Series and Parallel Cirellits

Transcription

1 CHAPTER' 23 Series and Parallel Cirellits... ~ Which Light Lights Lighter? A 60-watt light bulb and a 100-watt light bulb are connected in series with a 120-volt source. If the circuit is closed, which bulb will glow brighter? Chapter Outline 23.1 SIMPLE CIRCUITS Series Circuits Voltage Drops in a Series Circuit Parallel Circuits 23.2 APPLICATIONS OF CIRCUITS Safety Devices Combined Series-Parallel Circuits Ammeters and Voltmeters The electric circuits introduced in Chapter 22 had one source of electric energy and one device that used energy. Often many devices must be connected to one source. In this chapter, you will explore the ways in which devices can be connected in electric circuits. The photograph on the opposite page shows two light bulbs connected in series. Do you think the results would be the same if the two bulbs were connected in parallel? If devices are connected in series, all the current travels through each device. A parallel connection allows the current to split and travel through several devices at once. A closer look will show how these connections work and how they can be used. {concept Check The following terms or concepts from earlier chapters are important for a good understanding of this chapter. If you are not familiar with them, you should review them before studying this chapter. potential difference, Chapter 21. electric circuits, resistance, Chapter

2 Objectives describe a series connection; state its important characteristics. calculate current, voltage drops, and equivalent resistance when devices are connected in series. describe a voltage divider and solve problems involving one. describe a parallel connection; state its important characteristics. calculate the voltage drop, currents, and equivalent resistance when devices are connected in parallel. In a series circuit there is only one current path. The current in a series circuit is the same in each element. The total potential drop in a series circuit is the sum of the individual drops SIMPLE CIRCUITS Have you ever driven along a single-lane road with no turn-offs? If there is an accident or if the road is being repaired, all traffic stops. There is only one path over which cars can move. This is like the path of electric charges in a series connection. In the case where there is a second road, however, cars can use either path and avoid road blocks. A parallel connection has the same characteristics as this second case. Series Circuits When resistors or other electrical devices are connected in a series circuit, all current travels through each resistor. It moves through one resistor after the other. Consider the flow of water in a river as a model of an electric current. At the river's source, the gravitational potential of the water is highest. Rapids are like resistors. A river having one rapid after another is like a series connection. All the water in the river passes through each of the rapids but loses some potential energy at each one. In Figure 23-1, current passesthrough the generator, the ammeter, and each lamp in succession. The current is the same in each device and in the generator. To find the current, we first find the potential drops around the circuit. If a charge is carried completely around the circuit, its potential returns to the original value. Thus the increase in potential across the generator is equal to the sum of the potential drops around the remainder of the circuit. The total distance the river drops is the sum of the vertical drops through each of the rapids. In the electrical circuit, there is almost no drop across the ammeter or the wires. The potential difference across the three lamps in series, V, is equal to the sum of the potential drops across the individual lamps, V = V 1 + V 2 + V3- son isn isn 2A R = R, + R2 + R3 = 30fl + 15fl + 15fl = 60fl FIGURE A series circuit can be represented both pictorially and schematically. The total resistance of a series circuit is equal to the sum of the individual resistances. 120V 470 Seriesand Parallel Circuits

3 The resistance of each lamp is defined as R = VII. Thus, the potential drop across the first lamp is V 1 = IR 1 Therefore, V = IR1 + IR2 + IR3, or V = I(R1 + R2 + R3)' The current is given by V R1 + R2 + R3 The same current would exist in a circuit with a single resistor, R, that has a resistance equal to the sum of the three lamp resistances. Such a resistance is called the equivalent resistance of the circuit. For resistors in series, the equivalent resistance is the sum of all the resistances, F. Y. I. Henry Cavendish used the direct approach to measure the strength of an electric current. Lacking the appropriate instruments, he instead shocked himself with the current and then estimated the pain. I R = R1 + R2 + R3 1 Notice that the equivalent resistance is larger than any single resistance. The current through a series circuit is most easily found by first calculating the equivalent resistance, R, and then using the equation V 1= R' An equivalent resistance is the value of a single resistor that could replace all resistors in a circuit and not change the current. The equivalent resistance in a series circuit is the sum of the individual resistances. Example Problem Current in a Series Circuit Four 15-0 resistors are connected in series with a 45-V battery, Figure What is the current in the circuit? R = R1 + R2 + R3 + R4 = = 60 0 Using the definition of resistance in the form I V 45 V I = R = 60 0 = 0.75 A VIR gives Practice Problems Assume batteries and generators provide the stated potential difference for any current that flows. 1. There are three 20-0 resistors connected in series across a 120-V generator. a. What is the effective resistance of the circuit? b. What is the current in the circuit? 2. A 10-0 resistor, a 15-0 resistor, and a 5-0 resistor are connected in series across a 90- V battery. a. What is the equivalent resistance of the circuit? b. What is the current in the circuit? =- 45V FIGURE Use with the Example Problem Simple Circuits 471

4 POCKET LAB SERIES RESISTANCE Hook up a power supply, a resistor, and an ammeter in a series circuit. Predict what will happen to the current in the circuit when a second, identical resistor is added in series to the circuit. Predict the new currents when the circuit contains three and four resistors in series. Explain your prediction. Try it. Make a data table to show your results. Write a sentence that explains your results. (Hint: Include the idea of resistance.) The total change in potential around a closed circuit is zero. A voltage divider is a simple series circuit. + v=- FIGURE Consider a 9-V battery in a circuit with three resistors connected in series. a. If the resistance of one of the devices increases, how will the series resistance change? b. What will happen to the current? c. Will there be any change in the battery voltage? ~ 4. Ten Christmas tree bulbs connected in series have equal resistances. When connected to a 120-V outlet, the current through the bulbs is 0.06 A. a. What is the equivalent resistance of the circuit? b. What is the resistance of each bulb? Voltage Drops in a Series Circuit The net change in potential going around a circuit must be zero. A battery or generator raises the potential. While passing through the resistors, the potential drops an amount equal to the increase. The net change is zero. The potential drop across each device in a series circuit can be calculated by rewriting the equation that defines resistance, R = VII, as V = IR. First find the equivalent resistance, R, in the circuit. Then use the equivalent resistance to find the current. Find I = VIR, where V is the potential drop. Once the current in the circuit has been determined, multiply I by the resistance of the individual device to find the potential drop across that device. An important application of series resistors is the voltage divider. Suppose you have a 9-V battery but need a 5-V potential source. A voltage divider can supply this voltage. Consider the circuit in Figure Two resistors, R1 and R 2, are connected in series across a battery of magnitude V. The equivalent resistance of the circuit is R = R1 + R 2 The current I is given by I = VIR = V/(R 1 + R 2 ). The desired voltage, 5 V, is the voltage drop, V 2, across resistor R 2 V 2 = IR2 Replacing I by the equation above gives V 2 = IR2 = ( V ). R2 R1 + R2 VR 2 R1 + R2 Voltage dividers are often used with sensors like photoresistors. The resistance of a photoresistor depends on the amount of Iight that strikes it. Photoresistors are made of semiconductors like selenium or cadmium sulfide. A typical photoresistor can have a resistance of 400 fl when light strikes it, but fl when in the dark. The output voltage of a voltage divider that uses a photoresistor depends on the amount 472 Seriesand Parallel Circuits

5 Jjl9ht CdS lnsulatinq ~ supporting ba~ ~ Electrode c Lead ~ Lead Dry -=. cells -=. ~' Sensitivity adjustment (potentiometer) Light CdS cell a b FIGURE A light meter (a) and schematic diagram (b). of light striking the photoresistor sensor. This circuit can be used as a light meter, Figure In this device, an electronic circuit detects the potential difference and converts it to a measurement of illuminance that can be read on the digital display. Example Problem Voltage Drops in a Series Circuit A 45.0-V potential difference is placed across a 5.0-D resistor and a 10.0-D resistor connected in series, Figure a. What is the equivalent resistance of the circuit? b. What is the current through the circuit? c. What is the voltage drop across each resistor? d. What is the total voltage drop across the circuit? Given: R, = 5.0 D Unknowns: R, I, V 1, V 2 R2 = 10.0 D Basic equation: V = IR V = 45.0 V Solution: a. R = R, + R2 = 5.0 D D = 15.0 D b.1 = VIR = (45.0 V)/(15.0 D) = 3.00 A c. The voltage drop across R, is V 1 = IRl = (3.00 A)(5.0 D) = 15 V. The voltage drop across R2 is V 2 = IR2 = (3.00 A)(1O.0 D) = 30.0 V. d. V= V 1 + V 2 = 15 V V = 45 V FIGURE Use with the Example Problem Simple Circuits 473

6 ExampleProblem Voltage Divider 9.0V'=:' FIGURE23-6. Use with the Example Problem. IiCALcuLAtoR Since multiplication is the inverse of division, you can use the inverse key, 1/x, to perform calculation without having to reenter numbers. (9 V) (500 il) 400 il il Keys [llijg I. I I. 0l1J0CIJwwG Answer nnn,uu -03 FIGURE23-7. A typical parallel circuit is shown using both its pictorial and schematic representations. The total resistance of a parallel circuit is smaller than any of its individual resistances. s A 9.0-V battery and two resistors, R, = and R2 = 500 0, are connected as a voltage divider, Figure What is the voltage across R2? Given: V = 9.0 V R, = 4000 R2 = 5000 I. VR 2 So ution: V 2 = ----'=-- R, + R2 Practice Problems Unknown: V 2 Basic equation: V 2 (9.0 V)(500 0) A resistor and a resistor are connected in series and placed across a 120-V potential difference. a. What is the equivalent resistance of the circuit? b. What is the current in the circuit? c. What is the voltage drop across each resistor? d. What is the voltage drop across the two resistors together? 6. Three resistors of 3.0 ko (3.0 X ), 5.0 ko, and 4.0 ko are connected in series across a 12-V battery. a. What is the equivalent resistance? b. What is the current through the resistors? c. What is the voltage drop across each resistor? d. Find the total voltage drop across the three resistors. 7. A student makes a voltage divider from a 45-V battery, a 475-kO (475 x ) resistor, and a 235-kO resistor. The output voltage is measured across the smaller resistor. What is the voltage? () 5V Generator 474 Seriesand Parallel Circuits

7 FIGURE In a parallel circuit, the reciprocal of the total resistance is equal to the sum of the reciprocals of the individual resistances. Here you see three 60-fi resistors connected in parallel. For each resistor, the voltage drop is 120 volts and the current is 2 amperes. ~ 8. A photoresistor is used in a voltage divider as R 2. V = 9.0 V and u, = 500 O. a. What is the output voltage, V 2, across R 2, when a bright light strikes the photoresistor and R2 = 475 O? b. When the light is dim, R2 = 4.0 ko. What is V2? c. When the photoresistor is in total darkness, R MO (0.40 x ). What is V 2? Parallel Circuits In a parallel circuit there are several current paths. A typical parallel circuit is shown in Figure The circuit contains a voltage source and three resistors. The three resistors are connected in parallel; their left-hand ends are connected together and their right-hand ends are connected together. The current flows through three paths in this parallel circuit. Rapids in a wide river can illustrate a parallel circuit. The water going through rapids often divides into several channels. Some channels may have a large flow of water, others a small flow. The sum of the flows, however, is equal to the total flow of water in the river. In addition, no matter which channel the water follows, the loss in gravitational potential will be the same. All the water starts at the same level and comes together again at the same level below the rapids. In the same way, in a parallel electrical circuit, the total current is the sum of the currents through each current path, and the potential difference across each path is the same. A schematic diagram of three identical resistors connected in parallel is shown in Figure The 120-V generator provides the source of potential difference for the circuit. Each current path from A to B is part of a closed circuit including the generator, resistors, and ammeters. Each path acts as if the other paths were not present. What is the current through one resistor? There is a 120-V potential difference across each resistor. The current through a 6-0 resistor across a 120- V potential difference is V 120 V I = - = -- = 20 A. R 60 There is more than one current path in a parallel circuit. POCKET LAB PARALLEL RESISTANCE Hook up a power supply, a resistor, and an ammeter in a series circuit. Predict what will happen to the current in the circuit when a second, identical resistor is added in parallel to the first. Predict the new currents when the circuit contains three and four resistors in parallel. Explain your prediction. Try it. Make a data table to show your results. Write a sentence that explains your results. (Hint: Include the idea of resistance.) 23.1 Simple Circuits 475

8 The equivalent resistance of parallel resistors is smaller than any resistor. The current flow through each path from A to B is 20 A. The total current between the two points is 60 A. The potential difference across the two points is 120 V. A single resistor that would permit 60-A current with 120-V potential difference has a resistance given by V 120 V R=-=-- 2 D. 60 A In a parallel circuit, the total current is the sum of the currents in each path. In a parallel circuit, all the potential drops are the same. The equivalent resistance can be replaced by this single resistor. Placing a resistor in parallel with an existing resistor always decreases the equivalent resistance of the circuit. The resistance decreases because each new resistor provides an additional path for current to flow between points A and B. Notice that the equivalent resistance of the circuit is less than the resistance of any resistor in the circuit. To find the equivalent resistance of a parallel circuit, start by noting that the total current in the circuit is the sum of the currents through the branches of the circuit. Recall also that the potential across each is the same. If I is the total current, and 11, 12, and 13 are the currents through each of the branches, then = The current through each resistor, for example R 1, can be found from 11 = VIR1 The total current through the equivalent resistance, R, of the circuit is given by I = VIR, but all potential drops in a parallel circuit are the same. Dividing both sides of the equation by V then gives an equation for the equivalent resistance of three parallel resistors, I, 12 R, 60.0 R FIGURE Use with the Example Problem. 476 Series and Parallel Circuits

9 Example Problem Total Resistanceand Current in a Parallel Circuit Three resistors of , , and are connected in parallel across a 90.0-V difference in potential, Figure a. Find the equivalent resistance of the circuit. b. Find the current in the entire circuit. c. Find the current through each branch of the circuit. Given: R1 = Unknowns: R, I, 1 1, 12, 13 R2 = Basic equations: V = IR R3 = V = 90.0 V R R1 R2 R3 Solution: a. R. = ~ =.l. +.I. +.l, 1 6 R R = _ ~ _ 90.0 V b 9.00 A I - R c. The voltage drop across each resistor is 90.0 V. The currents are: V 90.0 V (for R 1 ) 11 R = 1.50 A V 90.0 V == - == R = 3.00 A V 90.0 V R = 4.50 A ELECTRIC POVVER PLANT OPERATOR A utility company needs high school graduates who have the interest and ability to learn detailed information about the production and distribution of electric power. Your high school record of reliability, effort, and cooperation are most important. This entry-level position includes extensive training. Success in this career field is dependent on quality work, continued training, and the ability to contribute positively to the work environment. For information Utility Workers of America, th Street, NW., Washington, DC Practice Problems 9. Three 15-0 resistors are connected in parallel and placed across a 30-V potential difference. a. What is the equivalent resistance of the parallel circuit? b. What is the current through the entire circuit? c. What is the current through each branch of the parallel circuit? 10. A resistor and a resistor are connected in parallel and placed across the terminals of a 15.0-V battery. a. What is the equivalent resistance of the parallel circuit? b. What is the current through the entire circuit? c. What is the current through each branch of the parallel circuit? 23.1 Simple Circuits 477

10 11. A resistor, a resistor, and a resistor are connected in parallel and placed across a potential difference of 12.0 V. a. What is the equivalent resistance of the parallel circuit? b. What is the current through the entire circuit? c. What is the current through each of the branches of the parallel circuit? ~ 12. Suppose the resistor in Practice Problem 10 is replaced by a resistor. a. Does the equivalent resistance become smaller, larger, or remain the same? b. Does the amount of current through the entire circuit change? in what way? c. Does the amount of current through the resistor change? in what way? You learned in Chapter 22 that the resistance of a 60-watt bulb is greater than the resistance of a 100-watt bulb. The brightness of a bulb is directly proportional to the power dissipated by the filament of the bulb. In a series circuit, the same current flows through all elements in the circuit. Therefore, according to P =,2R, the 60-watt bulb has a higher resistance, and for this reason, will dissipate more power and glow more brightly. A FIGURE Use with Concept Review 1.4. R R... CONCEPT REVIEW 1.1 Are car headlights connected in series or parallel? Draw on your experience. 1.2 Lamp dimmers often contain variable resistors. a. Would a dimmer be hooked in series or in parallel with the lamp to be controlled? Why? b. Should the resistance of the dimmer be increased or decreased to dim the lamp? 1.3 A switch is connected in series with a 7S-W bulb to a source of 120 V. a. What is the potential difference across the switch when it is closed or on? b. What is the potential difference across it when it is open or off? Explain. 1.4 Critical Thinking: The circuit in Figure has four identical resistors. Suppose a wire is connected between points A and B. Answer each question, explaining your reasoning. a. What current flows through the wire? b. What happens to the current through each resistor? c. What happens to the current drawn from the battery? d. What happens to the potential difference across each resistor? 478 Series and Parallel Circuits

11 P H Y SIC slab. : Circuits Purpose To construct series, parallel, and combination circuits. Materials V power supply. 4 small lamps with sockets and connecting wires Procedure 1. Turn the dial on the power supply to the off position. Use caution while working with line current electricity to prevent shocks. 2. Build a circuit with 2 lamps. Turn the power supply on so that both bulbs glow brightly. Record the setting on the power supply. This will be called the original setting. 3. Loosen one bulb from its socket and observe what happens to the other bulb. Turn the power supply off. If the other bulb goes out, you have made a series circuit. Go to the section marked SERIESCIRCUITS. If the other bulb stayed on you have made a parallel circuit. Go to the section marked PARALLEL CIRCUITS. Series Circuits 1. Make a series circuit with three bulbs. Turn the power supply dial to the original setting and notice the brightness of the bulbs. Turn the power supply off. 2. Make a series circuit with four bulbs. Turn the power supply dial to the original setting and notice the brightness of the bulbs. 3. Try loosening one of the middle bulbs. What happens to the others? Sketch the circuit. 5. If you have not done the parallel circuits, try to make a circuit with 2 lamps so that when you loosen one bulb the other stays on. Then go to PARALLEL CIRCUITS. Parallel Circuits 1. Make a parallel circuit with three bulbs. 2. Loosen one of the bulbs to verify that the other 2 stay on. Relight the bulbs. 3. Compare the brightness of the bulbs to the original brightness. Sketch the circuit. 4. If you have not already done series circuits, try to design a circuit with 2 bulbs so that when one is loosened the other will also go out. Then go to SERIESCIRCUITS. Observations and Data 1. Describe what happens to the brightness of the bulbs as more are added in a series circuit and a parallel circuit. Analysis The brightness of each bulb depends on the current passing through the bulb. 1. Write a statement that compares the current through each bulb in the same series circuit. 2. What happens to the current as more lamps are added to the series circuit? Applications 1. Design and construct a circuit with three lamps so that 1 is bright and 2 are dim. Draw the circuit Simple Circuits 479

12 Objectives describe a combination seriesparallel circuit. calculate the equivalent resistance in a combination circuit. distinguish between voltmeters and ammeters and state the important characteristics of each. describe how each is wired into a circuit and explain why such a method is used. A fuse or circuit breaker protects against dangerously large currents. A ground-fault interrupter opens the circuit if a shock hazard exists. FIGURE If too much current flows through the bimetallic strip, it will bend down, releasing the latch. The handle will move to the "off" position, opening the switch contacts. The handle will not stay in the "on" position with the switch contacts closed unless the bimetallic strip is cool and straight APPLICATIONS OF CIRCUITS Household wiring circuits are examples of the ideas we have examined in the earlier part of this chapter. We need to understand the requirements and limitations of these systems. Above all, it is important that we are aware of the safety measuresthat must be practiced to prevent accidents. Safety Devices In an electric circuit, fuses and circuit breakers are switches that act as safety devices. They prevent circuit overloads that can occur when too many appliances are turned on at the same time or a short circuit occurs in one appliance. When appliances are connected in parallel, each additional appliance placed in operation reduces the equivalent resistance in the circuit and causes more current to flow through the wires. The additional current may produce enough thermal energy (at the rate P =,2R) to melt insulation on the wires, causing a short circuit in the wires or even a fire. A fuse is a short piece of metal that melts from the heating effect of the current. The thickness of the metal is adjusted to determine the current needed to melt the fuse. A circuit breaker, Figure 23-11, is an automatic switch that opens when the current reaches some set value. If current greater than the set value flows in the circuit, it will be overloaded. The circuit breaker will open stopping all current flow. A ground-fault interrupter is often required by law in electrical outlets in bathrooms. Current flows both through the wire bringing charges from the power source and in the wire returning the charges. If there is no other current path, the magnitude of these two currents is the same. Sometimes, when an appliance, such as a hair dryer, is used, the appliance or user might touch a cold water pipe or a sink full of water. Another current path, through the user, could exist-as little as 5 ma is serious. The ground-fault interrupter contains an electronic circuit that detects small differences in current caused by the extra current path and opens the circuit, preventing dangerous shocks. Circuit breaker Bimetallic strip Current in from /' central switch 480 Seriesand ParallelCircuits

13 J120V Appliances FIGURE The wiring diagram for this house indicates the parallel nature of the circuit. This wiring arrangement will permit the use of one or more appliances. Electric wiring in a home uses parallel circuits, Figure 23-12, so that the current in anyone circuit does not depend on the current in the other circuits. The current in a device that dissipates power, P, when connected to a voltage source, V, is given by I = PIV. Figure shows a schematic diagram of the house circuit. Suppose, first, that a 240-W television is plugged into a 120-V outlet. The current that flows is given by I = (240 W)/(120 V) = 2 A. Then a 720-W curling iron is plugged in. The current through the iron is I = (720 W)/(120 V) = 6 A. Finally, a 1440-W hair dryer is added. The current through the hair dryer is I = (1440 W)/(120 V) = 12 A. The current through these three appliances can be found by considering them as a parallel circuit where the currents are independent of each other. Each resistor represents an appliance. The value of the resistance is found by first calculating the current the appliance draws as above and then using the equation R = VI/' The equivalent resistance of the three appliances is = =--=- R R = 6 O. The ls-a fuse is connected in series with the power source so the entire current flows through it. The current flowing through the fuse is V 120 V I = - = -- = 20 A. R 60 The 20-A current exceeds the rating of the ls-a fuse, so the fuse will melt, or "blow," cutting off current to the entire circuit. A short circuit occurs when a circuit is formed that has a very low resistance. The low resistance causes the current to be very large. If there were no fuse or circuit breaker, such a large current could easily Home wiring uses parallel circuits. --Fe 120V 10n FIGURE A 60-0, a 20-0, and a 10-0 resistor are connected in parallel across a 120-V source. The current through the circuit will cause the fuse to melt Applications of Circuits 481

14 R,8fl R,8fl 1 R2 1 R3 60V 12 son zon v, 13.L [J FIGURE Many circuits contain a combination of series and parallel connections. or I ~T R,8fl I R2.312fl ti start a fire. A short circuit can occur if the insulation on a lamp cord becomes old and brittle. The two wires in the cord could accidentally touch. The resistance of the wire might be only n. When placed across 120 V, this resistance would result in a current of V 120 V I = - = = A. R n Such a current would cause a fuse or a circuit breaker to open the circuit immediately, preventing the wires from becoming hot enough to start a fire. A short circuit is a very low resistance in a location where the resistance should be very high. F.Y.I. The largest voltages in your home are in your television set, where V to V are common. The largest currents are likely to be the 40 A in an electric range. In your car, voltages as high as V are used in the spark plugs, and currents as large as 300 A are needed to turn the started motor. Combined Series-Parallel Circuits Circuits often consist of combinations of series and parallel circuits, such as the three resistors in Figure The circuits can be understood by analyzing them in steps. 1. If any resistors are connected in parallel, calculate the single equivalent resistance that can replace them. 2. If any equivalent resistances are now connected in series, calculate a single new equivalent resistance that can replace them. 3. By repeating steps 1 and 2, you can reduce the circuit to a single resistance. The total circuit current can now be found. The voltage drops and currents through individual resistors can then be calculated. PROBLEM SOLVING STRATEGY 1. How do you figure out which resistors are in parallel? Resistors in parallel have separate current paths. They are in parallel only if they have the same potential differences across them. 2. How do you figure out which resistors are in series? Remember that a simple series circuit has only one current path. Resistors are in series only if there is one and only one current path through them. 482 Seriesand ParallelCircuits

15 Example Problem Series-ParallelCircuit In Figure 23-14, a , resistor is connected in parallel with a , resistor. The parallel connection is placed in series with an 8.0-0, resistor, and the entire circuit is placed across a 60.0-V difference of potential. a. What is the equivalent resistance (R 2,3) of the parallel portion of the circuit? b. What is the equivalent resistance of the entire circuit? c. What is the current in the entire circuit? d. What is the voltage drop across the 8.0-0, resistor? e. What is the voltage drop across the parallel portion of the circuit? f. What is the current in each line of the parallel portion of the circuit? Given: See Figure Unknowns: R 2,3, R, I, V 1, V p, 1 2, 13 Basic equations: V = IR R = R, + R2 +. l!r = l!rl + 1!R2 +. Solution: a. R2 and R3 are connected in parallel. Their equivalent resistance is =-+-= + =--- R2,3 R2 R a a R2,3 = b. The circuit is equivalent to a series circuit with an 8.0-0, resistor and a , resistor in series, Figure R = R, + R 2,3 = 8.0 a a = 20.0 a c. The current in the circuit is V 60.0 V 1 = - = = 3.00 A. R 20.0 a d. The voltage drop across the resistor is V 1 = IRl = (3.00 A)(8.0 a) = 24 V. e. The parallel branch (R 2 and R 3 ) behaves as a resistor, with the sum of the currents through R2 and R3 flowing through it. Therefore, the voltage drop across it is Vp = IR 2,3 = (3.00 A)(12.0 a) = 36.0 V. f. The 36.0-V drop across the parallel portion of the circuit is the same across all parts of the parallel circuit. Therefore, the current through the , resistor is ECONOMICS CONNECTION Thomas Edison did not invent the light bulb, but he did develop the first bulb that could be used in homes, and even more important, a commercially viable distribution system for electricity. Edison started with the basic assumption that each customer should be able to turn on and off his own light bulbs without affecting the other bulbs connected in the circuit. This meant that bulbs must be connected in parallel. Since only part of the circuit current flowed through each bulb, Edison had to design a high-resistance filament that required only a small current for a large heating effect. His assistants tried over 1600 materials including paper, cloth, thread, fishline, fiber, celluloid, boxwood, coconut shells, spruce, hickory, hay, maple shavings, rosewood, punk, cork, flax, bamboo, and the hair out of a redheaded Scotsman's beard before he developed a high vacuum-sealed bulb with a carbonized cotton. 12 = ~ = 36.0 V 1.20 A. R The current through the resistor is V 36.0 V 13 = - = = 1.80 A. R a The current through the parallel part of the circuit 1.20 A A, or 3.00 A. This value agrees with the value for the current calculated in Part c. is 23.2 Applications of Circuits 483

16 POCKET AMMETER LAB RESISTANCE Design an experiment using a power supply, a voltmeter, an ammeter, a resistor, and some wires to determine the resistance of the ammeter. Make a sketch of your set up and include measurements and equations. Be prepared to present your experiment to the rest of the class. FIGURE An ammeter measures current (a), and a standard laboratory voltmeter (b) measure potential differences. Ammeters are always placed in series within a circuit (c) and voltmeters are placed in parallel (d). Practice Problems 13. Two 60-0 resistors are connected in parallel. This parallel arrangement is connected in series with a 30-0 resistor. The combination is then placed across a 120-V potential difference. a. Draw a diagram of the circuit. b. What is the equivalent resistance of the parallel portion of the circuit? c. What single resistance could replace the three original resistors? d. What is the current in the circuit? e. What is the voltage drop across the 30-0 resistor? f. What is the voltage drop across the parallel portion of the circuit? g. What is the current in each branch of the parallel portion of the circuit? 14. Three 15-0 resistors are connected in parallel. This arrangement is connected in series with a 10-0 resistor. The entire combination is then placed across a 45-V difference in potential. a. Draw a diagram of the circuit. b. What is the equivalent resistance of the parallel portion of the circuit? c. What is the equivalent resistance of the entire circuit? d. What is the current in the entire circuit? e. What is the voltage drop across the 10-0 resistor? ~ 15. Suppose you are given three 68-0 resistors. You can use them in a series, parallel, or series-parallel circuit. Find the three different resistances you can produce in the circuit. a b ~ "I s..,,~'~:\ \ _I'l<::'. :,, \ -.: ~'." < _. c d =-20 V Ammeter R, =- 20 V 0.OH = R Series and Parallel Circuits

17 Ammeters and Voltmeters An ammeter is used to measure the current in any branch or part of a circuit. If, for example, we want to measure the current through a resistor, we would place an ammeter in series with the resistance. This requires opening a current path and inserting an ammeter. The use of an ammeter should not change the current in the circuit we are trying to measure. Because current would decrease if the ammeter increased the resistance in the circuit, the resistance of an ammeter should be as low as possible. Figure 23-15c shows a real ammeter as a meter placed in parallel with a O.Ol-D resistor. The ammeter resistance is much smaller than the values of the resistors. The current decrease would be from 1.0 A to A, too small to notice. A voltmeter is used to measure the voltage drop across some part of a circuit. To measure the potential drop across a resistor, connect the voltmeter in parallel with the resistor. To cause the smallest change in currents or voltages in the circuit, a voltmeter should have a very high resistance. Consider the circuit in Figure 23-15d. A typical real voltmeter consists of a meter in series with a lo-kd resistor. When it is connected in parallel with R 2, the equivalent resistance of the combination is smaller than R2 alone. Thus, the total resistance of the circuit decreases, increasing the current. R, has not changed, but the current through it has increased, increasing the potential drop across it. The battery, however, holds the potential drop across Rl and R2 constant. Thus, the potential drop across R2 must decrease. The result of connecting a voltmeter across a resistor is to lower the potential drop across it. The higher the resistance of the voltmeter, the smaller the voltage change. Using a voltmeter with a 10 OOO-Dresistance changes the voltage across R2 from 10 V to V, too small a change to detect. Modern electronic voltmeters, Figure 23-16, have even higher resistances, 10 7 D, and so produce even smaller changes.... CONCEPT REVIEW 2.1 Home fires were often caused by people who replaced fuses with pennies. Why would this cause a problem? 2.2 Consider the circuit in Figure made with identical bulbs. a. Compare the brightness of the three bulbs. b. What happens to the brightness of each bulb when bulb 1 is unscrewed from its socket? What happens to the three currents? c. Bulb 1 is screwed in again and bulb 3 is unscrewed. What happens to the brightness of each bulb? What happens to the three currents? 2.3 Consider again the circuit in Figure a. What happens to the brightness of each bulb if a wire is connected between points Band C? b. A fourth bulb is connected in parallel with bulb 3 alone. What happens to the brightness of each bulb? 2.4 Critical Thinking: In the same circuit as 2.3, the wire at point C is broken and a small resistor is inserted in series with bulbs 2 and 3. What happens to the brightness of the two bulbs? Explain. FIGURE A modern digital multi meter can be used to measure a wide range of currents, voltages, and resistances in electrical circuits. An ammeter circuit to be measured. is put in series with the A voltmeter is connected in parallel with the element to be measured. FIGURE Use with Concept Review 2.2, 2.3, and Applications of Circuits 485

18 P hysicsand technology ROBOTS AND OHM'S LAW In1854, the year that Georg Simon Ohm died, Lord Kelvin discovered that the resistance of a wire changes when it is strained (stretched). Engineers first used this phenomenon to analyze stresses in structural surfaces. For example, by monitoring the current through a fine metal wire mounted alongside a bridge, machines can indicate strain on the bridge based on the wire's varying resistance. Such a device is called a strain gauge. Researchers seeking ways to imitate the complex capabilities of the human hand have used a version of a strain gauge to give robots a sense of touch. As recently as 1985, robotic hands, equipped with superhuman strength, had the dexterity to gently crack an egg into a mixing bowl. However, a robot still needs some kind of tactile sensor in order to "see" with its hands. One of the most promising sensor designs consists of an electronic chip printed with a fine metallic network and covered by a thin sheet of conducting rubber. Pressure on the rubber changes the resistance of the chip, transmitting a "picture" of the strain on the robotic skin.. What might be some other valuable appl ications for strain gauge "sensors"? CHAPTER 23 REVIEW ~=-=---~~~~~~=-~~ SUMMARY 23.1 Simple Circuits The current is the same everywhere in a simple series circuit. The equivalent resistance of a series circuit is the sum of the resistances of its parts. The sum of the voltage drops across resistors in series is equal to the potential difference applied across the combination. A voltage divider is a series circuit used to produce a voltage source from a higher voltage battery. The voltage drops across all branches of a parallel circuit are the same. In a parallel circuit, the total current is equal to the sum of the currents in the branches. The reciprocal of the equivalent resistance of parallel resistors is equal to the sum of the reciprocals of the individual resistances. 486 Seriesand Parallel Circuits If any branch of a parallel circuit is opened, there is no current in that branch. The current in the other branches is unchanged Applications of Circuits A fuse or circuit breaker, placed in series with appliances, creates an open circuit when dangerously high currents flow. A complex circuit is often a combination of series and parallel branches. Any parallel branch is first reduced to a single equivalent resistance. Then any resistors in series are replaced by a single resistance. An ammeter is used to measure the current in a branch or part of a circuit. An ammeter always has a low resistance and is connected in series. A voltmeter measures the potential difference (voltage) across any part or combination of parts of a circuit. A voltmeter always has a high resistance and is connected in parallel with the part of the circuit being measured.

19 KEY TERMS series circuit equivalent resistance voltage divider parallel circuit fuse REVIEWING CONCEPTS ground-fault interrupter short circuit ammeter voltmeter 1. Why is it frustrating when one bulb burns out on a string of Christmas tree lights connected in series? 2. Why does the equivalent resistance decrease as more resistors are added to a parallel circuit? 3. Several resistors with different values are connected in parallel. The equivalent resistance will be less than which resistor? 4. Compare the amount of current entering a junction in a parallel circuit with that leaving the junction. Note: a junction is a point where three or more conductors are joined. 5. Why is household wiring done in parallel instead of in series? 6. Why is there a difference in total resistance between three 60-0 resistors connected in series and three 60-0 resistors connected in parallel? 7. Explain the function of a fuse in an electric circuit. 8. What is a short circuit? Why is a short circuit dangerous? 9. Why does an ammeter have a very low resistance? 10. Why does a voltmeter have a very high resistance? APPLYING CONCEPTS 1. What happens to the current in the other two lamps if one lamp in a three-lamp series circuit burns out? 2. In the voltage divider in Figure 23-3, the resistor R1 is a variable resistor. What happens to the voltage output, V 2 of the voltage divider if the resistance of the variable resistor is increased? 3. Circuit A contains three 60-0 resistors in series. Circuit B contains three 60-0 resistors in parallel. How does the current in the second 60-0 resistor change if a switch cuts off the current to the first 60-0 resistor in a. circuit A? b. circuit B? 4. What happens to the current in the other two lamps if one lamp in a three-lamp parallel circuit burns out? 5. An engineer needs a 10-0 resistor and a 15-0 resistor. But there are only 30-0 resistors in stock. Must new resistors be purchased? Explain. 6. If you have a 6-V battery and many 1.5-V bulbs, how could you connect them so they light but do not have more than 1.5-V across each bulb? 7. Two lamps have different resistances, one larger than the other. a. If they are connected in parallel, which is brighter (dissipates more power)? b. When connected in series, which is brighter? 8. Two 100-W light bulbs are connected in series to a 12-V battery and another two 100-W light bulbs are connected in parallel to a separate 12-V battery. Which battery will run out of energy first? Why? 9. For each part of this question, write the form of circuit that applies: series or parallel. a. The current is the same throughout. b. The total resistance is equal to the sum of the individual resistances. c. The voltage drop is the same across each resistor. d. The voltage drop is proportional to the resistance. e. Adding a resistor decreases the total resistance. f. Adding a resistor increases the total resistance. g. If the current through one resistor goes to zero there is no current in the entire circuit. h. If the current through one resistor goes to zero, the current through all other resistors remains the same. i. This form is suitable for house wiring. 10. Why is it dangerous to replace a 15-A fuse in a circuit with one of 30 A? Chapter 23 Review 487

20 PROBLEMS 23.1 Simple Circuits 1. A 20.0-D lamp and a 5.0-D lamp are connected in series and placed across a potential difference of 50 V. What is a. the equivalent resistance of the circuit? b. the current in the circuit? c. the voltage drop across each lamp? d. the power dissipated in each lamp? 2. Three identical lamps are connected in series to a 6.0-V battery. What is the voltage drop across each lamp? 3. The load across a 12- V battery consists of a series combination of three resistors of 15 D, 21 D, and 24 D. a. What is the total resistance of the load? b. What is the current in the circuit? 4. The load across a battery consists of two resistors, with values of 15 D and 45 D connected in series. a. What is the total resistance of the load? b. What is the voltage of the battery if the current in the circuit is 0.10 A? 5. A lamp having a resistance of 10 D is connected across a 15- V battery. a. What is the current through the lamp? b. What resistance must be connected in series with the lamp to reduce the current to 0.50 A? 6. A string of eighteen identical Christmas tree lights are connected in series to a 120- V source. The string dissipates 64.0 W. a. What is the equivalent resistance of the light string? b. What is the resistance of a single light? c. What power is dissipated by each lamp? 7. One of the bulbs in the previous problem burns out. The lamp has a wire that shorts out the lamp filament when it burns out. This drops the resistance of the lamp to zero. a. What is the resistance of the light string now? b. Find the power dissipated by the string. c. Did the power go up or down when a bulb burned out? 8. A 75.0-W bulb is connected to a 120-V source. a. What is the current through the bulb? b. What is the resistance of the bulb? c. A lamp dimmer puts a resistance in series with the bulb. What resistance would be needed to reduce the current to A? 9. In the previous problem, you found the resistance of a lamp and a dimmer resistor. a. Assuming the resistances are constant, find the voltage drops across the lamp and the resistor. b. Find the power dissipated by the lamp. c. Find the power dissipated by the dimmer. 10. Amy needs 5.0 V for some integrated circuit experiments. She uses a 6.0-V battery and two resistors to make a voltage divider. One resistor is 330 D. She decides to make the other resistor smaller. What value should it have? 11. Pete is designing a voltage divider using a 12.0-V battery and a D resistor as R 2. What resistor should be used as R1 if the output voltage across R2 is to be 4.00 V? ~ 12. A typical television dissipates 275 W when plugged into a 120- V outlet. a. Find the resistance of the television. b. The resistance of the wires between the outlet and fuse is 2.5 D. The television and wires form a series circuit that works like a voltage divider. Find the voltage drop across the television. c. A 12-D hair dryer is now plugged into the same outlet. Find the equivalent resistance of the two appliances. d. Find the voltage drop across the television and hair dryer combination. The lower voltage explains why the television picture sometimes shrinks when another appliance is turned on. 13. Three identical lamps are connected in parallel to each other and then connected to a 6.0-V battery. What is the voltage drop across each lamp? 14. A 16.0-D and a 20.0-D resistor are connected in parallel. A difference in potential of 40.0 V is applied to the combination. a. Compute the equivalent resistance of the parallel circuit. b. What is the current in the circuit? c. How large is the current through the D resistor? 15. During a laboratory exercise, you are supplied with the following appartus: a battery of potential difference V, two heating elements of low resistance that can be placed in water, an 488 Seriesand Parallel Circuits

21 ammeter of negligible resistance, a voltmeter of extremely high resistance, wires of negligible resistance, a beaker that is well-insulated and has negligible heat capacity, g of water at 25 C. a. By means of a diagram using standard symbols, show how these components should be connected to heat the water as rapidly as possible. b. If the voltmeter reading holds steady at 50.0 V and the ammeter reading holds steady at 5.0 A, estimate the time in seconds required to completely vaporize the water in the beaker. Use 4200 J/kg Co as the specific heat of water and 2.3 x 10 6 J/kg as the heat of vaporization of water Applications of Circuits 16. A circuit contains six 240-D lamps, 60-W bulbs, and a 10.0-D heater connected in parallel. The voltage across the circuit is 120 V. What is the current in the circuit a. when four lamps are turned on? b. when all lamps are on? c. if six lamps and the heater are operating? d. If the circuit has a fuse rated at 12 A, will it melt if everything is on? 17. Determine the reading of each ammeter and each voltmeter in Figure b. Find the current to the bulb. c. Find the power rating of the bulb. d. Since the current in the bulb is 0.50 A, the current in the lead lines must also be 0.50 A. Calculate the voltage drop due to the two leads. Kitchen light 240 n FIGURE Use with Problem 19 and 20. ~ 20. A power saw is operated by an electric motor. When electric motors are first turned on, they have a very low resistance. Suppose that a kitchen light discussed in the previous problem is on and a power saw is turned on. The saw, light, and the lead lines have an initial total resistance of 6.0 D. a. Compute the equivalent resistance of the light-saw parallel circuit. b. What current flows to the light? c. What is the total voltage drop across the two leads to the light? d. What voltage remains to operate the light? Will this cause the light to dim temporarily? USING LAB SKILLS FIGURE Use with Problem 17 and Determine the power used by each resistance shown in Figure A typical home circuit is diagrammed in Figure Note that the lead lines to the kitchen lamp each have very low resistances of 0.25 D. The lamp has a resistance of D. Although the circuit is a parallel circuit, the lead lines are in series with each of the components of the circuit. a. Compute the equivalent resistance of the circuit consisting of just the light and the lead lines to and from the light. 1. When you did the Pocket Lab on page 472, you found that adding more resistors in a series circuit caused the current to decrease. What happens to the voltage across each resistor as more are added in series? Use a voltmeter to find out. Describe your results. THINKING PHYSIC-LY A W three-way bulb has three terminals on its base. Explain how to connect 120 V to the terminals to obtain a low, medium, and a high brightness. Chapter 23 Review 489