buku physics11

EXERCISESConceptual Questions 9. Using an electric field map, how can you tell the difference between regions where there is a1. Using atomic theory, explain why electrons weak electric field and a strong electric field? move from one substance to another, such as when acetate is rubbed with silk. Problems2. A new solid material developed for the 15.2 The Basis of Electric Charge International Space Station is being tested for its electrostatic properties. In a series of steps, 10. Which parts of the atom are represented by describe how you would test this material to positive signs and by negative signs? determine its rightful place in an updated electrostatic series. 11. What is the charge on each of the following? a) A normal atom of oxygen3. A computer technician is always warned to b) An electron touch the metal body of a computer before c) A nucleus touching any electronic parts. What does d) A neutron touching the computer body do to prevent e) A proton static electric damage to the circuits? 12. Several items are listed below along with the4. We often “ground” our electric appliances by circumstances that could have left them with attaching a wire from the appliance to the a charge. For each example, state which item ground so that excess charge can be deposited would have a positive or a negative charge there. How does this safety measure work? because of a deficit or excess of electrons. Why do we not have to worry about Earth a) A piece of rubber rubbed with silk picking up a static electric charge? b) The silk from part a) c) An acetate sheet rubbed with cat’s fur5. Why are we more susceptible to building up d) Glass rubbed with wool electrostatic charges when the weather is very dry? What time of year would static shocks be 13. A piece of amber is rubbed with fur. most frequent? a) What type of charge is on the amber? b) What kinds of particles are transferred to6. When you refuel your boat at a marina, the the amber? attendant usually touches the fuel delivery nozzle to the metal edge of the boat’s fuel 14. A glass rod is suspended from a string and inlet before adding any gasoline. Why? charged by rubbing it with a piece of silk. a) After rubbing, what type of charge is on7. If an electric field is set up by charging a each material? single negative point charge, how would you b) What should happen if the silk is brought describe the field shape around this charge? close to the glass rod? Sketch this field shape. 15. At a birthday party, a balloon is rubbed on8. If you were to double the charge value on a someone’s hair and brought close to a ribbon test object that is used to map out an electric on a party decoration. The ribbon was field, what would happen to the strength of repelled by the balloon. What type of charge the electric field at each test point? is on the ribbon? chapter 15: Electrostatics 539

16. State whether each of the following is an 23. With a deficit of 4.0 ϫ 1011 electrons, what is electric conductor or an insulator and give the charge on an electroscope? a reason why. a) Plastic food wrap 24. A glass rod with a charge of 5.4 ϫ 108 elec- b) A lightning rod trons touches another insulator so that all of c) Your plastic comb the excess electrons are shared equally. What d) A party balloon stuck to a wall is the final charge on the glass rod? e) A car’s tire if you are caught in a lightning storm 25. An atom is known to have a nucleus with a f) The rubber belt on a Van de Graaff electro- positive charge of 2.4 ϫ 10Ϫ12 C. How many static generator electrons does this atom have?17. The CN Tower in Toronto is a prime location 26. Draw two circles in your notebook, about for lightning strikes. Why? What did the engi- 5 cm apart, and label them both with (Ϫ) neers have to do to protect people and prop- negative signs. Use the concept of placing test erty from lightning damage? charges on the page to map what the electric field would look like. 15.3 The Creation and Transfer of Charge 27. How would the field map change if the charge on the left was tripled?18. Summarize the steps involved in charging a balloon positively by induction. 28. Draw two parallel lines in your notes to rep- resent two parallel metal plates, one positive 15.4 Measurement of Charge and the other negative, about 2 cm apart. Map the electric field between the plates.19. What is the function of an electroscope? 15.6 Applications of Electrostatics20. A metal leaf electroscope is touched by a posi- and Charge Transfer tively charged strip. a) What type of charge will register on the 29. Assume that you are asked to spraypaint a electroscope? What causes this process? metal fence. How can static electricity help b) What will happen to the leaf (or leaves) of you avoid wasting paint? the electroscope? Why? c) What will happen to the leaves of the elec- 30. The ease with which we can photocopy has troscope if the overall system is grounded? environmental consequences. Discuss some of the ways that the extensive use of photo-21. A wire passes a charge of 15.0 C. How many copiers has affected our lives. You might con- electrons pass through the wire? sider land fill, recycling programs, and indoor air quality.22. Small charges, such as those passed between people when a static electric shock is mistak- enly given, are measured in small units called ␮C (microcoulombs). A shock of 1.1 ␮C was passed from one student to another in a dry physics classroom. How many electrons does this amount represent?540 u n i t e : E l e c t r i c i ty a n d M a g n et i s m

15.1 The Law of Electric Charges LABORATORY EXERCISESPurpose troscope in each case (attraction or repulsion). Also predict the type of charge on the testTo examine the attraction and repulsion of elec- material in each case.trostatic charges. 3. Set up an electroscope by taking a polyethyl- ene strip and rubbing it with wool. Then placeSafety Consideration it on a watch glass, as shown in Fig. Lab.15.1. 4. Rub another strip of polyethylene with woolNote: Use two small pieces of putty or masking and use it as a “test material.” Bring it close totape to secure any ebonite or glass rods to the one end of the electroscope and observe whatwatch glass so they do not roll off and break. happens. The electroscope should rotate one way or another on the desk, showing a forceEquipment of attraction or repulsion. 5. Rub the polyethylene with wool again, thenTwo small watch glasses, Two polyethylene strips use the wool as a test material.Two acetate strips, Two ebonite rods 6. Rub an acetate strip with cotton (or wool) andTwo glass rods, Woolen cloth bring it close to the polyethylene electroscope.Cotton cloth, Silk cloth, Fur 7. Create another electroscope from an acetate strip rubbed with cotton.Fig.Lab.15.1 Charged 8. Repeat the experiment using the test materi- als in the order listed in the table. Watch glass polyethylene rod 9. Repeat the entire experiment twice more, using ebonite, glass, fur, and silk as shown in Polyethylene the table. stripProcedure1. Prepare a data chart similar to Table Lab.15.1.2. Use the electrostatic series table in the text- book to predict what will happen to the elec- Table Lab.15.1 Testing the Electrostatic SeriesElectroscope material Charged test material Observation Charge on test material (watch glass) repulsion/attraction (ϩ) or (Ϫ)Polyethylene (wool) Prediction Actual Prediction ActualPolyethylene (wool) observation observationPolyethylene (wool)Acetate (cotton/wool) Polyethylene (wool)Acetate (cotton/wool) Wool (polyethylene)Acetate (cotton/wool) Acetate (cotton/wool)Ebonite (fur) Acetate (cotton/wool)Ebonite (fur) Cotton/wool (acetate)Ebonite (fur) Polyethylene (wool)Glass (silk) Ebonite (fur)Glass (silk) Fur (ebonite)Glass (silk) Glass (silk) Glass (silk) Silk (glass) Ebonite (fur) chapter 15: Electrostatics 541

LABORATORY EXERCISES Discussion Conclusion 1. What effect do like charges have on one The electrostatic series is based on Benjamin another? Franklin declaring that amber became negative when rubbed with fur. In a brief paragraph, 2. What effect do unlike charges have on one describe the implications on this experiment if another? Franklin had set the convention by declaring that glass becomes negative when rubbed with silk. 3. a) State which objects became positively charged and the circumstances of the change. b) State which objects became negatively charged and the circumstances of the change. c) Did any objects remain neutral? 4. What problems occurred when your analysis involved an attraction of the electroscope?542 u n i t e : E l e c t r i c i ty a n d M a g n et i s m

Current Electricity 16and Electric Circuits Chapter Outline For image see student 16.1 Introduction 16.2 Current text. 16.3 Electrical Potential 16.4 Supplying Electrical Potential Energy 16.5 Resistance—Ohm’s Law 16.6 Series and Parallel Circuits 16.7 The Circuit Analysis Game 16.8 Power in Electric Circuits 16.9 The Cost of Electricity ST S E Electric Current and the Human Body 16.1 Circuit Analysis—Electrical Theory in PracticeBy the end of this chapter, you will be able to• explain how current, voltage, and resistance deliver electrical power to a load• summarize the ways in which electrical energy is supplied from various sources• use Kirchhoff’s laws to quantitatively predict how current and 543 voltage are distributed in series and parallel circuits

16.1 Introduction Electrons in a static state have energy, but they are far more useful when they are made to transfer their energy. Electric current involves electrons repelling one another and passing through a conductor. Energized electrons, directed by a conductor, allow energy to be used to power many devices.Fig.16.1 A water-electricity circuit analogy 16.2 Current In Fig. 16.1, the pumping station provides the water with gravitational potential energy, pumping it up into the water tower. The water loses this energy as it is piped through the sprinkler and back to its point of origin. In an electric circuit, an energy source provides elec- trons with energy. Conductors transport the electrons to where the electron energy is transferred, then back to the source to be re-energized. Figure 16.2 illustrates that a conductor carries charge in a way similar to the way in which pipes carry water. The flow of charge is called electric current. In this fig- ure, electrons flow through a conductor.Fig.16.2 Electron flow in a conductor Current is the rate of charge flow and is given the symbol I. Current is the total amount of charge movingis like water flow in a pipe past a point in a conductor divided by the time taken. I ϭ ᎏQᎏ tProton Electron where I is the current in amperes (A), Q is the charge in coulombs (C), and t is the time in seconds. The base unit for current is C/s, which is given the derived name of ampere (A), after André Marie Ampère (1775–1836). One ampere is one coulomb of charge moving through a point in a conductor every second. As with any equation, it can be used in various ways to calculate certain cir- cuit parameters. e x a m p l e 1 Current and charge How much current flows through a hair dryer if 1400 C of charge pass through it in 2.25 minutes? Solution and Connection to Theory Given t ϭ 2.25 min Iϭ? Q ϭ 1400 C544 u n i t e : E l e c t r i c i ty a n d M a g n et i s m

΂ ΃t ϭ 2.25 min ᎏ16m0ᎏisn ϭ 135 sI ϭ ᎏQᎏ ϭ ᎏ140ᎏ0 C ϭ 10.4 ᎏCᎏt 135 s sTherefore, the current through the hair dryer is 10.4 A.e x a m p l e 2 Charge and currentYour night light uses a 7 W light that draws about 6.0 ϫ 10Ϫ2 A of cur-rent. How much charge passes through this bulb in 8.0 hours?Solution and Connection to TheoryGiven t ϭ 8.0 hours Qϭ?I ϭ 6.0 ϫ 10Ϫ2 A ϭ 6.0 ϫ 10Ϫ2 ᎏCᎏ s΂ ΃΂ ΃T ϭ 8.0 h ᎏ60 mᎏin ᎏ60ᎏs ϭ 28 800 s 1 h 1 minTo convert hours to seconds, you can also multiply by 3600 s/h.Q ϭ It ϭ 6.0 ϫ 10Ϫ2 ᎏCᎏ (28 800 s) ϭ 1.7 ϫ 103 C sTherefore, the charge through the light bulb is 1.7 ϫ 103 C.e x a m p l e 3 Electron flowHow many electrons have passed through the night light in Example 2? EXPONENT MATHThe charge on one electron is 1.602 ϫ 10Ϫ19 C. Taking the multiplier and powerSolution and Connection to Theory parts separately, you get ᎏ1.ᎏ7 ϭ 1.1Given 1.602Q ϭ 1.7 ϫ 103 C e ϭ 1.602 ϫ 10Ϫ19 C ᎏ1100Ϫ319 ϭ 103Ϫ(Ϫ19) ϭ 1022Q ϭ NeN ϭ ᎏQeᎏ ϭ ᎏ1.610.72 ϫϫᎏ11003ϪC19 C ϭ 1.1 ϫ 1022 electronsDirection of Current Flow Benjamin Franklin thought that the positive terminal had an excess ofIt is now well understood that current is a flow of negatively charged electrons electricity and the negative terminalrepelling one another. Historically, current flow was thought to move from the had a deficit of electrons. From thispositive (ϩ) terminal to the negative (Ϫ) terminal of any power supply. The perspective, it only made sense thatmodel of positive charge flow is called conventional current. For the remain- current flowed from a positive to ader of this chapter, we will consider current as the flow of electrons. negative region. In Chapters 17 and 18, we will use the concept of posi- tive (ϩ) charge flow. chapter 16: Current Electricity and Electric Circuits 545

To keep track of the direction of electron current flow, a coloured wiring convention is used in which black represents the negative terminal and red represents the positive terminal of any power supply. Therefore, we will think of current as flowing from the black negative (Ϫ) terminal to the red positive (ϩ) terminal. Measurement of Current The measurement of current, as we will discover in our labs, can be quite tricky. Like a turnstile that counts the number of passengers entering the subway, our device to measure current must be an integral part of the circuit. An ammeter (a current-measuring device) must be wired so that all current flows through it. In all cases, the conductor must be rewired so that the cur- rent runs serially through the ammeter. The ammeter must be an excellent conductor so that no energy is lost due to its addition to the circuit. In DC or direct current, the current flows in a single direction from the power supply through the conductor to a load, such as a light bulb or other device that uses energy, and back to the power supply, as shown in Fig. 16.3. In alternating current (AC), the electrons periodically reverse the direction of their flow. This reversal is carried out with the help of elec- tric and magnetic forces, which we will discuss in Chapters 17 and 18. For electric current to flow, it must have a complete path from the neg- ative side of the power supply to the positive side. This path of current is called a circuit and is required for any electrical device to work properly. The latest digital ammeters are very forgiving if you forget about the cor- rect red–black wiring convention or push too much current through them. Table 16.1 lists some of the problems and the safe use of these ammeters.Fig.16.3 A typical series circuit (a) (a)and schematic (b) BatterySafety tip. When you are discon- (b)necting wires in the lab, be careful!High current causes sparks. A ؊ ؉546 u n i t e : E l e c t r i c i ty a n d M a g n et i s m

Table 16.1 Ammeter SafetyProblem Analog meter Digital meter Safe solutionToo much current passes Wires, other internal An error or a null display. Check setting of meter beforethrough the meter for its mechanisms may burn out. There may be no damage. A powering up the circuit.current setting. The fuse may burn out. fuse may burn out. Always choose a higher current setting than required. Fig.16.4A Fig.16.4B For image For image see student see student text. text.Meter leads are connected in Analog needle is \"buried\" below A negative sign is displayed in Follow coloured wiringreverse. conventions. Electrons out zero on the meter. Internal front of the current value. of (Ϫ) and into (ϩ). damage may result.Drawing CircuitsTo help us understand circuits, we need a way to illustrate them in a quickand simple fashion. Figure 16.5 is a legend of many circuit symbols andwhat they represent. We will begin to draw circuits using the items in thistable as a standard.Source of electrical potential Fig.16.5 Circuit symbols؊؉ Cell Battery ؉؊ DC ؉ DC generator ؊ Cells in series AC generator ؊؊؊ Cells in parallel ؉؉؉Electrical loads Resistor (fixed) Resistor (variable) or rheostat Lamp CoilM Motor TransformerElectric meters A Ammeter V Voltmeter G Galvanometer Fuse Switch (open) Ground Switch (closed) chapter 16: Current Electricity and Electric Circuits 547

gpplyin 1. How is current different from a static charge build-up? What causes Co electron flow? If you have already studied energy (Chapters 7 and a the 8), use principles discussed there to explain why charges move. tsncep 2. Calculate the current produced for a) a light bulb with 1.0 ϫ 105 C of charge passing through the fila- ment for 2.5 hours. b) an electroscope with a surplus of 2.2 ϫ 1010 electrons that dis- charges completely in 0.6 s through a person’s finger. 3. a) How long would it take to pass 700 C of charge through a toaster drawing 10 A of current? b) How many electrons pass through the toaster during this time? 4. What is the difference between conventional and electron current flow?Fig.16.6 Gravitational and 16.3 Electrical Potentialelectrical potential energy Charge does not flow on its own. We saw previously that excess of any one charge causes a force of repulsion. A complete circuit also allows the excessenergy charge to “see” a region of charge deficit at the power supply.of Decrease In the same way that the bicycle in Fig. 16.6 possesses gravitational potential energy at different heights in a gravitational field (recall force at aIncrease of distance), an electric charge has a certain amount of electrical potential supply energy because of the electric field set up by the power supply. energyPower Load Work had to be done on the bicycle to increase its gravitational poten- tial energy. Similarly for the charge in the circuit. Work is done by the power supply to increase the electrical potential energy of each coulomb of charge from a low to a high value. As the charge flows through the load, its energy decreases.In some textbooks, the equation is The electrical potential energy for each coulomb of charge in a circuit isV ϭ ᎏWᎏ, where W is the work. called the electric potential difference (V), Q V ϭ ᎏQEᎏ where E is the energy required to increase the electric potential of a charge, Q. Potential difference is often called voltage. The unit for electric potential difference is the volt, named after Count Alessandro Volta (1745–1827). One volt (V) is the electric potential difference between two points if one joule of work (J) is required to move one coulomb (C) of charge between the points.548 u n i t e : E l e c t r i c i ty a n d M a g n et i s m

e x a m p l e 4 Potential difference and energyWhat is the potential difference across an air conditioner if 72 C of chargetransfer 8.5 ϫ 103 J of energy to the fan and compressor?Solution and Connection to TheoryGivenQ ϭ 72 C E ϭ 8.5 ϫ 103 J Vϭ?V ϭ ᎏQEᎏV ϭ ᎏ8.5 ϫᎏ103 J 72 CV ϭ 1.2 ϫ 102 VTherefore, the potential difference or voltage in the air conditioner is1.2 ϫ 102 V.e x a m p l e 5 Energy and potential differenceA static electric shock delivered to a student from a “friend” transfers1.5 ϫ 101 J of electrical energy through a potential difference of 500 V.What is the quantity of charge transferred in the spark?Solution and Connection to TheoryGivenE ϭ 1.5 ϫ 101 J V ϭ 500 V Qϭ?V ϭ ᎏEᎏ І Q ϭ ᎏEᎏ QVQ ϭ ᎏ1.5 ϫᎏ101 J 500 VQ ϭ 0.03 CTherefore, the charge transfer between the “friends” is 0.03 C. chapter 16: Current Electricity and Electric Circuits 549

Fig.16.7 Delivering electricalenergyRecall that A practical analogy may help you to understand how electric current,V ϭ ᎏEᎏ and I ϭ ᎏQᎏ energy, and potential are related. In the product delivery system in Fig. 16.7, each cell in the loading line adds a package of energy to a coulomb “truck.” Qt Just as trucks deliver products to a store over a road, so a circuit deliversso E ϭ VQ and Q ϭ It energy to a load. In electricity, the carrier of the energy is a coulomb ofTherefore, E ϭ VQ charge, which delivers its energy cargo in joules per coulomb. A store requires a certain amount of goods to be delivered. The amount delivered ϭ VIt depends on the size of the trucks’ loads and their speed. So, too, for an elec- tric circuit. The energy delivered to the load depends on the potential550 (energy per charge) and the rate at which the charge is delivered (the cur- rent). The energy transferred by charge flow is E ϭ VIt where E is the energy in joules, V is the potential difference in volts, I is the current in amperes, and t is the time in seconds. e x a m p l e 6 Electrical energy One 1.5 V (AA) battery runs a portable MP3 player that draws 5.7 ϫ 10Ϫ3 A of current for about six hours before it runs out. How much energy does the battery transfer? Solution and Connection to Theory Given V ϭ 1.5 V I ϭ 5.7 ϫ 10Ϫ3 A tϭ6h Eϭ? ΂ ΃t ϭ 6 h ᎏ3610ᎏh0 s ϭ 21 600 s E ϭ VIt ϭ 1.5 V(5.7 ϫ 10Ϫ3 A)(21 600 s) ϭ 185 J Therefore, the battery transfers 185 J of energy. unit e: Electricity and Magnetism

e x a m p l e 7 Potential difference and energyA coffee maker draws about 5.0 A of current for 270 s using 1.6 ϫ 105 Jof energy. What is the potential difference across the coffee maker?Solution and Connection to TheoryGiven t ϭ 270 s E ϭ 1.6 ϫ 105 J Vϭ?I ϭ 5.0 AE ϭ VItRearrange the equation for V.V ϭ ᎏEᎏ V ϭ ᎏ1.6 ϫᎏ105 J V ϭ 119 V It 5.0 A(270 s)The potential difference across the coffee maker is 119 V.Measuring Potential DifferencePotential difference between any two points can be measured using avoltmeter. As illustrated in Fig. 16.8, a voltmeter must be connected inparallel with a load in the circuit in order to compare the potential beforeand after the load. The voltmeter must have a large resistance—that is, it must be a muchpoorer conductor than the load to which it is connected, so that the meas-urement by the voltmeter will divert a minimal current from the circuit. Fig.16.8 The connection of a voltmeter in a circuit(a) A V ؊ ؉(b) chapter 16: Current Electricity and Electric Circuits 551

gpplyin 1. Explain the difference between current and voltage.Co 2. You go to a store to buy a 12 V car battery. All the batteries are 12 V,a the tsncep but they differ in cranking amps. What should you look for in a winter battery? 3. a) A 12 V car battery delivers 1.3 ϫ 104 J of energy to the starter motor. How much charge does it deliver? b) How many electrons does the battery transfer? c) Given that you keep the key turned for 2.5 s (time to turn the motor over in order to start the car), how many amps are deliv- ered to the starter motor? 4. Lightning transfers charge between a charged cloud and the ground. If the voltage difference between the two is 1.3 ϫ 108 V and 3.2 ϫ 109 J of energy is transferred, find a) the total charge moved between the two potential energy surfaces. b) the number of electrons making up the charge. c) the current delivered if the lightning stroke takes 25 microsec- onds to hit the ground. Career Alert! Meteorologists (people who study weather) require a large amount of physics training. The study of storms and their formations requires knowledge about electricity, sound (acoustics to describe thunder pro- duction), energy (transfer of energy and convection systems), forces (describing the effects of lightning), and pressure. 16.4 Supplying Electrical Potential Energy Electrical energy always originates from some other form of energy. Electrical energy is a kind of medium of exchange for obtaining energy from easily produced forms to the forms we need. There are many different devices for converting energy from some other form to electrical potential energy. The original form may be chemical, mechanical, thermal, or light energy. Table 16.2 lists some of the common sources of electrical energy.552 u n i t e : E l e c t r i c i ty a n d M a g n et i s m

Table 16.2 Sources of Electrical EnergySource process Common applicationsVoltaic cells Chemical potential energy is released during a reaction as electrons are driven between two different metals. E.g., common dry cell batteries for portable electric devicesPiezo-electricity Quartz and Rochelle salt crystals create small electric potential when mechanical force or stress is applied to them. E.g., phonograph cartridges, barbecue spark startersThermoelectricity Two pieces of different metals joined together and subjected to temperature differentials transfer thermal energy to electric potential energy in what is called a thermocouple. E.g., gas appliance pilot light safety system: current keeps gas valve open as long as flame is litPhoto electricity Light energy absorbed by electrons of certain metals causes charge flow. E.g., satellite and international space station power supply, calculator power supplyElectromagnetic Kinetic energy of water or steam forces conductors to rotate in ainduction in generators magnetic field.(see Chapter 18) E.g., hydro-, nuclear-, fossil-fuel-powered electric generators16.5 Resistance—Ohm’s LawThe amount of current flow in a circuit, and therefore the Fig.16.9 The pathway makes a differenceamount of energy transferred to any useful device,depends on two things: (1) the potential difference of the Ϫ Ϫ Ϫpower supply (the amount of push) and (2) the nature of Ϫ Ϫthe pathway through the loads that are using the electric Ϫpotential energy. We have already discussed potential Loadenergy in detail, so we must now turn our attention to the Ϫelectrical characteristics of the pathway through the load. Ϫ ϪϪ Ϫ Compare the two simple circuits in Fig. 16.9. The (a) Ϫ Ϫ Ϫ Ϫonly difference between the two in the electrical sense Ϫis that the broader path in (a) allows the current to passthrough the load much more easily than the narrower Ϫ Ϫ Ϫpath in (b). Ϫ Ϫ Ϫ The push on the charge (potential difference) is the Load Ϫsame, but the pathways are very different. The overallresult is that the more difficult the path, the more oppo- Ϫ ϪϪsition there is to flow. The measure of this oppositionto flow is called electrical resistance. To define resist- (b) Ϫ ϪϪance more quantitatively, we must experiment with thecircuit by measuring the quantities of potential differ-ence across the load and the current passing through it.This way, we can indirectly develop a quantitativemeasure of resistance. chapter 16: Current Electricity and Electric Circuits 553

Fig.16.10 1.5 Votlage V (V) 1.0(a) A 0.5 V 0 0.1 0.2 0.3 0.4 0.5 0.6 ؊(b) ؉ (c) Current I (A)I (A) Table 16.3 ᎏVᎏ (⍀) The circuit illustrated in Fig. 16.10(a) and schematized in 16.10(b) I allows us to experiment with the electrical potential across a load. The idea0 V (V) is to see how much current flows over a load when the potential difference0.12 – is varied. Figure 16.10(c) shows that the graph of voltage vs. current is a0.25 0 2.6 sthtreaisglhopt elinanedanthdethᎏVIᎏersaltoiopemoufstthreepgrraepsehn, tththeeᎏVIᎏrersaitsitoa,nicsecoofnsthtaenlto. aTdhbeerceafoursee,0.40 0.31 2.6 the resistance remained unchanged in the experiment.0.57 0.64 2.6 1.04 2.6 R ϭ ᎏVIᎏ, 1.47 where R is the resistance in volts/ampere, which is given the derived unit of ohm (⍀), after Georg Simon Ohm (1787–1854).⍀ is the capital Greek letter Omega. V is the potential difference in volts (V) and I is the resulting current in amperes (A). In general, Ohm found that the ᎏVIᎏ ratio was constant for a particular resistor. This ratio is called Ohm’s law. The steps in the process of develop- ing Ohm’s law (or any other linear equation) are shown in Fig. 16.11. Fig.16.11 Generating the Linear Equation for Ohm’s Law from Data m etho Measure data Plot data s of I (A) V (V) do ces V (V) Find slope pr of straight line, which is Create equation a constant Rϭ V I (A) I554 u n i t e : E l e c t r i c i ty a n d M a g n et i s m

The amount of current flowing through a resistor varies directly asthe amount of potential difference applied across the resistor as long asother variables, such as temperature, are controlled. There is a resistanceof 1 ⍀ when 1 A of current flows with a potential difference of 1 V acrossa resistor.e x a m p l e 8 Calculate currentAn electric stove element is connected to a 240 V supply and has aknown resistance of 19.8 ⍀. What current will this element draw?Solution and Connection to TheoryGiven R ϭ 19.8 ⍀ Iϭ?V ϭ 240 V І I ϭ ᎏRVᎏV ϭ IR I ϭ ᎏ1294.08ᎏV⍀I ϭ 12.1 ATherefore, the current that the element will draw is 12.1 A.e x a m p l e 9 Calculate resistanceWhat is the resistance of a 1200 W hair dryer that draws about 10 A froma 120 V circuit?Solution and Connection to TheoryGiven V ϭ 120 V Rϭ?I ϭ 10 A І R ϭ ᎏVIᎏV ϭ IRR ϭ ᎏVIᎏ ϭ ᎏ120ᎏV 10 A ϭ 12 ⍀Therefore, the hair dryer has a resistance of about 12 ⍀. chapter 16: Current Electricity and Electric Circuits 555

e x a m p l e 1 0 Electrical cautionAn electric belt sander that was designed for a 120 V circuit to drawabout 11.0 A was accidentally plugged into a 240 V line. The sander oper-ated for about five seconds until it burned out. (a) What is the resistanceof the sander? (b) What current did the sander draw from the 240 V line?Solution and Connection to Theorya) GivenV1 ϭ 120 V I1 ϭ 11.0 A R1 ϭ ?R1 ϭ ᎏVᎏ1 I1 ϭ ᎏ1112.00ᎏAV ϭ 10.9 ⍀The sander has a resistance of 10.9 ⍀.b) GivenV2 ϭ 240 V R1 ϭ 10.9 ⍀ I2 ϭ ?I2 ϭ ᎏVᎏ2 R1 ϭ ᎏ1204.09ᎏV⍀ ϭ 22.0 AA current of 22.0 A is twice the capacity of the wiring in the sander.Factors that Determine ResistanceYou have seen that a thinner wire has a larger resistance than a thicker one.Other properties of conductors also affect their resistance. The resistance ofa conductor depends on its length, cross-sectional area, the material it ismade of, and its temperature. Table 16.4 describes how these four factorsaffect resistance. The formulas show the ratio of the resistances of two con-ductors with different values for each factor.556 u n i t e : E l e c t r i c i ty a n d M a g n et i s m

Table 16.4 Factors that Affect ResistanceFactor Description ProportionalityLength The longer the conductor, the greater If the length is doubled, the resistance. then the resistance is ᎏRᎏ1 .ᎏLᎏ1 doubled R2 ϭ L2Cross-sectional area The larger the cross-sectional area or If the cross-sectional area thickness of the conductor, the less resistant it has to charge flow. is doubled, the resistance goes to half of its original ᎏRᎏ1 .ᎏAᎏ2 value R2 ϭ A1Type of material Some materials are better conductors If the resistivity (␳) isTemperature than others. The general measure of the resistance of a substance is called the doubled, then the resistivity. Resistivity has units ⍀·m. resistance is also doubled. Since moving charge is impeded by molecules, greater molecular motion at ϭᎏRᎏ1 ᎏ␳␳ᎏ21 higher temperatures tends to increase the resistance. R2 An increase in temperature of the conductor usually contributes to an increase in the resistance, but not for all substances.e x a m p l e 1 1 Resistance of aluminum wireA 200 m piece of aluminum wire has a resistance of 1.7 ⍀. What is theresistance of a 50 m portion cut from it?Solution and Connection to TheoryGivenL1 ϭ 200 m R1 ϭ 1.7 ⍀ L2 ϭ 50 m R2 ϭ ?ᎏRᎏ1 ϭ ᎏLᎏ1R2 L2΂ ΃R2 ϭ R1 ᎏLᎏ2 L1 ᎏ50ᎏm 200 m΂ ΃R2 ϭ 1.7 ⍀ ϭ 0.42 ⍀Therefore, the resistance of the 50 m piece of wire is 0.42 ⍀.e x a m p l e 1 2 Resistance and cross-section How would the cross-sectional area of a 50 m piece of aluminum have to be changed to give it the same resistance as the 200 m piece in Example 11? chapter 16: Current Electricity and Electric Circuits 557

Solution and Connection to Theory Given R1 ϭ 0.42 ⍀ A1 ϭ A1 R2 ϭ 1.7 ⍀ A2 ϭ ? ᎏRᎏ1 ϭ ᎏAᎏ2 R2 A1 ΂ ΃A2ϭA1ᎏRᎏ1 R2 ΂ ΃A2 ϭ A1 ᎏ01.4.72ᎏ⍀⍀ ϭ 0.25A1 Since area ϭ ␲r2, the 50 m wire would have to be about half the radius (͙ෆ0.25 ϭ 0.5) in order to make its resistance the same as that of the 200 m wire. g pplyin 1. Use the values V ϭ 120 V, I ϭ 10.6 A, and R ϭ 11.3 ⍀ to create threeCo the problems, using each variable as the unknown.a ncep ts 2. Calculate the new resistance of a copper wire if558 a) the wire is cut from a length of 200 m (R ϭ 1.1 ⍀) to a length of 35 m. b) if the cross-section of the wire is changed by a factor of 0.24. 3. By connecting an extension cord to an electrical load, such as a heater, you increase the total resistance of the system. Suggest why the resist- ance in the extension cord should be small, and possible problems that could arise from using one. Should the wire have a large or small cross- section in order to keep the resistance down to a minimum? The gauge number of a wire indicates its cross-sectional area. A wire that has a small gauge number has a large cross-sectional area. Similarly, a small cross-section has a large gauge number. Superconductivity Part 1: Transmission Lines Superconductivity is the ability of a material to conduct electricity without heat loss due to electrical resistance. The first superconductors became superconducting only at low temperatures (near 0 K or Ϫ270°C). In recent years, high-temperature superconducting materials, known as HTSs, have exhibited these properties at temperatures as high as 140 K (Ϫ133°C). This property saves on cooling costs to keep the material at the required temperature. The promise of superconductors is transmission lines that carry elec- tricity without energy loss. This ability would reduce the cost of produc- ing electricity as well as the space occupied by supply lines (Fig. 16.12). Currently, there is a global race to produce a wire material that can carry currents of 100–1000 A (called the critical current, Ic) at the highest pos- sible temperature without losing its superconductive properties. unit e: Electricity and Magnetism

Fig.16.12 Existing overhead load Source Superconducting cable4. Search the Web to find the highest temperature reached by a super- conducting material to date. Does it currently have any practical uses?5. Research the possible improvements in computer technology as a result of superconductivity. 16.6 Series and Parallel Circuits Fig.16.13 Simple series and parallel circuitsNow that we know how to relate current, potential difference, and ؊resistance in simple circuits, we can examine the different ways in ؉which those circuits can be combined. The two simplest ways toconnect conductors and loads are illustrated in Fig. 16.13. In a ؊series circuit, the loads are connected one after another in a sin- ؉gle path, whereas in a parallel circuit, they are side by side. Each arrangement affects the way in which potential differenceand current act in the various parts of the circuit. Gustav RobertKirchhoff (1824–1887) studied the way each of the circuit parame-ters behaved in series and parallel circuits. His research led to thepublication of the laws of both current and voltage, calledKirchhoff’s laws. Kirchhoff’s current law The total amount of current into a junction point of a circuit equals the total current that flows out of that same junction. In Fig. 16.14, three branches meet at one junction point and two Fig.16.14 An illustration ofbranches leave another junction point so that I1 ϩ I2 ϩ I3 ϭ IT ϭ I4 ϩ I5. Kirchhoff’s current law Kirchhoff’s voltage law The total of all electrical potential decreases in any complete circuit loop is equal to any potential increases in that circuit loop. l1 lT l4 l2 Junction points l5 l3 chapter 16: Current Electricity and Electric Circuits 559

Fig.16.15 An illustration of Kirchhoff’s voltage law In Fig. 16.15, the potential increase, VT is equivalent to the sum of all the potential losses so that VT ϭ V1 ϩ V2 ϩ V3. V1 Kirchhoff’s laws are particular applications of the laws of ؊ V2 conservation of electric charge and the conservation of energy. In other words, in any circuit, there is no net gain orVT loss of electric charge or energy. ؉ V3 e x a m p l e 1 3 Kirchhoff’s laws in a series circuit Figure 16.16 shows a simple series circuit. Use Kirchhoff’s voltage and current laws to find the values of the missing voltage (V2) and current (I3). Fig.16.16 A simple series circuit 10.0 A 30 V R2 10.0 A V2 ؊ 10.0 A R1 100 V R3 ؉ l3 30 V Solution and Connection to Theory Voltage According to Kirchhoff’s voltage law, this series circuit has one voltage increase of 100 V. This voltage must be distributed so that the sum of all voltage drops for each individual series resistor must equal this value. Therefore, VT ϭ V1 ϩ V2 ϩ V3 so V2 ϭ VT Ϫ V1 Ϫ V3 ϭ 100 V Ϫ 30 V Ϫ 30 V ϭ 40 V Current According to Kirchhoff’s current law, this series circuit has no real junction point, so it has only one path of flow. Therefore, IT ϭ I1 ϭ I2 ϭ I3 ϭ 10 A.560 u n i t e : E l e c t r i c i ty a n d M a g n et i s m

e x a m p l e 1 4 Kirchhoff’s laws in a parallel circuitFigure 16.17 shows a simple parallel circuit. Use Kirchhoff’s voltage andcurrent laws to find the values of the missing voltage (V2) and current (I3).Solution and Connection to Theory Fig.16.17 A simple parallel circuitVoltage According to Kirchhoff’s voltage law, this 9 A Junction pointsparallel circuit has one voltage increase of 30 V.This increase must be met by the same decrease for ؊ 30 V 3 A R1 R2 3 A V2 R3 l3 30 Veach of the three different parallel resistor paths.Therefore, VT ϭ V1 ϭ V2 ϭ V3, so the voltage drop 30 Vacross any of the three parallel resistors is 30 V, nomatter what their resistances are. ؉ Junction pointsCurrent According to Kirchhoff’s current law, this parallel circuit hasfour junction points (circled in Fig. 16.17), one at the top and bottom ofeach branch, to resistors 1 and 2. The sum of the current entering thesejunctions must equal the sum of the current exiting them.Therefore, IT ϭ I1 ϩ I2 ϩ I3 ϭ 9A,so I3 ϭ IT Ϫ I1 Ϫ I2 ϭ 9 A Ϫ 3A Ϫ 3A ϭ 3A In Examples 13 and 14, we had a simple case of a series or a parallel cir-cuit with three resistors. In practice, however, circuits are more complicatedcombinations of both series and parallel elements. We can determine theoverall resistance of series or parallel circuit elements by applying bothOhm’s law for resistance and both of Kirchhoff’s laws.Resistances in SeriesReferring to Fig. 16.16 again, we find that all current must first passthrough resistor 1, then 2, and so on. The voltage drops across each resistor.The sum of the voltage drops gives the overall voltage drop in the circuit. From Kirchhoff’s law, VT ϭ V1 ϩ V2 ϩ V3 From Ohm’s law, ITRT ϭ I1R1 ϩ I2R2 ϩ I3R3 But from Kirchhoff’s law, IT ϭ I1 ϭ I2 ϭ I3 ϭ I The currents factor out; IRT ϭ IR1 ϩ IR2 ϩ IR3 therefore, RT ϭ R1 ϩ R2 ϩ R3chapter 16: Current Electricity and Electric Circuits 561

For the more general case where you have more than three resistors, the equation can be generalized as RT ϭ R1 ϩ R2 ϩ R3... ϩ RN where N is the total number of series resistors in the circuit.If the values of all the resistors in a e x a m p l e 1 5 Resistors in seriesseries circuit are the same, the overallresistance can be determined by What is the series equivalent resistance of 10 ⍀, 20 ⍀, and 30 ⍀ resistors connected in series? RT ϭ NR Solution and Connection to Theorywhere N is the total number of resis- RT ϭ R1 ϩ R2 ϩ R3tors and R is the resistance of each Therefore, RT ϭ 10 ⍀ ϩ 20 ⍀ ϩ 30 ⍀ ϭ 60 ⍀individual resistor. Resistances in Parallel Referring to Fig. 16.17 again, we find that the total current must split and distribute itself among all of the available circuit paths. From Kirchhoff’s law, IT ϭ I1 ϩ I2 ϩ I3 From Ohm’s law, ᎏVᎏT ϭ ᎏVᎏ1 ϩ ᎏVᎏ2 ϩ ᎏVᎏ3 RT R1 R2 R3 But from Kirchhoff’s law, VT ϭ V1 ϭ V2 ϭ V3 ϭ V The voltages factor out; ᎏVᎏ ϭ ᎏVᎏ ϩ ᎏVᎏ ϩ ᎏVᎏ therefore, RT R1 R2 R3 ᎏ1ᎏ ϭ ᎏ1ᎏ ϩ ᎏ1ᎏ ϩ ᎏ1ᎏ RT R1 R2 R3If the values of all the resistors in a For the more general case where you have more than three resistors, the equation can be generalized toparallel circuit are the same, the over-all resistance can be determined by ᎏ1ᎏ ϭ ᎏNᎏ ᎏ1ᎏ ϭ ᎏ1ᎏ ϩ ᎏ1ᎏ ϩ ᎏ1ᎏ ϩ ᎏ1ᎏ RT R RT R1 R2 R иии RN 3 ᎏRᎏІ RT ϭ N where N is the total number of parallel resistors in the circuit.where N is the total number of resis-tors and R is the resistance of eachindividual resistor.562 u n i t e : E l e c t r i c i ty a n d M a g n et i s m

e x a m p l e 1 6 Resistors in parallelWhat is the parallel equivalent resistance for a 25 ⍀, 40 ⍀, and a 10 ⍀ Example 16 can be done much fasterresistor, wired in parallel? ᎏ1ᎏ with the X key on your calculator. Simply key each parallel resistor value ᎏ1ᎏSolution and Connection to Theory followed by the X key as you add them up. After you press the equal ᎏ1ᎏ ϭ ᎏ1ᎏ ϩ ᎏ1ᎏ ϩ ᎏ1ᎏ (ϭ) sign for the sum, don’t forget to RT R1 R2 R3 press the ᎏX1ᎏ key one more time for the final answer. ᎏ1ᎏ ϭ ᎏ1ᎏ ϩ ᎏ1ᎏ ϩ ᎏ1ᎏ RT 25 ⍀ 40 ⍀ 10 ⍀ ᎏ1ᎏ ϭ ᎏ(40 ⍀)(1ᎏ0 ⍀)(ϩ25ᎏ(2⍀5)(⍀40)(ᎏ1⍀0)(⍀10) ϩ⍀ᎏ)(25 ⍀)ᎏ(40 ⍀) ϭ ᎏ11060500ᎏ0⍀⍀23 RTІ RT ϭ ᎏ10 00ᎏ0 ⍀ ϭ 6.1 ⍀ 1650Note how the units cancel correctly.1. Calculate the total resistance for the following: gpplyin a) Three resistors, each 20 ⍀, connected in series Co b) Three resistors, each 20 ⍀, connected in parallel a the c) Three resistors, each 20 ⍀, connected in parallel, which are then tsncep connected to three resistors, each 20 ⍀, connected in series.2. For resistors of 10 ⍀, 15 ⍀, and 20 ⍀, two of the resistors are con- nected in parallel and one resistor is connected in series. Calculate the possible total resistance values.3. a) What happens to a string of light bulbs connected in series if one light bulb goes out? b) What happens to a string of light bulbs connected in parallel if one light bulb goes out?4. A 1.0 ⍀ resistor is hooked up to a 1.0 ϫ 106 ⍀ resistor in a) series b) parallel. For each situation, calculate the total resistance and explain the dominance of one resistor in the total value.Three-way Light BulbsThree-way light bulbs, which have three different light intensity set-tings, have two filaments of different resistance connected in parallel.Each filament can be turned on separately, or they can both be turnedon at the same time, thus producing three different light intensities. chapter 16: Current Electricity and Electric Circuits 563

16.7 The Circuit Analysis Game The distribution of current and electric potential energy loss in a circuit is determined by the way the circuit is designed and the overall resistance of cer- tain circuit branches. The study of how the circuit parameters of current and voltage are distributed in a circuit is called circuit analysis. What follows is a simple set of rules to a game, similar to a crossword, but even more fun. The game board consists of a circuit (Fig. 16.18) with numerous unknown parameters.Fig.16.18 A sample circuit V1for analysis lT l1564 ؊ R1 ϭ 5 ⍀ V2 l2 R2 ϭ 10 ⍀ l3 V3 RT ϭ ? R3 ϭ 15 ⍀ 120 V ؉ R4 ϭ 20 ⍀ l4 V4 The object of the game is to find all the missing parameters in the cir- cuit, such as the following: Total current, resistance, and voltage (IT, RT, VT) The voltage across each resistor (V1, V2, V3, V4) The current through each resistor (I1, I2, I3, I4) The resistance of each resistor (R1, R2, R3, R4) The score card for the game is provided in Table 16.5. We will fill it in as we go. The initial given values are marked in bold. Table 16.5 Score Card for Circuit Analysis Game Circuit Item V (V) I (A) R (⍀) Resistor (R1) 619.4 53.87 1(a)5 Resistor (R2) 723.2 82.32 1(a)10 Resistor (R3) 723.2 81.55 1(a)15 Resistor (R4) 677.4 53.87 1(a)20 Total Resistance (RT) 1(a)120 43.87 331 Note: The superscripted numbers represent the order in which the calculations are to be performed. unit e: Electricity and Magnetism

The winner of the game is the first person to complete the score cardwith written proof of all entries.Rules and Strategies of the Circuit Analysis Game Fig.16.19 A series-equivalent1. a) Fill in all known quantities. circuit b) Fill in any “buried” known quantities. These quantities are the ones you can assume at a glance from Kirchhoff’s laws, like the fact that V1 any resistor in line with IT has the same value. l12. Reduce any parallel element to a series equivalent. R1ᎏ1ᎏ ϭ ᎏ1ᎏ ϩ ᎏ1ᎏ ؊ R2-3 l2-3 V2-3R2Ϫ3 R2 R3 R4 VT l4ᎏ1ᎏ ϭ ᎏ101ᎏ⍀ ϩ ᎏ151ᎏ⍀ ϭ ᎏ12550ᎏ⍀⍀2R2Ϫ3 ؉ V4 R2Ϫ3 ϭ ᎏ15205ᎏ⍀ ϭ 6 ⍀ IP-16-20T3. Draw the circuit as a series-equivalent circuit (Fig. 16.19).Calculate the total resistance of the circuit by adding all resistors in theseries equivalent (if all individual resistances are given). RT ϭ R1 ϩ R2Ϫ3 ϩ R4 RT ϭ 5 ⍀ ϩ 6 ⍀ ϩ 20 ⍀ ϭ 31 ⍀4. Use V ϭ IR to find any missing variable if you are given any two out of three parameters in a row. We now have a VT and an RT.IT ϭ ᎏVᎏT ϭ ᎏ13210ᎏ⍀V ϭ 3.87 A RT5. According to Kirchhoff’s current law, all series resistors have the same current,І I1 ϭ I4 ϭ I2Ϫ3 ϭ IT ϭ 3.87 A6. If you have two out of three knowns in a row on the score card,V1 ϭ I1(R1) ϭ 3.87 A(5 ⍀) ϭ 19.4 VV4 ϭ I4(R4) ϭ 3.87 A(20 ⍀) ϭ 77.4 V7. Voltages (Kirchhoff’s voltage law for series and parallel circuits) Here, V2 ϭ V3 by Kirchhoff’s voltage law. The sum of all individual volt- ages must equal the total voltage.І V2 ϭ V3 ϭ V2Ϫ3 ϭ VT – V1 Ϫ V4 ϭ 120 V Ϫ 19.4 V Ϫ 77.4 V ϭ 23.2 V chapter 16: Current Electricity and Electric Circuits 565

8. Two out of three knowns in a row. Check after every step. I2 ϭ ᎏVᎏ2 ϭ ᎏ2130.2ᎏ⍀V ϭ 2.32 A R2 I3 ϭ ᎏVᎏ3 ϭ ᎏ2135.2ᎏ⍀V ϭ 1.55 A R3 You just won the game! You can be sure by going back and double- checking the answers on the score card.Fig.16.20 RA Other strategies that you could use IT IT 1. Current splitting (with parallel elements): If the current that is entering the parallel resistors (usually the total RB current, IT) and the resistor values are known, try this method to see RB how the current is split between the two resistors. RA ϩ RB΂ ΃IA ϭ IT Given two resistors, R2 ϭ 10 ⍀ and R3 ϭ 15 ⍀, more of the total cur- rent, IT, would flow through the smaller resistor, R2. The current would be split between the two resistors proportionally, based on their resist- ance values, as shown below (see also Fig. 16.20). I2 ϭ IT (the resistance ratio) ᎏRᎏ3 ϭ 3.87 A ᎏ10 ⍀15ϩᎏ⍀15 ⍀ ϭ 3.87 A ᎏ12ᎏ55 R2 ϩ R3 ΂ ΃ ΂ ΃ ΂ ΃I2 ϭIT ϭ 2.32 ATips from the coach Don’t let the ΂ ΃The resistance ratio for any resistor is ϭ ᎏopposᎏing resᎏistorrounding “throw you off your game.” sum of both resistorsSomeone else may have slightly differ-ent answers, especially if he or she Of course, I3 would be easily found by subtracting from the total:used a different strategy than you did. I3 ϭ IT Ϫ I2 ϭ 3.87 A Ϫ 2.32 A ϭ 1.55 A 2. Series-equivalent resistors Find the voltage drop across any parallel element by using the series- equivalent resistance and the total current. V2Ϫ3 ϭ V2 ϭ V3 ϭ I2Ϫ3(R2Ϫ3) V2Ϫ3 ϭ V2 ϭ V3 ϭ 3.87 A(6 ⍀) V2Ϫ3 ϭ V2 ϭ V3 ϭ 23.2 VTips from the coach Remember Like any game, it is not who wins that is important, but rather how thethat using current-splitting or series- game is played. This attitude is also true for circuit analysis. The mostequivalent resistors will take you on important object is to play the game by trying out the individual strategiesa different route to win the game. to finish the problem. Like any game, you will need lots of practice. The practice for these problems appears at the end of the chapter.566 u n i t e : E l e c t r i c i ty a n d M a g n et i s m

Fig.16.21 Summary of Series and Parallel Circuits Simple series Current splitting guttin Add all series resistors RA it all in a branch To geth RA RB IT IT p er RA ϭ RA ϩ RB ϩ • • • ϩ RN RB RB RA ϩ RB ΂ ΃IA ϭ IT Current (I) remains the same throughout Combination RAB Series / parallel IT ϭ I1 ϭ I2 ϭ I3 ϭ • • • ϭ INOriginal RA RB Both RABC Series Voltage drops add to circuit strategies RAB equivalent total voltage increase RC circuit VT ϭ V1 ϩ V2 ϩ V3 ϩ • • • ϩ VN Simple parallel Resistances add to total RT ϭ R1 ϩ R2 ϩ R3 ϩ • • • ϩ RN Add all parallel resistors in a branch RA “2 out of 3” Ohm’s law over sub-branches RA RB VAB IAB ϭ VAB RAB I I I I RAB ϭ RA ϩ RB ϩ ••• RN RB1. A 5.0 V power supply is hooked up to resistors with values of 10 ⍀, gpplyin 15 ⍀, and 20 ⍀. Calculate the voltage across each resistor and the Co current through it for the following resistor combinations: a the a) All the resistors are in series. tsncep b) All the resistors are in parallel. c) Two of the resistors are connected in parallel with one con- nected in series. Do all the combinations.2. Explain Kirchhoff’s laws using your results from Question 1. 16.8 Power in Electric Circuits From Chapter 7, recall that P ϭ ᎏEᎏ tRecall that power is the rate at which work is done. In the case of electriccircuits, power is the rate at which electrical energy is passed on to various and from Section 16.3, E ϭ VItcircuit loads. For each particular load, a coulomb (1 C) of charge is the energy Then, P ϭ ᎏEᎏ ϭ ᎏVIᎏtcarrier, and the electric potential is the amount of energy (1 V ϭ 1 J/C) thatthe charge is carrying. Therefore, the amount of power dissipated in the load ttdepends on how fast the charge arrives at the load. P ϭ IV where I is the current in amperes (A), V is the potential difference in volts (V), and P is the power in watts (W). chapter 16: Current Electricity and Electric Circuits 567

There are two other formulas for power: P = IV, but V = IR P ϭ IV, but V ϭ IR. Therefore, I ϭ ᎏVᎏ І P = I(IR) R = I2R І P ϭ IV ϭ ᎏVᎏ(V) І P ϭ ᎏVᎏ2 R RThe unit for power is the watt, which We now have three tools to calculate power, depending on the givens inmust come out of the units for both your problem:current and voltage: P ϭ IV P ϭ IV P ϭ ᎏVᎏ2In unit format, RP ϭ A (current) ϫ V (voltage) P ϭ I2R ϭ ᎏCᎏ (current) ϫ ᎏJᎏ (voltage) e x a m p l e 1 7 Household power sC What is the maximum power that can be drawn from a standard 120 V ϭ ᎏJᎏ ϭ W (power) household circuit that has a circuit breaker (or fuse) of 15 A? s Solution and Connection to Theory Given I ϭ 15 A Pϭ? V ϭ 120 V P ϭ IV P ϭ (15 A)(120 V) P ϭ 1800 W Therefore, the maximum power on any 15 A fused/breakered circuit is 1800 W. e x a m p l e 1 8 Power to the amp Calculate the power rating of a stereo amplifier (not the speaker power) if it is plugged into a standard 120 V outlet and has a resistance of 120 ⍀. Solution and Connection to Theory Given R ϭ 120 ⍀ Pϭ? V ϭ 120 V P ϭ ᎏVᎏ2 R568 u n i t e : E l e c t r i c i ty a n d M a g n et i s m

P ϭ ᎏ(120ᎏV)2 120 ⍀P ϭ 120 WTherefore, the power rating of the stereo is 120 W.e x a m p l e 1 9 Power in the currentHow much power is dissipated in a circuit with a 15 ⍀ resistor that drawsa current of 10.0 A?Solution and Connection to TheoryGiven I ϭ 10.0 A Pϭ?R ϭ 15 ⍀P ϭ I2RP ϭ (10.0 A)2(15 ⍀)P ϭ 1500 WTherefore, the power dissipated in the circuit is 1500 W.1. Calculate the power produced by a 12 V battery charger if it delivers gpplyin a) 10 A in the fast-charge mode. b) 2 A in trickle-charge mode. Co the a ncep2. a) A portable heater, rated at 1 kW, is plugged into a 120 V outlet. ts How many amperes of current will it draw? 569 b) Will it burn out if it is plugged into a 220 V circuit?3. a) For Question 2, calculate the amount of charge delivered in 1.0 minutes. b) Calculate the number of electrons in that amount of charge. c) Find the energy delivered by the heater. d) What is the resistance of the heater? 16.9 The Cost of ElectricityEvery month, we buy electrical energy from an electric power company, suchas BC Hydro or Ontario Power Generation. The SI unit for energy is thejoule, but this unit is so small, that buying energy by the joule would notmake much sense. Imagine that your monthly electricity bill stated that youused 4 500 000 000 J at $ 0.000 000 022 per joule. The bill would be for $99,but you can see how silly this unit of energy is in this case. As a result, a chapter 16: Current Electricity and Electric Circuits