20.0ºC? (b) What is its resistance at 150ºC? Exercise 2.41 Assuming a constant temperature coefficient of resistivity, what is the maximum percent decrease in the resistance of a constantan wire starting at 20.0ºC? 0.06% Exercise 2.42 A wire is drawn through a die, stretching it to four times its original length. By what factor does its resistance increase? Exercise 2.43 A copper wire has a resistance of Ω at 20.0ºC, and an iron wire has a resistance of Ω at the same temperature. At what temperature are their resistances equal? 17ºC Exercise 2.44 (a) Digital medical thermometers determine temperature by measuring the resistance of a semiconductor device called a thermistor (which has α = / ºC ) when it is at the same temperature as the patient. What is a patient s temperature if the thermistor s resistance at that temperature is 82.0% of its value at 37.0ºC (normal body temperature)? (b) The negative value for α may not be maintained for very low temperatures. Discuss why and whether this is the case here. (Hint: Resistance can t become negative.) Exercise 2.45 Integrated Concepts (a) Redo Exercise 2.32 taking into account the thermal expansion of the tungsten filament. You may assume a thermal expansion coefficient of / ºC. (b) By what percentage does your answer differ from that in the example? (a) 4.7 Ω (total) (b) 3.0% decrease

36 36 Chapter 2 Circuit Elements Exercise 2.46 Unreasonable Results (a) To what temperature must you raise a resistor made of constantan to double its resistance, assuming a constant temperature coefficient of resistivity? (b) To cut it in half? (c) What is unreasonable about these results? (d) Which assumptions are unreasonable, or which premises are inconsistent? Glossary capacitance: amount of charge stored per unit volt capacitor: dielectric: a device that stores electric charge an insulating material dielectric strength: the maximum electric field above which an insulating material begins to break down and conduct ohm: the unit of resistance, given by 1Ω = 1 V/A ohmic: a type of a material for which Ohm's law is valid Ohm s law: an empirical relation stating that the current I is proportional to the potential difference V, V; it is often written as I = V/R, where R is the resistance parallel plate capacitor: two identical conducting plates separated by a distance polar molecule: a molecule with inherent separation of charge resistance: resistivity: by ρ the electric property that impedes current; for ohmic materials, it is the ratio of voltage to current, R = V/I an intrinsic property of a material, independent of its shape or size, directly proportional to the resistance, denoted simple circuit: a circuit with a single voltage source and a single resistor temperature coefficient of resistivity: an empirical quantity, denoted by α, which describes the change in resistance or resistivity of a material with temperature This OpenStax book is available for free at

37 Chapter 3 Circuits 37 3 CIRCUITS 3.1 Electric Power and Energy Power in Electric Circuits Power is associated by many people with electricity. Knowing that power is the rate of energy use or energy conversion, what is the expression for electric power? Power transmission lines might come to mind. We also think of lightbulbs in terms of their power ratings in watts. Let us compare a 25-W bulb with a 60-W bulb. (See Figure 3.1(a).) Since both operate on the same voltage, the 60-W bulb must draw more current to have a greater power rating. Thus the 60-W bulb s resistance must be lower than that of a 25-W bulb. If we increase voltage, we also increase power. For example, when a 25-W bulb that is designed to operate on 120 V is connected to 240 V, it briefly glows very brightly and then burns out. Precisely how are voltage, current, and resistance related to electric power? Figure 3.1 (a) Which of these lightbulbs, the 25-W bulb (upper left) or the 60-W bulb (upper right), has the higher resistance? Which draws more current? Which uses the most energy? Can you tell from the color that the 25-W filament is cooler? Is the brighter bulb a different color and if so why? (credits: Dickbauch, Wikimedia Commons; Greg Westfall, Flickr) (b) This compact fluorescent light (CFL) puts out the same intensity of light as the 60-W bulb, but at 1/4 to 1/10 the input power. (credit: dbgg1979, Flickr) Electric energy depends on both the voltage involved and the charge moved. This is expressed most simply as PE = qv, where q is the charge moved and V is the voltage (or more precisely, the potential difference the charge moves through). Power is the rate at which energy is moved, and so electric power is P = PE t = qv t. (3.1) Recognizing that current is I = q / t (note that Δt = t here), the expression for power becomes P = IV. (3.2) Electric power ( P ) is simply the product of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the joule, power has units of joules per second, or watts. Thus, 1 A V = 1 W. For example, cars often have one or more auxiliary power outlets with which you can charge a cell phone or other electronic devices. These outlets may be rated at 20 A, so that the circuit can deliver a maximum power P = IV = (20 A)(12 V) = 240 W. In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes ( 1 ka V = 1 kw ). To see the relationship of power to resistance, we combine Ohm s law with P = IV. Substituting I = V/R gives P = (V / R)V = V 2 /R. Similarly, substituting V = IR gives P = I(IR) = I 2 R. Three expressions for electric power are listed together here for convenience: P = IV (3.3)

38 38 Chapter 3 Circuits P = V 2 R P = I 2 R. (3.4) (3.5) Note that the first equation is always valid, whereas the other two can be used only for resistors. In a simple circuit, with one voltage source and a single resistor, the power supplied by the voltage source and that dissipated by the resistor are identical. (In more complicated circuits, P can be the power dissipated by a single device and not the total power in the circuit.) Different insights can be gained from the three different expressions for electric power. For example, P = V 2 / R implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in P = V 2 / R, the effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about 100 W, burning it out. If the bulb s resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too. Example 3.1 Calculating Power Dissipation and Current: Hot and Cold Power (a) Consider the examples given in Ohm s Law: Resistance and Simple Circuits ( m42344/latest/) and Resistance and Resistivity ( Then find the power dissipated by the car headlight in these examples, both when it is hot and when it is cold. (b) What current does it draw when cold? Strategy for (a) For the hot headlight, we know voltage and current, so we can use P = IV to find the power. For the cold headlight, we know the voltage and resistance, so we can use P = V 2 / R to find the power. for (a) Entering the known values of current and voltage for the hot headlight, we obtain P = IV = (2.50 A)(12.0 V) = 30.0 W. (3.6) The cold resistance was Ω, and so the power it uses when first switched on is P = V 2 R = (12.0 V) Ω = 411 W. (3.7) Discussion for (a) The 30 W dissipated by the hot headlight is typical. But the 411 W when cold is surprisingly higher. The initial power quickly decreases as the bulb s temperature increases and its resistance increases. Strategy and for (b) The current when the bulb is cold can be found several different ways. We rearrange one of the power equations, P = I 2 R, and enter known values, obtaining I = P R = 411 W Ω = 34.3 A. (3.8) Discussion for (b) The cold current is remarkably higher than the steady-state value of 2.50 A, but the current will quickly decline to that value as the bulb s temperature increases. Most fuses and circuit breakers (used to limit the current in a circuit) are designed to tolerate very high currents briefly as a device comes on. In some cases, such as with electric motors, the current remains high for several seconds, necessitating special slow blow fuses. The Cost of Electricity The more electric appliances you use and the longer they are left on, the higher your electric bill. This familiar fact is based on the relationship between energy and power. You pay for the energy used. Since P = E / t, we see that E = Pt (3.9) is the energy used by a device using power P for a time interval t. For example, the more lightbulbs burning, the greater P used; the longer they are on, the greater t is. The energy unit on electric bills is the kilowatt-hour ( kw h ), consistent with the relationship E = Pt. It is easy to estimate the cost of operating electric appliances if you have some idea of their power consumption rate in watts or kilowatts, the time they are on in hours, and the cost per kilowatt-hour for your electric utility. Kilowatt-hours, like all other specialized energy units such as food calories, can be converted to joules. You can prove to yourself This OpenStax book is available for free at

39 Chapter 3 Circuits 39 that 1 kw h = J. The electrical energy ( E ) used can be reduced either by reducing the time of use or by reducing the power consumption of that appliance or fixture. This will not only reduce the cost, but it will also result in a reduced impact on the environment. Improvements to lighting are some of the fastest ways to reduce the electrical energy used in a home or business. About 20% of a home s use of energy goes to lighting, while the number for commercial establishments is closer to 40%. Fluorescent lights are about four times more efficient than incandescent lights this is true for both the long tubes and the compact fluorescent lights (CFL). (See Figure 3.1(b).) Thus, a 60-W incandescent bulb can be replaced by a 15-W CFL, which has the same brightness and color. CFLs have a bent tube inside a globe or a spiral-shaped tube, all connected to a standard screw-in base that fits standard incandescent light sockets. (Original problems with color, flicker, shape, and high initial investment for CFLs have been addressed in recent years.) The heat transfer from these CFLs is less, and they last up to 10 times longer. The significance of an investment in such bulbs is addressed in the next example. New white LED lights (which are clusters of small LED bulbs) are even more efficient (twice that of CFLs) and last 5 times longer than CFLs. However, their cost is still high. Making Connections: Energy, Power, and Time The relationship E = Pt is one that you will find useful in many different contexts. The energy your body uses in exercise is related to the power level and duration of your activity, for example. The amount of heating by a power source is related to the power level and time it is applied. Even the radiation dose of an X-ray image is related to the power and time of exposure. Example 3.2 Calculating the Cost Effectiveness of Compact Fluorescent Lights (CFL) If the cost of electricity in your area is 12 cents per kwh, what is the total cost (capital plus operation) of using a 60-W incandescent bulb for 1000 hours (the lifetime of that bulb) if the bulb cost 25 cents? (b) If we replace this bulb with a compact fluorescent light that provides the same light output, but at one-quarter the wattage, and which costs $1.50 but lasts 10 times longer (10,000 hours), what will that total cost be? Strategy To find the operating cost, we first find the energy used in kilowatt-hours and then multiply by the cost per kilowatt-hour. for (a) The energy used in kilowatt-hours is found by entering the power and time into the expression for energy: E = Pt = (60 W)(1000 h) = 60,000 W h. (3.10) In kilowatt-hours, this is Now the electricity cost is E = 60.0 kw h. cost = (60.0 kw h)($0.12/kw h) = $7.20. (3.11) (3.12) The total cost will be $7.20 for 1000 hours (about one-half year at 5 hours per day). for (b) Since the CFL uses only 15 W and not 60 W, the electricity cost will be $7.20/4 = $1.80. The CFL will last 10 times longer than the incandescent, so that the investment cost will be 1/10 of the bulb cost for that time period of use, or 0.1($1.50) = $0.15. Therefore, the total cost will be $1.95 for 1000 hours. Discussion Therefore, it is much cheaper to use the CFLs, even though the initial investment is higher. The increased cost of labor that a business must include for replacing the incandescent bulbs more often has not been figured in here. Making Connections: Take-Home Experiment Electrical Energy Use Inventory 1) Make a list of the power ratings on a range of appliances in your home or room. Explain why something like a toaster has a higher rating than a digital clock. Estimate the energy consumed by these appliances in an average day (by estimating their time of use). Some appliances might only state the operating current. If the household voltage is 120 V, then use P = IV. 2) Check out the total wattage used in the rest rooms of your school s floor or building. (You might need to assume the long fluorescent lights in use are rated at 32 W.) Suppose that the building was closed all weekend and that these lights were left on from 6 p.m. Friday until 8 a.m. Monday. What would this oversight cost? How about for an entire year of weekends? Section Summary Electric power P is the rate (in watts) that energy is supplied by a source or dissipated by a device.

40 40 Chapter 3 Circuits Three expressions for electrical power are and P = IV, P = V 2 R, P = I 2 R. The energy used by a device with a power P over a time t is E = Pt. (3.13) (3.14) (3.15) 3.2 Kirchhoff's Principles This section is adapted from The University of Maryland BERG group ( Kirchhoff%27s%20principles%20%282013%29). The basic ideas that we have developed about how electric charges move in matter serve as a basis for analyzing a wide variety of electric circuits and devices and for modeling the electrical behavior of biological systems. But these circuits, devices, and models can quickly become quite complex. It becomes useful to establish a set of foothold ideas -- principles that we can hold on to and refer back to in order to organize our thinking in a complex situations -- to provide a "stake in the ground" that we can trust and use to support our safety net of coherent and linked ideas. The foothold principles for understanding electric currents were developed by the 19th century German physicist, Gustav Kirchhoff (yes, two "h"s) are called Kirchhoffs laws (or principles). (He also formulated laws of spectroscopy and thermochemistry.) The (idealized) context for Kirchhoff's principles Kirchhoff's principles are restrictions of more general electromagnetic laws (Maxwell's equations, conservation of charge) to standard situations in electrical circuits. We'll talk about them and use them in the context of analyzing connected networks of electrical devices -- batteries, resistors, capacitors, and wires. Here's how we will represent and idealize them: This OpenStax book is available for free at

41 Chapter 3 Circuits 41 Table 3.1 Batteries -- devices that maintain a constant electrical pressure difference (voltage) across their terminals (like a water pump that raises water to a certain height). We use the symbol shown at the right with the longer line corresponding to the end of the battery with the higher potential. Resistors -- devices that have significant drag and oppose current. Pressure will drop across them when current is flowing through them. The resistor shown in the figure is of the kind used in hand-built electrical circuits. The stripes color-code the size of the resistor and its precision. We indicate them with the zig-zag symbol shown on the right.

42 42 Chapter 3 Circuits Capacitors -- devices that can maintain a charge separation in response to a pressure differential (voltage) applied across its plates. The symbol for a capacitor is the pair of parallel lines shown on the right. Be sure to distinguish between the symbol for a capacitor (parallel lines of equal length) and a battery (parallel lines of different length)! Wires -- conductors that connect the other devices. We treat them as perfect conductors -- having zero resistance. This approximation only works when there are other resistors in the circuit having significant resistance, compared to which the resistance of the wires are negligible. The symbol for a wire is just a line and it could be straight or curved -- makes no difference. Kirchhoff's 1st (Flow) Principle The first principle is basically a combination of two ideas: conservation of charge (the total amount of positive charge minus the total amount of negative charge is a constant) in electrical circuits, due to the strong repulsive forces between like charges, electrical elements remain neutral -- there is no build-up of charge anywhere. The principle is often called "the flow rule" and is stated as follows: The total amount of current flowing into any volume in an electrical network equals the amount flowing out. From our analysis of how a capacitor and a resistor both work, we know that this idea doesn't hold when things are just getting started. For example, when we charge a capacitor, charge is flowing into one side of the capacitor and out of the other: charge (of opposite sign) is building up on each plate of the capacitor in violation of the flow rule. But if we put a box around the capacitor and don't look inside, the rule works. It also works when the system is in the steady state and things have stabilized. A similar thing holds for a resistor. When a current just starts to build up through the resistor, a build-up of equal and opposite This OpenStax book is available for free at

43 Chapter 3 Circuits 43 charges at the two ends of the resistor are what is responsible for establishing the electric field in the resistor (creating a potential drop across the resistor) that keeps the charges moving through the drag of the resistor at a constant velocity, consistent with Newton's laws of motion. In both these cases, when we hook up these devices to a circuit, the first principle violations that take place in the interior of the device happen fast -- in nanoseconds or less. And if we consider the whole device instead of just a part of it, the principle still works even on that time scale. Kirchhoff's 2nd (Resistance) principle The second principle tells what happens when there is a current in a resistor -- there is a potential drop in the direction of the current which is proportional to the current times a property of the resistor. This is just Ohm's law and it hold for any device in which the drag resisting the flow is proportional to the velocity. (See Resistive electric flow: Ohm's law.) We can even stretch its validity by letting the resistance be a function of the current. Mostly, we wont need to do this. ΔV = IR (3.16) Note that we have to be a bit careful here. This is ONLY true for a resistor -- not necessarily for a cluster of connected resistors. If we want to deal with a cluster of resistors, and put a box around it pretending it is a single resistor, we have to figure out what the effective resistance of the combination is using the combined principles. For capacitors we have the analogous result: Kirchhoff's 3rd (Loop) principle Q = CΔV Where Kirchhoff's first principle controls the current in an electrical network, the second deals with the voltage drops in the network. We can understand it by using the water analogy. The electric potential is analogous in the water model to the height that the water has been raised. One of the things we know about heights is that if you make a loop and come back to the same point, you will be at the same height from which you started. Whatever drops (descents) you made had to be cancelled by and equal sum of rises (climbs) in order to get back to your starting point. The same thing is true of electric potential (voltage). As we travel through a circuit, we may have rises, say if we go through a battery from its low end to its high end, and we may have drops, say if we go through a resistance in the direction of the current flow. Kirchhoff's third principle states: Following around any loop in an electrical network the potential has to come back to the same value (sum of drops = sum of rises). (3.17) This can be a bit tricky to apply! Just as if you go up a hill you are rising, but if you walk down that same hill you are descending, whether you have a rise or a drop in electric potential as you go through a device depends on which way you are following your loop If you go through a battery from the positive end to the negative end, it gives you a drop! If you go through a resistor in the direction opposite the direction a current is flowing you get a rise! Useful heuristics Applying Kirchhoff's principles to a complex circuit is sometimes complicated. There are two variables to be solved for -- the voltage (electrical pressure) and the current. These are independent variables. They affect each other, but your intuitions as to what is happening sometimes refers to one, sometimes to the other -- but it's easy to get confused! A useful way to think of the voltage throughout the circuit is as analogous to pressure (in the air flow model) or height (in the water flow model). Moving throughout the circuit there are different values of this variable -- the voltage (electric pressure) -- but it doesn't move or change. it is the difference between voltages (say at opposite ends of a resistor) that drives current through a resistor. One of the best ways to start analyzing an electrical network is by figuring out what you know about the voltage. And here's a corollary to Ohm's law that helps a great deal: A conductor in a circuit that can be treated as having 0 resistance, e.g., a wire, is an equipotential (has the same value of the potential everywhere along it) even if there is current flowing through it, since for that wire, the voltage drop across the wire is given by Ohm's law: ΔV=IR=0, even if I is NOT zero. The best advice in handling the current in an electrical circuit problem is to choose some directions for the directions you think

44 44 Chapter 3 Circuits the currents are flowing in and take those as positive. Then just apply Kirchhoff's principles to generate relationships (equations) among the various variables. If you have chosen wrong the signs will come out negative. No problem! It just tells you that your initial assumption was wrong and that the current is flowing in the opposite direction from the one you expected. Conceptual Questions Exercise 3.1 Why do incandescent lightbulbs grow dim late in their lives, particularly just before their filaments break? Exercise 3.2 The power dissipated in a resistor is given by P = V 2 / R, which means power decreases if resistance increases. Yet this power is also given by P = I 2 R, which means power increases if resistance increases. Explain why there is no contradiction here. title Exercise 3.3 What is the power of a MV lightning bolt having a current of A? W Exercise 3.4 What power is supplied to the starter motor of a large truck that draws 250 A of current from a 24.0-V battery hookup? Exercise 3.5 A charge of 4.00 C of charge passes through a pocket calculator s solar cells in 4.00 h. What is the power output, given the calculator s voltage output is 3.00 V? (See Figure 3.2.) Figure 3.2 The strip of solar cells just above the keys of this calculator convert light to electricity to supply its energy needs. (credit: Evan-Amos, Wikimedia Commons) Exercise 3.6 How many watts does a flashlight that has C pass through it in h use if its voltage is 3.00 V? This OpenStax book is available for free at

45 Chapter 3 Circuits 45 Exercise 3.7 Find the power dissipated in each of these extension cords: (a) an extension cord having a Ω resistance and through which 5.00 A is flowing; (b) a cheaper cord utilizing thinner wire and with a resistance of Ω. (a) 1.50 W (b) 7.50 W Exercise 3.8 Verify that the units of a volt-ampere are watts, as implied by the equation P = IV. Exercise 3.9 Show that the units 1 V 2 / Ω = 1W, as implied by the equation P = V 2 / R. V 2 Ω = V/A V2 = AV = C s C J = s J = 1 W Exercise 3.10 Show that the units 1 A 2 Ω = 1 W, as implied by the equation P = I 2 R. Exercise 3.11 Verify the energy unit equivalence that 1 kw h = J. 1 kw h= J 1 s (1 h) h s = J Exercise 3.12 Electrons in an X-ray tube are accelerated through kv and directed toward a target to produce X-rays. Calculate the power of the electron beam in this tube if it has a current of 15.0 ma. Exercise 3.13 An electric water heater consumes 5.00 kw for 2.00 h per day. What is the cost of running it for one year if electricity costs 12.0 cents/kw h? See Figure 3.3. Figure 3.3 On-demand electric hot water heater. Heat is supplied to water only when needed. (credit: aviddavid, Flickr)

46 46 Chapter 3 Circuits $438/y Exercise 3.14 With a 1200-W toaster, how much electrical energy is needed to make a slice of toast (cooking time = 1 minute)? At 9.0 cents/kw h, how much does this cost? Exercise 3.15 What would be the maximum cost of a CFL such that the total cost (investment plus operating) would be the same for both CFL and incandescent 60-W bulbs? Assume the cost of the incandescent bulb is 25 cents and that electricity costs 10 cents/kwh. Calculate the cost for 1000 hours, as in the cost effectiveness of CFL example. $6.25 Exercise 3.16 Some makes of older cars have 6.00-V electrical systems. (a) What is the hot resistance of a 30.0-W headlight in such a car? (b) What current flows through it? Exercise 3.17 Alkaline batteries have the advantage of putting out constant voltage until very nearly the end of their life. How long will an alkaline battery rated at 1.00 A h and 1.58 V keep a 1.00-W flashlight bulb burning? 1.58 h Exercise 3.18 A cauterizer, used to stop bleeding in surgery, puts out 2.00 ma at 15.0 kv. (a) What is its power output? (b) What is the resistance of the path? Exercise 3.19 The average television is said to be on 6 hours per day. Estimate the yearly cost of electricity to operate 100 million TVs, assuming their power consumption averages 150 W and the cost of electricity averages 12.0 cents/kw h. $3.94 billion/year Exercise 3.20 An old lightbulb draws only 50.0 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what factor is its diameter reduced, assuming uniform thinning along its length? Neglect any effects caused by temperature differences. Exercise gauge copper wire has a diameter of mm. Calculate the power loss in a kilometer of such wire when it carries A W Exercise 3.22 Integrated Concepts This OpenStax book is available for free at

47 Chapter 3 Circuits 47 Cold vaporizers pass a current through water, evaporating it with only a small increase in temperature. One such home device is rated at 3.50 A and utilizes 120 V AC with 95.0% efficiency. (a) What is the vaporization rate in grams per minute? (b) How much water must you put into the vaporizer for 8.00 h of overnight operation? (See Figure 3.4.) Figure 3.4 This cold vaporizer passes current directly through water, vaporizing it directly with relatively little temperature increase. Exercise 3.23 Integrated Concepts (a) What energy is dissipated by a lightning bolt having a 20,000-A current, a voltage of MV, and a length of 1.00 ms? (b) What mass of tree sap could be raised from 18.0ºC to its boiling point and then evaporated by this energy, assuming sap has the same thermal characteristics as water? (a) J (b) 769 kg Exercise 3.24 Integrated Concepts What current must be produced by a 12.0-V battery-operated bottle warmer in order to heat 75.0 g of glass, 250 g of baby formula, and g of aluminum from 20.0ºC to 90.0ºC in 5.00 min? Exercise 3.25 Integrated Concepts How much time is needed for a surgical cauterizer to raise the temperature of 1.00 g of tissue from 37.0ºC to 100ºC and then boil away g of water, if it puts out 2.00 ma at 15.0 kv? Ignore heat transfer to the surroundings s Exercise 3.26 Integrated Concepts Hydroelectric generators (see Figure 3.5) at Hoover Dam produce a maximum current of A at 250 kv. (a) What is the power output? (b) The water that powers the generators enters and leaves the system at low speed (thus its kinetic energy does not change) but loses 160 m in altitude. How many cubic meters per second are needed, assuming 85.0% efficiency?

48 48 Chapter 3 Circuits Figure 3.5 Hydroelectric generators at the Hoover dam. (credit: Jon Sullivan) Exercise 3.27 Integrated Concepts (a) Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 750-kg electric car be able to supply: (a) To accelerate from rest to 25.0 m/s in 1.00 min? (b) To climb a m -high hill in 2.00 min at a constant 25.0-m/s speed while exerting friction? (c) To travel at a constant 25.0-m/s speed, exerting a See Figure N of force to overcome air resistance and N force to overcome air resistance and friction? Figure 3.6 This REVAi, an electric car, gets recharged on a street in London. (credit: Frank Hebbert) (a) 343 A (b) A (c) A Exercise 3.28 Integrated Concepts A light-rail commuter train draws 630 A of 650-V DC electricity when accelerating. (a) What is its power consumption rate in kilowatts? (b) How long does it take to reach 20.0 m/s starting from rest if its loaded mass is kg, assuming 95.0% efficiency and constant power? (c) Find its average acceleration. (d) Discuss how the acceleration you found for the light-rail train compares to what might be typical for an automobile. Exercise 3.29 Integrated Concepts This OpenStax book is available for free at