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.5×10−3 5.0×10−3 3.93×10−3 4.3×10−3 Manganin (Cu, Mn, Ni alloy) 0.000×10−3 Constantan (Cu, Ni alloy) Mercury Nichrome (Ni, Fe, Cr alloy) Semiconductors Carbon (pure) Germanium (pure) Silicon (pure) 0.002×10−3 0.89×10−3 0.4×10−3 −0.5×10−3 −50×10−3 −70×10−3 Note also that is negative for the semiconductors listed in Table 20.2, meaning that their resistivity decreases with increasing temperature. They become better conductors at higher temperature, because increased thermal agitation increases the number of free charges available to carry current. This property of decreasing with temperature is also related to the type and amount of impurities present in the semiconductors. 2. Values at 20°C. 882 Chapter 20 \| Electric Current, Resistance, and Ohm's Law The resistance of an object also depends on temperature, since 0 is directly proportional to. For a cylinder we know = /, and so, if and do not change greatly with temperature, will have the same temperature dependence as. (Examination of the coefficients of linear expansion shows them to be about two orders of magnitude less than typical temperature coefficients of resistivity, and so the effect of temperature on and is about two orders of magnitude less than on.) Thus, = 0(1 + Δ) (20.24) is the temperature dependence of the resistance of an object, where 0 is the original resistance and is the resistance after a temperature change Δ. Numerous thermometers are based on the effect of temperature on resistance. (See Figure 20.15.) One of the most common is the thermistor, a semiconductor crystal with a strong temperature dependence, the resistance of which is measured to obtain its temperature. The device is small, so that it quickly comes into thermal equilibrium with the part of a person it touches. Figure 20.15 These familiar thermometers are based on the automated measurement of a thermistor's temperature-dependent resistance. (credit: Biol, Wikimedia Commons) Example 20.6 Calculating Resistance: Hot-Filament Resistance Although caution must be used in applying = 0(1 + Δ) and = 0(1 + Δ) for temperature changes greater than 100ºC, for tungsten the equations work reasonably well for
very large temperature changes. What, then, is the resistance of the tungsten filament in the previous example if its temperature is increased from room temperature ( 20ºC ) to a typical operating temperature of 2850ºC? Strategy This is a straightforward application of = 0(1 + Δ), since the original resistance of the filament was given to be 0 = 0.350 Ω, and the temperature change is Δ = 2830ºC. Solution The hot resistance is obtained by entering known values into the above equation: = 0(1 + Δ) = (0.350 Ω)[1 + (4.510–3 / ºC)(2830ºC)] = 4.8 Ω. (20.25) Discussion This value is consistent with the headlight resistance example in Ohm's Law: Resistance and Simple Circuits. PhET Explorations: Resistance in a Wire Learn about the physics of resistance in a wire. Change its resistivity, length, and area to see how they affect the wire's resistance. The sizes of the symbols in the equation change along with the diagram of a wire. Figure 20.16 Resistance in a Wire (http://cnx.org/content/m55357/1.2/resistance-in-a-wire_en.jar) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 883 Applying the Science Practices: Examining Resistance Using the PhET Simulation “Resistance in a Wire”, design an experiment to determine how different variables – resistivity, length, and area – affect the resistance of a resistor. For each variable, you should record your results in a table and then create a graph to determine the relationship. 20.4 Electric Power and Energy Learning Objectives By the end of this section, you will be able to: • Calculate the power dissipated by a resistor and the power supplied by a power supply. • Calculate the cost of electricity under various circumstances. The information presented in this section supports the following AP® learning objectives and science practices: • 5.B.9.8 The student is able to translate between graphical and symbolic representations of experimental data describing relationships among power, current, and potential difference across a resistor. (S.P. 1.5) Power in Electric Circuits Power is associated by many people with electricity. Knowing that power
is the rate of energy use or energy conversion, what is the expression for electric power? Power transmission lines might come to mind. We also think of lightbulbs in terms of their power ratings in watts. Let us compare a 25-W bulb with a 60-W bulb. (See Figure 20.17(a).) Since both operate on the same voltage, the 60-W bulb must draw more current to have a greater power rating. Thus the 60-W bulb's resistance must be lower than that of a 25-W bulb. If we increase voltage, we also increase power. For example, when a 25-W bulb that is designed to operate on 120 V is connected to 240 V, it briefly glows very brightly and then burns out. Precisely how are voltage, current, and resistance related to electric power? Figure 20.17 (a) Which of these lightbulbs, the 25-W bulb (upper left) or the 60-W bulb (upper right), has the higher resistance? Which draws more current? Which uses the most energy? Can you tell from the color that the 25-W filament is cooler? Is the brighter bulb a different color and if so why? (credits: Dickbauch, Wikimedia Commons; Greg Westfall, Flickr) (b) This compact fluorescent light (CFL) puts out the same intensity of light as the 60-W bulb, but at 1/4 to 1/10 the input power. (credit: dbgg1979, Flickr) Electric energy depends on both the voltage involved and the charge moved. This is expressed most simply as PE =, where is the charge moved and is the voltage (or more precisely, the potential difference the charge moves through). Power is the rate at which energy is moved, and so electric power is = =. (20.26) Recognizing that current is = / (note that Δ = here), the expression for power becomes = (20.27) 884 Chapter 20 \| Electric Current, Resistance, and Ohm's Law Electric power ( ) is simply the product of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the joule, power has units of joules per second, or watts. Thus, 1 A ⋅ V = 1 W. For example, cars often have one or more auxiliary power outlets with which you can charge a cell phone or other electronic devices. These outlets may be rated at
20 A, so that the circuit can deliver a maximum power = = (20 A)(12 V) = 240 W. In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes ( 1 kA ⋅ V = 1 kW ). To see the relationship of power to resistance, we combine Ohm's law with =. Substituting = gives = ( / ) = 2 /. Similarly, substituting = gives = () = 2. Three expressions for electric power are listed together here for convenience: = = 2 = 2. Note that the first equation is always valid, whereas the other two can be used only for resistors. In a simple circuit, with one voltage source and a single resistor, the power supplied by the voltage source and that dissipated by the resistor are identical. (In more complicated circuits, can be the power dissipated by a single device and not the total power in the circuit.) (20.28) (20.29) (20.30) Making Connections: Using Graphs to Calculate Resistance As ∝ 2 and ∝ 2, the graph for power versus current or voltage is quadratic. An example is shown in the figure below. Figure 20.18 The figure shows (a) power versus current and (b) power versus voltage relationships for simple resistor circuits. (a) (b) Using equations (20.29) and (20.30), we can calculate the resistance in each case. In graph (a), the power is 50 W when current is 5 A; hence, the resistance can be calculated as = / 2 = 50/ 52 = 2 Ω. Similarly, the resistance value can be calculated in graph (b) as = 2/ = 102/ 50 = 2 Ω Different insights can be gained from the three different expressions for electric power. For example, = 2 / implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in = 2 /, the effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about 100 W, burning it out. If the bulb's resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too. Example 20.7 Calculating Power Dissipation and Current: Hot and Cold Power (a) Consider the examples
given in Ohm's Law: Resistance and Simple Circuits and Resistance and Resistivity. Then find the power dissipated by the car headlight in these examples, both when it is hot and when it is cold. (b) What current does it draw when cold? Strategy for (a) For the hot headlight, we know voltage and current, so we can use = to find the power. For the cold headlight, we know the voltage and resistance, so we can use = 2 / to find the power. Solution for (a) Entering the known values of current and voltage for the hot headlight, we obtain This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law The cold resistance was 0.350 Ω, and so the power it uses when first switched on is = = (2.50 A)(12.0 V) = 30.0 W. = 2 = (12.0 V)2 0.350 Ω = 411 W. 885 (20.31) (20.32) Discussion for (a) The 30 W dissipated by the hot headlight is typical. But the 411 W when cold is surprisingly higher. The initial power quickly decreases as the bulb's temperature increases and its resistance increases. Strategy and Solution for (b) The current when the bulb is cold can be found several different ways. We rearrange one of the power equations, = 2, and enter known values, obtaining = = 411 W 0.350 Ω = 34.3 A. (20.33) Discussion for (b) The cold current is remarkably higher than the steady-state value of 2.50 A, but the current will quickly decline to that value as the bulb's temperature increases. Most fuses and circuit breakers (used to limit the current in a circuit) are designed to tolerate very high currents briefly as a device comes on. In some cases, such as with electric motors, the current remains high for several seconds, necessitating special “slow blow” fuses. The Cost of Electricity The more electric appliances you use and the longer they are left on, the higher your electric bill. This familiar fact is based on the relationship between energy and power. You pay for the energy used. Since = /, we see that = (20.34) is the energy used by a device using power for a time interval. For
example, the more lightbulbs burning, the greater used; the longer they are on, the greater is. The energy unit on electric bills is the kilowatt-hour ( kW ⋅ h ), consistent with the relationship =. It is easy to estimate the cost of operating electric appliances if you have some idea of their power consumption rate in watts or kilowatts, the time they are on in hours, and the cost per kilowatt-hour for your electric utility. Kilowatt-hours, like all other specialized energy units such as food calories, can be converted to joules. You can prove to yourself that 1 kW ⋅ h = 3.6106 J. The electrical energy ( ) used can be reduced either by reducing the time of use or by reducing the power consumption of that appliance or fixture. This will not only reduce the cost, but it will also result in a reduced impact on the environment. Improvements to lighting are some of the fastest ways to reduce the electrical energy used in a home or business. About 20% of a home's use of energy goes to lighting, while the number for commercial establishments is closer to 40%. Fluorescent lights are about four times more efficient than incandescent lights—this is true for both the long tubes and the compact fluorescent lights (CFL). (See Figure 20.17(b).) Thus, a 60-W incandescent bulb can be replaced by a 15-W CFL, which has the same brightness and color. CFLs have a bent tube inside a globe or a spiral-shaped tube, all connected to a standard screw-in base that fits standard incandescent light sockets. (Original problems with color, flicker, shape, and high initial investment for CFLs have been addressed in recent years.) The heat transfer from these CFLs is less, and they last up to 10 times longer. The significance of an investment in such bulbs is addressed in the next example. New white LED lights (which are clusters of small LED bulbs) are even more efficient (twice that of CFLs) and last 5 times longer than CFLs. However, their cost is still high. Making Connections: Energy, Power, and Time The relationship = is one that you will find useful in many different contexts. The energy your body uses in exercise is related to the power level and duration of your activity, for example. The amount of heating by a power source is related to the power level and time it is applied
. Even the radiation dose of an X-ray image is related to the power and time of exposure. Example 20.8 Calculating the Cost Effectiveness of Compact Fluorescent Lights (CFL) If the cost of electricity in your area is 12 cents per kWh, what is the total cost (capital plus operation) of using a 60-W incandescent bulb for 1000 hours (the lifetime of that bulb) if the bulb cost 25 cents? (b) If we replace this bulb with a compact fluorescent light that provides the same light output, but at one-quarter the wattage, and which costs $1.50 but lasts 10 times longer (10,000 hours), what will that total cost be? Strategy 886 Chapter 20 \| Electric Current, Resistance, and Ohm's Law To find the operating cost, we first find the energy used in kilowatt-hours and then multiply by the cost per kilowatt-hour. Solution for (a) The energy used in kilowatt-hours is found by entering the power and time into the expression for energy: = = (60 W)(1000 h) = 60,000 W ⋅ h. In kilowatt-hours, this is Now the electricity cost is = 60.0 kW ⋅ h. cost = (60.0 kW ⋅ h)($0.12/kW ⋅ h) = $7.20. The total cost will be $7.20 for 1000 hours (about one-half year at 5 hours per day). Solution for (b) (20.35) (20.36) (20.37) Since the CFL uses only 15 W and not 60 W, the electricity cost will be $7.20/4 = $1.80. The CFL will last 10 times longer than the incandescent, so that the investment cost will be 1/10 of the bulb cost for that time period of use, or 0.1($1.50) = $0.15. Therefore, the total cost will be $1.95 for 1000 hours. Discussion Therefore, it is much cheaper to use the CFLs, even though the initial investment is higher. The increased cost of labor that a business must include for replacing the incandescent bulbs more often has not been figured in here. Making Connections: Take-Home Experiment—Electrical Energy Use Inventory 1) Make a list of the power ratings on a range of appliances in your home or room.
Explain why something like a toaster has a higher rating than a digital clock. Estimate the energy consumed by these appliances in an average day (by estimating their time of use). Some appliances might only state the operating current. If the household voltage is 120 V, then use =. 2) Check out the total wattage used in the rest rooms of your school's floor or building. (You might need to assume the long fluorescent lights in use are rated at 32 W.) Suppose that the building was closed all weekend and that these lights were left on from 6 p.m. Friday until 8 a.m. Monday. What would this oversight cost? How about for an entire year of weekends? 20.5 Alternating Current versus Direct Current Learning Objectives By the end of this section, you will be able to: • Explain the differences and similarities between AC and DC current. • Calculate rms voltage, current, and average power. • Explain why AC current is used for power transmission. Alternating Current Most of the examples dealt with so far, and particularly those utilizing batteries, have constant voltage sources. Once the current is established, it is thus also a constant. Direct current (DC) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit. Most well-known applications, however, use a time-varying voltage source. Alternating current (AC) is the flow of electric charge that periodically reverses direction. If the source varies periodically, particularly sinusoidally, the circuit is known as an alternating current circuit. Examples include the commercial and residential power that serves so many of our needs. Figure 20.19 shows graphs of voltage and current versus time for typical DC and AC power. The AC voltages and frequencies commonly used in homes and businesses vary around the world. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 901 Figure 20.38 This NASA scientist and NEEMO 5 aquanaut's heart rate and other vital signs are being recorded by a portable device while living in an underwater habitat. (credit: NASA, Life Sciences Data Archive at Johnson Space Center, Houston, Texas) PhET Explorations: Neuron Figure 20.39 Neuron (http://cnx.org/content/m55361/1.2/neuron_en.jar) Stimulate a neuron
and monitor what happens. Pause, rewind, and move forward in time in order to observe the ions as they move across the neuron membrane. Glossary AC current: current that fluctuates sinusoidally with time, expressed as I = I0 sin 2πft, where I is the current at time t, I0 is the peak current, and f is the frequency in hertz AC voltage: voltage that fluctuates sinusoidally with time, expressed as V = V0 sin 2πft, where V is the voltage at time t, V0 is the peak voltage, and f is the frequency in hertz alternating current: (AC) the flow of electric charge that periodically reverses direction ampere: (amp) the SI unit for current; 1 A = 1 C/s bioelectricity: electrical effects in and created by biological systems direct current: (DC) the flow of electric charge in only one direction drift velocity: the average velocity at which free charges flow in response to an electric field electric current: the rate at which charge flows, I = ΔQ/Δt electric power: the rate at which electrical energy is supplied by a source or dissipated by a device; it is the product of current times voltage electrocardiogram (ECG): especially in the heart usually abbreviated ECG, a record of voltages created by depolarization and repolarization, microshock sensitive: a condition in which a person's skin resistance is bypassed, possibly by a medical procedure, rendering the person vulnerable to electrical shock at currents about 1/1000 the normally required level nerve conduction: the transport of electrical signals by nerve cells ohm: the unit of resistance, given by 1Ω = 1 V/A Ohm's law: an empirical relation stating that the current I is proportional to the potential difference V, ∝ V; it is often written as I = V/R, where R is the resistance ohmic: a type of a material for which Ohm's law is valid resistance: the electric property that impedes current; for ohmic materials, it is the ratio of voltage to current, R = V/I 902 resistivity: by ρ an intrinsic property of a material, independent of its shape or size, directly proportional to the resistance, denoted Chapter 20 \| Electric Current, Resistance, and Ohm's Law rms current: the root mean square of the current, rms = 0 / 2, where I0
is the peak current, in an AC system rms voltage: the root mean square of the voltage, rms = 0 / 2, where V0 is the peak voltage, in an AC system semipermeable: property of a membrane that allows only certain types of ions to cross it shock hazard: when electric current passes through a person short circuit: also known as a “short,” a low-resistance path between terminals of a voltage source simple circuit: a circuit with a single voltage source and a single resistor temperature coefficient of resistivity: an empirical quantity, denoted by α, which describes the change in resistance or resistivity of a material with temperature thermal hazard: a hazard in which electric current causes undesired thermal effects Section Summary 20.1 Current • Electric current is the rate at which charge flows, given by = Δ Δ, where Δ is the amount of charge passing through an area in time Δ. • The direction of conventional current is taken as the direction in which positive charge moves. • The SI unit for current is the ampere (A), where 1 A = 1 C/s. • Current is the flow of free charges, such as electrons and ions. • Drift velocity d is the average speed at which these charges move. • Current is proportional to drift velocity d, as expressed in the relationship = d. Here, is the current through a wire of cross-sectional area. The wire's material has a free-charge density, and each carrier has charge and a drift velocity d. • Electrical signals travel at speeds about 1012 times greater than the drift velocity of free electrons. 20.2 Ohm’s Law: Resistance and Simple Circuits • A simple circuit is one in which there is a single voltage source and a single resistance. • One statement of Ohm's law gives the relationship between current, voltage, and resistance in a simple circuit to be =. • Resistance has units of ohms ( Ω ), related to volts and amperes by 1 Ω = 1 V/A. • There is a voltage or drop across a resistor, caused by the current flowing through it, given by =. 20.3 Resistance and Resistivity • The resistance of a cylinder of length and cross-sectional area is =, where is the resistivity of the material. • Values of in Table 20.1 show that materials fall into three groups—conductors, semiconductors, and insulators. • Temperature affects resistivity; for relatively small temperature
changes Δ, resistivity is = 0(1 + Δ), where 0 is the original resistivity and α is the temperature coefficient of resistivity. • Table 20.2 gives values for, the temperature coefficient of resistivity. • The resistance of an object also varies with temperature: = 0(1 + Δ), where 0 is the original resistance, and is the resistance after the temperature change. 20.4 Electric Power and Energy • Electric power is the rate (in watts) that energy is supplied by a source or dissipated by a device. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 903 • Three expressions for electrical power are = = 2, and • The energy used by a device with a power over a time is =. = 2. 20.5 Alternating Current versus Direct Current • Direct current (DC) is the flow of electric current in only one direction. It refers to systems where the source voltage is constant. • The voltage source of an alternating current (AC) system puts out = 0 sin 2, where is the voltage at time, • 0 is the peak voltage, and In a simple circuit, = and AC current is = 0 sin 2, where is the current at time, and 0 = 0 is the peak current. is the frequency in hertz. • The average AC power is ave = 1 20 0. • Average (rms) current rms and average (rms) voltage rms are rms = 0 2 and rms = 0 2, where rms stands for root mean square. • Thus, ave = rmsrms. • Ohm's law for AC is rms = rms. • Expressions for the average power of an AC circuit are ave = rmsrms, ave = rms 2, and ave = rms 2, analogous to the expressions for DC circuits. 20.6 Electric Hazards and the Human Body • The two types of electric hazards are thermal (excessive power) and shock (current through a person). • Shock severity is determined by current, path, duration, and AC frequency. • Table 20.3 lists shock hazards as a function of current. • Figure 20.28 graphs the threshold current for two hazards as a function of frequency. 20.7 Nerve Conduction–Electrocardiograms • Electric potentials in neurons
and other cells are created by ionic concentration differences across semipermeable membranes. • Stimuli change the permeability and create action potentials that propagate along neurons. • Myelin sheaths speed this process and reduce the needed energy input. • This process in the heart can be measured with an electrocardiogram (ECG). Conceptual Questions 20.1 Current 1. Can a wire carry a current and still be neutral—that is, have a total charge of zero? Explain. 2. Car batteries are rated in ampere-hours ( A ⋅ h ). To what physical quantity do ampere-hours correspond (voltage, charge,...), and what relationship do ampere-hours have to energy content? 3. If two different wires having identical cross-sectional areas carry the same current, will the drift velocity be higher or lower in the better conductor? Explain in terms of the equation d =, by considering how the density of charge carriers relates to whether or not a material is a good conductor. 4. Why are two conducting paths from a voltage source to an electrical device needed to operate the device? 5. In cars, one battery terminal is connected to the metal body. How does this allow a single wire to supply current to electrical devices rather than two wires? 6. Why isn't a bird sitting on a high-voltage power line electrocuted? Contrast this with the situation in which a large bird hits two wires simultaneously with its wings. 20.2 Ohm’s Law: Resistance and Simple Circuits 904 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 7. The drop across a resistor means that there is a change in potential or voltage across the resistor. Is there any change in current as it passes through a resistor? Explain. 8. How is the drop in a resistor similar to the pressure drop in a fluid flowing through a pipe? 20.3 Resistance and Resistivity 9. In which of the three semiconducting materials listed in Table 20.1 do impurities supply free charges? (Hint: Examine the range of resistivity for each and determine whether the pure semiconductor has the higher or lower conductivity.) 10. Does the resistance of an object depend on the path current takes through it? Consider, for example, a rectangular bar—is its resistance the same along its length as across its width? (See Figure 20.40.) Figure 20.40 Does current taking two different paths through the same
object encounter different resistance? 11. If aluminum and copper wires of the same length have the same resistance, which has the larger diameter? Why? 12. Explain why = 0(1 + Δ) for the temperature variation of the resistance of an object is not as accurate as = 0(1 + Δ), which gives the temperature variation of resistivity. 20.4 Electric Power and Energy 13. Why do incandescent lightbulbs grow dim late in their lives, particularly just before their filaments break? 14. The power dissipated in a resistor is given by = 2 /, which means power decreases if resistance increases. Yet this power is also given by = 2, which means power increases if resistance increases. Explain why there is no contradiction here. 20.5 Alternating Current versus Direct Current 15. Give an example of a use of AC power other than in the household. Similarly, give an example of a use of DC power other than that supplied by batteries. 16. Why do voltage, current, and power go through zero 120 times per second for 60-Hz AC electricity? 17. You are riding in a train, gazing into the distance through its window. As close objects streak by, you notice that the nearby fluorescent lights make dashed streaks. Explain. 20.6 Electric Hazards and the Human Body 18. Using an ohmmeter, a student measures the resistance between various points on his body. He finds that the resistance between two points on the same finger is about the same as the resistance between two points on opposite hands—both are several hundred thousand ohms. Furthermore, the resistance decreases when more skin is brought into contact with the probes of the ohmmeter. Finally, there is a dramatic drop in resistance (to a few thousand ohms) when the skin is wet. Explain these observations and their implications regarding skin and internal resistance of the human body. 19. What are the two major hazards of electricity? 20. Why isn't a short circuit a shock hazard? 21. What determines the severity of a shock? Can you say that a certain voltage is hazardous without further information? 22. An electrified needle is used to burn off warts, with the circuit being completed by having the patient sit on a large butt plate. Why is this plate large? 23. Some surgery is performed with high-voltage electricity passing from a metal scalpel through the tissue being cut. Considering the nature of electric fields at the surface of conductors, why would you expect most of the current to flow
from the sharp edge of the scalpel? Do you think high- or low-frequency AC is used? 24. Some devices often used in bathrooms, such as hairdryers, often have safety messages saying “Do not use when the bathtub or basin is full of water.” Why is this so? 25. We are often advised to not flick electric switches with wet hands, dry your hand first. We are also advised to never throw water on an electric fire. Why is this so? 26. Before working on a power transmission line, linemen will touch the line with the back of the hand as a final check that the voltage is zero. Why the back of the hand? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 905 27. Why is the resistance of wet skin so much smaller than dry, and why do blood and other bodily fluids have low resistances? 28. Could a person on intravenous infusion (an IV) be microshock sensitive? 29. In view of the small currents that cause shock hazards and the larger currents that circuit breakers and fuses interrupt, how do they play a role in preventing shock hazards? 20.7 Nerve Conduction–Electrocardiograms 30. Note that in Figure 20.31, both the concentration gradient and the Coulomb force tend to move Na+ prevents this? ions into the cell. What 31. Define depolarization, repolarization, and the action potential. 32. Explain the properties of myelinated nerves in terms of the insulating properties of myelin. 906 Chapter 20 \| Electric Current, Resistance, and Ohm's Law Problems & Exercises 20.1 Current 1. What is the current in milliamperes produced by the solar cells of a pocket calculator through which 4.00 C of charge passes in 4.00 h? 2. A total of 600 C of charge passes through a flashlight in 0.500 h. What is the average current? 3. What is the current when a typical static charge of 0.250 C moves from your finger to a metal doorknob in 1.00 s? 4. Find the current when 2.00 nC jumps between your comb and hair over a 0.500 - s time interval. 5. A large lightning bolt had a 20,000-A current and moved 30.0 C of charge
. What was its duration? 6. The 200-A current through a spark plug moves 0.300 mC of charge. How long does the spark last? 7. (a) A defibrillator sends a 6.00-A current through the chest of a patient by applying a 10,000-V potential as in the figure below. What is the resistance of the path? (b) The defibrillator paddles make contact with the patient through a conducting gel that greatly reduces the path resistance. Discuss the difficulties that would ensue if a larger voltage were used to produce the same current through the patient, but with the path having perhaps 50 times the resistance. (Hint: The current must be about the same, so a higher voltage would imply greater power. Use this equation for power: = 2.) Figure 20.41 The capacitor in a defibrillation unit drives a current through the heart of a patient. 8. During open-heart surgery, a defibrillator can be used to bring a patient out of cardiac arrest. The resistance of the path is 500 Ω and a 10.0-mA current is needed. What voltage should be applied? 9. (a) A defibrillator passes 12.0 A of current through the torso of a person for 0.0100 s. How much charge moves? (b) How many electrons pass through the wires connected to the patient? (See figure two problems earlier.) 10. A clock battery wears out after moving 10,000 C of charge through the clock at a rate of 0.500 mA. (a) How long did the clock run? (b) How many electrons per second flowed? This content is available for free at http://cnx.org/content/col11844/1.13 11. The batteries of a submerged non-nuclear submarine supply 1000 A at full speed ahead. How long does it take to move Avogadro's number ( 6.021023 rate? ) of electrons at this 12. Electron guns are used in X-ray tubes. The electrons are accelerated through a relatively large voltage and directed onto a metal target, producing X-rays. (a) How many electrons per second strike the target if the current is 0.500 mA? (b) What charge strikes the target in 0.750 s? 13. A large cyclotron directs a beam of He++ target with a beam current of 0.250 mA. (
a) How many He++ nuclei per second is this? (b) How long does it take for 1.00 C to strike the target? (c) How long before 1.00 mol of He++ nuclei strike the target? nuclei onto a 14. Repeat the above example on Example 20.3, but for a wire made of silver and given there is one free electron per silver atom. 15. Using the results of the above example on Example 20.3, find the drift velocity in a copper wire of twice the diameter and carrying 20.0 A. 16. A 14-gauge copper wire has a diameter of 1.628 mm. What magnitude current flows when the drift velocity is 1.00 mm/s? (See above example on Example 20.3 for useful information.) 17. SPEAR, a storage ring about 72.0 m in diameter at the Stanford Linear Accelerator (closed in 2009), has a 20.0-A circulating beam of electrons that are moving at nearly the speed of light. (See Figure 20.42.) How many electrons are in the beam? Figure 20.42 Electrons circulating in the storage ring called SPEAR constitute a 20.0-A current. Because they travel close to the speed of light, each electron completes many orbits in each second. 20.2 Ohm’s Law: Resistance and Simple Circuits 18. What current flows through the bulb of a 3.00-V flashlight when its hot resistance is 3.60 Ω? 19. Calculate the effective resistance of a pocket calculator that has a 1.35-V battery and through which 0.200 mA flows. 20. What is the effective resistance of a car's starter motor when 150 A flows through it as the car battery applies 11.0 V to the motor? 21. How many volts are supplied to operate an indicator light on a DVD player that has a resistance of 140 Ω, given that 25.0 mA passes through it? 22. (a) Find the voltage drop in an extension cord having a 0.0600- Ω resistance and through which 5.00 A is flowing. (b) A cheaper cord utilizes thinner wire and has a resistance of 0.300 Ω. What is the voltage drop in it when 5.00 A Chapter 20 \| Electric Current, Resistance, and Ohm's Law 907 flows? (c) Why is the voltage to whatever appliance is being used reduced by this amount? What
is the effect on the appliance? 23. A power transmission line is hung from metal towers with glass insulators having a resistance of 1.00109 Ω. What current flows through the insulator if the voltage is 200 kV? (Some high-voltage lines are DC.) low temperatures. Discuss why and whether this is the case here. (Hint: Resistance can't become negative.) 38. Integrated Concepts (a) Redo Exercise 20.25 taking into account the thermal expansion of the tungsten filament. You may assume a thermal expansion coefficient of 1210−6 / ºC. (b) By what percentage does your answer differ from that in the example? 20.3 Resistance and Resistivity 39. Unreasonable Results 24. What is the resistance of a 20.0-m-long piece of 12-gauge copper wire having a 2.053-mm diameter? 25. The diameter of 0-gauge copper wire is 8.252 mm. Find the resistance of a 1.00-km length of such wire used for power transmission. 26. If the 0.100-mm diameter tungsten filament in a light bulb is to have a resistance of 0.200 Ω at 20.0ºC, how long should it be? 27. Find the ratio of the diameter of aluminum to copper wire, if they have the same resistance per unit length (as they might in household wiring). 28. What current flows through a 2.54-cm-diameter rod of pure silicon that is 20.0 cm long, when 1.00 × 103 V is applied to it? (Such a rod may be used to make nuclearparticle detectors, for example.) 29. (a) To what temperature must you raise a copper wire, originally at 20.0ºC, to double its resistance, neglecting any changes in dimensions? (b) Does this happen in household wiring under ordinary circumstances? 30. A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.00% from its value at 20.0ºC. Over what temperature range can it be used? 31. Of what material is a resistor made if its resistance is 40.0% greater at 100ºC than at 20.0ºC? 32. An electronic device designed to operate at any temperature in the range from –10.0ºC to 55.0ºC contains pure carbon resistors. By what factor does
their resistance increase over this range? 33. (a) Of what material is a wire made, if it is 25.0 m long with a 0.100 mm diameter and has a resistance of 77.7 Ω at 20.0ºC? (b) What is its resistance at 150ºC? 34. Assuming a constant temperature coefficient of resistivity, what is the maximum percent decrease in the resistance of a constantan wire starting at 20.0ºC? 35. A wire is drawn through a die, stretching it to four times its original length. By what factor does its resistance increase? 36. A copper wire has a resistance of 0.500 Ω at 20.0ºC, and an iron wire has a resistance of 0.525 Ω at the same temperature. At what temperature are their resistances equal? 37. (a) Digital medical thermometers determine temperature by measuring the resistance of a semiconductor device called a thermistor (which has = – 0.0600 / ºC ) when it is at the same temperature as the patient. What is a patient's temperature if the thermistor's resistance at that temperature is 82.0% of its value at 37.0ºC (normal body temperature)? (b) The negative value for may not be maintained for very (a) To what temperature must you raise a resistor made of constantan to double its resistance, assuming a constant temperature coefficient of resistivity? (b) To cut it in half? (c) What is unreasonable about these results? (d) Which assumptions are unreasonable, or which premises are inconsistent? 20.4 Electric Power and Energy 40. What is the power of a 1.00×102 MV lightning bolt having a current of 2.00 × 104 A? 41. What power is supplied to the starter motor of a large truck that draws 250 A of current from a 24.0-V battery hookup? 42. A charge of 4.00 C of charge passes through a pocket calculator's solar cells in 4.00 h. What is the power output, given the calculator's voltage output is 3.00 V? (See Figure 20.43.) Figure 20.43 The strip of solar cells just above the keys of this calculator convert light to electricity to supply its energy needs. (credit: Evan-Amos, Wikimedia Commons) 43. How many watts does a flashlight that has 6.00×102 C pass through it in 0.500 h use if its voltage is
3.00 V? 44. Find the power dissipated in each of these extension cords: (a) an extension cord having a 0.0600 - Ω resistance and through which 5.00 A is flowing; (b) a cheaper cord utilizing thinner wire and with a resistance of 0.300 Ω. 45. Verify that the units of a volt-ampere are watts, as implied by the equation =. 46. Show that the units 1 V2 / Ω = 1W, as implied by the equation = 2 /. 47. Show that the units 1 A2 ⋅ Ω = 1 W, as implied by the equation = 2. 908 Chapter 20 \| Electric Current, Resistance, and Ohm's Law minute? (b) How much water must you put into the vaporizer for 8.00 h of overnight operation? (See Figure 20.45.) 48. Verify the energy unit equivalence that 1 kW ⋅ h = 3.60106 J. 49. Electrons in an X-ray tube are accelerated through 1.00×102 kV and directed toward a target to produce Xrays. Calculate the power of the electron beam in this tube if it has a current of 15.0 mA. 50. An electric water heater consumes 5.00 kW for 2.00 h per day. What is the cost of running it for one year if electricity costs 12.0 cents/kW ⋅ h? See Figure 20.44. Figure 20.44 On-demand electric hot water heater. Heat is supplied to water only when needed. (credit: aviddavid, Flickr) 60. Integrated Concepts Figure 20.45 This cold vaporizer passes current directly through water, vaporizing it directly with relatively little temperature increase. (a) What energy is dissipated by a lightning bolt having a 20,000-A current, a voltage of 1.00×102 MV, and a length of 1.00 ms? (b) What mass of tree sap could be raised from 18.0ºC to its boiling point and then evaporated by this energy, assuming sap has the same thermal characteristics as water? 61. Integrated Concepts What current must be produced by a 12.0-V battery-operated bottle warmer in order to heat 75.0 g of glass, 250 g of baby formula, and 3.00×102 g of aluminum from 20.0ºC to 90.0ºC in 5.00 min
? 62. Integrated Concepts How much time is needed for a surgical cauterizer to raise the temperature of 1.00 g of tissue from 37.0ºC to 100ºC and then boil away 0.500 g of water, if it puts out 2.00 mA at 15.0 kV? Ignore heat transfer to the surroundings. 63. Integrated Concepts Hydroelectric generators (see Figure 20.46) at Hoover Dam produce a maximum current of 8.00×103 A at 250 kV. (a) What is the power output? (b) The water that powers the generators enters and leaves the system at low speed (thus its kinetic energy does not change) but loses 160 m in altitude. How many cubic meters per second are needed, assuming 85.0% efficiency? 51. With a 1200-W toaster, how much electrical energy is needed to make a slice of toast (cooking time = 1 minute)? At 9.0 cents/kW · h, how much does this cost? 52. What would be the maximum cost of a CFL such that the total cost (investment plus operating) would be the same for both CFL and incandescent 60-W bulbs? Assume the cost of the incandescent bulb is 25 cents and that electricity costs 10 cents/kWh. Calculate the cost for 1000 hours, as in the cost effectiveness of CFL example. 53. Some makes of older cars have 6.00-V electrical systems. (a) What is the hot resistance of a 30.0-W headlight in such a car? (b) What current flows through it? 54. Alkaline batteries have the advantage of putting out constant voltage until very nearly the end of their life. How long will an alkaline battery rated at 1.00 A ⋅ h and 1.58 V keep a 1.00-W flashlight bulb burning? 55. A cauterizer, used to stop bleeding in surgery, puts out 2.00 mA at 15.0 kV. (a) What is its power output? (b) What is the resistance of the path? 56. The average television is said to be on 6 hours per day. Estimate the yearly cost of electricity to operate 100 million TVs, assuming their power consumption averages 150 W and the cost of electricity averages 12.0 cents/kW ⋅ h. 57. An old lightbulb draws only 50.0 W, rather than its original 60
.0 W, due to evaporative thinning of its filament. By what factor is its diameter reduced, assuming uniform thinning along its length? Neglect any effects caused by temperature differences. 58. 00-gauge copper wire has a diameter of 9.266 mm. Calculate the power loss in a kilometer of such wire when it carries 1.00×102 A. 59. Integrated Concepts Cold vaporizers pass a current through water, evaporating it with only a small increase in temperature. One such home device is rated at 3.50 A and utilizes 120 V AC with 95.0% efficiency. (a) What is the vaporization rate in grams per This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 909 (a) An immersion heater utilizing 120 V can raise the temperature of a 1.00×102 -g aluminum cup containing 350 g of water from 20.0ºC to 95.0ºC in 2.00 min. Find its resistance, assuming it is constant during the process. (b) A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance. 68. Integrated Concepts (a) What is the cost of heating a hot tub containing 1500 kg of water from 10.0ºC to 40.0ºC, assuming 75.0% efficiency to account for heat transfer to the surroundings? The cost of electricity is 9 cents/kW ⋅ h. (b) What current was used by the 220-V AC electric heater, if this took 4.00 h? 69. Unreasonable Results (a) What current is needed to transmit 1.00×102 MW of power at 480 V? (b) What power is dissipated by the transmission lines if they have a 1.00 - Ω resistance? (c) What is unreasonable about this result? (d) Which assumptions are unreasonable, or which premises are inconsistent? 70. Unreasonable Results (a) What current is needed to transmit 1.00×102 MW of power at 10.0 kV? (b) Find the resistance of 1.00 km of wire that would cause a 0.0100% power loss. (c) What is the diameter of a 1.00-km-long copper wire having this resistance? (d) What is unreasonable about these results? (e) Which
assumptions are unreasonable, or which premises are inconsistent? 71. Construct Your Own Problem Consider an electric immersion heater used to heat a cup of water to make tea. Construct a problem in which you calculate the needed resistance of the heater so that it increases the temperature of the water and cup in a reasonable amount of time. Also calculate the cost of the electrical energy used in your process. Among the things to be considered are the voltage used, the masses and heat capacities involved, heat losses, and the time over which the heating takes place. Your instructor may wish for you to consider a thermal safety switch (perhaps bimetallic) that will halt the process before damaging temperatures are reached in the immersion unit. 20.5 Alternating Current versus Direct Current 72. (a) What is the hot resistance of a 25-W light bulb that runs on 120-V AC? (b) If the bulb's operating temperature is 2700ºC, what is its resistance at 2600ºC? 73. Certain heavy industrial equipment uses AC power that has a peak voltage of 679 V. What is the rms voltage? 74. A certain circuit breaker trips when the rms current is 15.0 A. What is the corresponding peak current? 75. Military aircraft use 400-Hz AC power, because it is possible to design lighter-weight equipment at this higher frequency. What is the time for one complete cycle of this power? 76. A North American tourist takes his 25.0-W, 120-V AC razor to Europe, finds a special adapter, and plugs it into 240 V AC. Assuming constant resistance, what power does the razor consume as it is ruined? Figure 20.46 Hydroelectric generators at the Hoover dam. (credit: Jon Sullivan) 64. Integrated Concepts (a) Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 750-kg electric car be able to supply: (a) To accelerate from rest to 25.0 m/s in 1.00 min? (b) To climb a 2.00×102 -m high hill in 2.00 min at a constant 25.0-m/s speed while exerting 5.00×102 N of force to overcome air resistance and friction? (c) To travel at a constant 25.0-m/s speed, exerting a 5.00×102 N force to overcome air resistance and friction? See Figure 20.47. Figure 20
.47 This REVAi, an electric car, gets recharged on a street in London. (credit: Frank Hebbert) 65. Integrated Concepts A light-rail commuter train draws 630 A of 650-V DC electricity when accelerating. (a) What is its power consumption rate in kilowatts? (b) How long does it take to reach 20.0 m/s starting from rest if its loaded mass is 5.30104 kg, assuming 95.0% efficiency and constant power? (c) Find its average acceleration. (d) Discuss how the acceleration you found for the light-rail train compares to what might be typical for an automobile. 66. Integrated Concepts (a) An aluminum power transmission line has a resistance of 0.0580 Ω / km. What is its mass per kilometer? (b) What is the mass per kilometer of a copper line having the same resistance? A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance. 67. Integrated Concepts 910 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 90. (a) During surgery, a current as small as 20.0 μA applied directly to the heart may cause ventricular fibrillation. If the resistance of the exposed heart is 300 Ω, what is the smallest voltage that poses this danger? (b) Does your answer imply that special electrical safety precautions are needed? 91. (a) What is the resistance of a 220-V AC short circuit that generates a peak power of 96.8 kW? (b) What would the average power be if the voltage was 120 V AC? 92. A heart defibrillator passes 10.0 A through a patient's torso for 5.00 ms in an attempt to restore normal beating. (a) How much charge passed? (b) What voltage was applied if 500 J of energy was dissipated? (c) What was the path's resistance? (d) Find the temperature increase caused in the 8.00 kg of affected tissue. 93. Integrated Concepts A short circuit in a 120-V appliance cord has a 0.500- Ω resistance. Calculate the temperature rise of the 2.00 g of surrounding materials, assuming their specific heat capacity is 0.200 cal/g⋅ºC and that it takes 0.0500 s for a circuit breaker to interrupt the current. Is this likely to be damaging? 94. Construct Your Own
Problem Consider a person working in an environment where electric currents might pass through her body. Construct a problem in which you calculate the resistance of insulation needed to protect the person from harm. Among the things to be considered are the voltage to which the person might be exposed, likely body resistance (dry, wet, …), and acceptable currents (safe but sensed, safe and unfelt, …). 20.7 Nerve Conduction–Electrocardiograms 95. Integrated Concepts Use the ECG in Figure 20.37 to determine the heart rate in beats per minute assuming a constant time between beats. 96. Integrated Concepts (a) Referring to Figure 20.37, find the time systolic pressure lags behind the middle of the QRS complex. (b) Discuss the reasons for the time lag. 77. In this problem, you will verify statements made at the end of the power losses for Example 20.10. (a) What current is needed to transmit 100 MW of power at a voltage of 25.0 kV? (b) Find the power loss in a 1.00 - Ω transmission line. (c) What percent loss does this represent? 78. A small office-building air conditioner operates on 408-V AC and consumes 50.0 kW. (a) What is its effective resistance? (b) What is the cost of running the air conditioner during a hot summer month when it is on 8.00 h per day for 30 days and electricity costs 9.00 cents/kW ⋅ h? 79. What is the peak power consumption of a 120-V AC microwave oven that draws 10.0 A? 80. What is the peak current through a 500-W room heater that operates on 120-V AC power? 81. Two different electrical devices have the same power consumption, but one is meant to be operated on 120-V AC and the other on 240-V AC. (a) What is the ratio of their resistances? (b) What is the ratio of their currents? (c) Assuming its resistance is unaffected, by what factor will the power increase if a 120-V AC device is connected to 240-V AC? 82. Nichrome wire is used in some radiative heaters. (a) Find the resistance needed if the average power output is to be 1.00 kW utilizing 120-V AC. (b) What length of Nichrome wire, having a cross-sectional area of 5.00mm
2, is needed if the operating temperature is 500º C? (c) What power will it draw when first switched on? 83. Find the time after = 0 when the instantaneous voltage of 60-Hz AC first reaches the following values: (a) 0 / 2 (b) 0 (c) 0. 84. (a) At what two times in the first period following = 0 does the instantaneous voltage in 60-Hz AC equal rms? (b) −rms? 20.6 Electric Hazards and the Human Body 85. (a) How much power is dissipated in a short circuit of 240-V AC through a resistance of 0.250 Ω? (b) What current flows? 86. What voltage is involved in a 1.44-kW short circuit through a 0.100 - Ω resistance? 87. Find the current through a person and identify the likely effect on her if she touches a 120-V AC source: (a) if she is standing on a rubber mat and offers a total resistance of 300 k Ω ; (b) if she is standing barefoot on wet grass and has a resistance of only 4000 k Ω. 88. While taking a bath, a person touches the metal case of a radio. The path through the person to the drainpipe and ground has a resistance of 4000 Ω. What is the smallest voltage on the case of the radio that could cause ventricular fibrillation? 89. Foolishly trying to fish a burning piece of bread from a toaster with a metal butter knife, a man comes into contact with 120-V AC. He does not even feel it since, luckily, he is wearing rubber-soled shoes. What is the minimum resistance of the path the current follows through the person? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 911 Test Prep for AP® Courses 20.1 Current 1. Which of the following can be explained on the basis of conservation of charge in a closed circuit consisting of a battery, resistor, and metal wires? a. The number of electrons leaving the battery will be equal to the number of electrons entering the battery. b. The number of electrons leaving the battery will be less than the number of electrons entering the battery. c. The number of protons leaving the battery will be equal to the number of protons entering the battery. d
. The number of protons leaving the battery will be less than the number of protons entering the battery. 2. When a battery is connected to a bulb, there is 2.5 A of current in the circuit. What amount of charge will flow though the circuit in a time of 0.5 s? a. 0.5 C b. 1 C c. 1.25 C d. 1.5 C 3. If 0.625 × 1020 electrons flow through a circuit each second, what is the current in the circuit? 4. Two students calculate the charge flowing through a circuit. The first student concludes that 300 C of charge flows in 1 minute. The second student concludes that 3.125 × 1019 electrons flow per second. If the current measured in the circuit is 5 A, which of the two students (if any) have performed the calculations correctly? 20.2 Ohm’s Law: Resistance and Simple Circuits 5. If the voltage across a fixed resistance is doubled, what happens to the current? It doubles. It halves. It stays the same. a. b. c. d. The current cannot be determined. Figure 20.48 If the four wires are made from the same material, which of the following is true? Select two answers. a. Resistance of Wire 3 > Resistance of Wire 2 b. Resistance of Wire 1 > Resistance of Wire 2 c. Resistance of Wire 1 < Resistance of Wire 4 d. Resistance of Wire 4 < Resistance of Wire 3 10. Suppose the resistance of a wire is R Ω. What will be the resistance of another wire of the same material having the same length but double the diameter? a. R/2 b. 2R c. R/4 d. 4R 11. The resistances of two wires having the same lengths and cross section areas are 3 Ω and 11 Ω. If the resistivity of the 3 Ω wire is 2.65 × 10−8 Ω·m, find the resistivity of the 1 Ω wire. 12. The lengths and diameters of three wires are given below. If they all have the same resistance, find the ratio of their resistivities. Table 20.5 Wire Length Diameter Wire 1 2 m 1 cm Wire 2 1 m 0.5 cm Wire 3 1 m 1 cm 13. Suppose the resistance of a wire is 2 Ω. If the wire is stretched to three times its length, what will be its resistance? Assume that the volume
does not change. 20.4 Electric Power and Energy 6. The table below gives the voltages and currents recorded across a resistor. 14. Table 20.4 Voltage (V) 2.50 5.00 7.50 10.00 12.50 Current (A) 0.69 1.38 2.09 2.76 3.49 a. Plot the graph and comment on the shape. b. Calculate the value of the resistor. 7. What is the resistance of a bulb if the current in it is 1.25 A when a 4 V voltage supply is connected to it? If the voltage supply is increased to 7 V, what will be the current in the bulb? 20.3 Resistance and Resistivity 8. Which of the following affect the resistivity of a wire? length a. b. area of cross section c. material d. all of the above 9. The lengths and diameters of four wires are given as shown. Figure 20.49 The circuit shown contains a resistor R connected to a voltage supply. The graph shows the total energy E dissipated by the resistance as a function of time. Which of the following shows the corresponding graph for double resistance, i.e., if R is replaced by 2R? 912 Chapter 20 \| Electric Current, Resistance, and Ohm's Law a. Figure 20.50 b. Figure 20.51 c. Figure 20.52 d. Figure 20.53 15. What will be the ratio of the resistance of a 120 W, 220 V lamp to that of a 100 W, 110 V lamp? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 913 21 CIRCUITS, BIOELECTRICITY, AND DC INSTRUMENTS Figure 21.1 Electric circuits in a computer allow large amounts of data to be quickly and accurately analyzed.. (credit: Airman 1st Class Mike Meares, United States Air Force) Chapter Outline 21.1. Resistors in Series and Parallel 21.2. Electromotive Force: Terminal Voltage 21.3. Kirchhoff’s Rules 21.4. DC Voltmeters and Ammeters 21.5. Null Measurements 21.6. DC Circuits Containing Resistors and Capacitors Connection for AP® Courses Electric circuits are commonplace in our everyday lives. Some circuits are simple, such as those in flash
lights while others are extremely complex, such as those used in supercomputers. This chapter takes the topic of electric circuits a step beyond simple circuits by addressing both changes that result from interactions between systems (Big Idea 4) and constraints on such changes due to laws of conservation (Big Idea 5). When the circuit is purely resistive, everything in this chapter applies to both DC and AC. However, matters become more complex when capacitance is involved. We do consider what happens when capacitors are connected to DC voltage sources, but the interaction of capacitors (and other nonresistive devices) with AC sources is left for a later chapter. In addition, a number of important DC instruments, such as meters that measure voltage and current, are covered in this chapter. Information and examples presented in the chapter examine cause-effect relationships inherent in interactions involving electrical systems. The electrical properties of an electric circuit can change due to other systems (Enduring Understanding 4.E). More specifically, values of currents and potential differences in electric circuits depend on arrangements of individual circuit components (Essential Knowledge 4.E.5). In this chapter several series and parallel combinations of resistors are discussed and their effects on currents and potential differences are analyzed. In electric circuits the total energy (Enduring Understanding 5.B) and the total electric charge (Enduring Understanding 5.C) are conserved. Kirchoff’s rules describe both, energy conservation (Essential Knowledge 5.B.9) and charge conservation (Essential 914 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Knowledge 5.C.3). Energy conservation is discussed in terms of the loop rule which specifies that the potential around any closed circuit path must be zero. Charge conservation is applied as conservation of current by equating the sum of all currents entering a junction to the sum of all currents leaving the junction (also known as the junction rule). Kirchoff’s rules are used to calculate currents and potential differences in circuits that combine resistors in series and parallel, and resistors and capacitors. The concepts in this chapter support: Big Idea 4 Interactions between systems can result in changes in those systems. Enduring Understanding 4.E The electric and magnetic properties of a system can change in response to the presence of, or changes in, other objects or systems. Essential Knowledge 4.E.5 The values of currents and electric potential differences in an electric circuit are determined by the properties and arrangement of the individual circuit elements such as sources of em
f, resistors, and capacitors. Big Idea 5 Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.B The energy of a system is conserved. Essential Knowledge 5.B.9 Kirchhoff’s loop rule describes conservation of energy in electrical circuits. Enduring Understanding 5.C The electric charge of a system is conserved. Essential Knowledge 5.C.3 Kirchhoff’s junction rule describes the conservation of electric charge in electrical circuits. Since charge is conserved, current must be conserved at each junction in the circuit. Examples should include circuits that combine resistors in series and parallel. 21.1 Resistors in Series and Parallel By the end of this section, you will be able to: Learning Objectives • Draw a circuit with resistors in parallel and in series. • Use Ohm’s law to calculate the voltage drop across a resistor when current passes through it. • Contrast the way total resistance is calculated for resistors in series and in parallel. • Explain why total resistance of a parallel circuit is less than the smallest resistance of any of the resistors in that circuit. • Calculate total resistance of a circuit that contains a mixture of resistors connected in series and in parallel. The information presented in this section supports the following AP® learning objectives and science practices: • 4.E.5.1 The student is able to make and justify a quantitative prediction of the effect of a change in values or arrangements of one or two circuit elements on the currents and potential differences in a circuit containing a small number of sources of emf, resistors, capacitors, and switches in series and/or parallel. (S.P. 2.2, 6.4) • 4.E.5.2 The student is able to make and justify a qualitative prediction of the effect of a change in values or arrangements of one or two circuit elements on currents and potential differences in a circuit containing a small number of sources of emf, resistors, capacitors, and switches in series and/or parallel. (S.P. 6.1, 6.4) • 4.E.5.3 The student is able to plan data collection strategies and perform data analysis to examine the values of currents and potential differences in an electric circuit that is modified by changing or rearranging circuit elements, including sources of emf, resistors, and capacitors. (S.P. 2.2, 4
.2, 5.1) • 5.B.9.3 The student is able to apply conservation of energy (Kirchhoff’s loop rule) in calculations involving the total electric potential difference for complete circuit loops with only a single battery and resistors in series and/or in, at most, one parallel branch. (S.P. 2.2, 6.4, 7.2) Most circuits have more than one component, called a resistor that limits the flow of charge in the circuit. A measure of this limit on charge flow is called resistance. The simplest combinations of resistors are the series and parallel connections illustrated in Figure 21.2. The total resistance of a combination of resistors depends on both their individual values and how they are connected. Figure 21.2 (a) A series connection of resistors. (b) A parallel connection of resistors. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 915 Resistors in Series When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then 1 in Figure 21.2(a) could be the resistance of the screwdriver’s shaft, 2 the resistance of its handle, 3 the person’s body resistance, and 4 the resistance of her shoes. Figure 21.3 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubbersoled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.) Figure 21.3 Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right). To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 21.3. According to Ohm’s law, the voltage drop,, across a resistor when a current flows through it is calculated using
the equation =, where equals the current in amps (A) and is the resistance in ohms ( Ω ). Another way to think of this is that is the voltage necessary to make a current flow through a resistance. So the voltage drop across 1 is 1 = 1, that across 2 is 2 = 2, and that across 3 is 3 = 3. The sum of these voltages equals the voltage output of the source; that is, = 1 + 2 + 3. (21.1) This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation =, where is the electric charge and is the voltage. Thus the energy supplied by the source is, while that dissipated by the resistors is 1 + 2 + 3. (21.2) Connections: Conservation Laws The derivations of the expressions for series and parallel resistance are based on the laws of conservation of energy and conservation of charge, which state that total charge and total energy are constant in any process. These two laws are directly involved in all electrical phenomena and will be invoked repeatedly to explain both specific effects and the general behavior of electricity. These energies must be equal, because there is no other source and no other destination for energy in the circuit. Thus, = 1 + 2 + 3. The charge cancels, yielding = 1 + 2 + 3, as stated. (Note that the same amount of charge passes through the battery and each resistor in a given amount of time, since there is no capacitance to store charge, there is no place for charge to leak, and charge is conserved.) Now substituting the values for the individual voltages gives = 1 + 2 + 3 = (1 + 2 + 3). Note that for the equivalent single series resistance s, we have = s. This implies that the total or equivalent series resistance s of three resistors is s = 1 + 2 + 3. This logic is valid in general for any number of resistors in series; thus, the total resistance s of a series connection is s = 1 + 2 + 3 +..., (21.3) (21.4) (21.5) 916 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments as proposed. Since all of the current must pass through each resistor, it experiences the resistance of each, and resistances in series simply add up. Example 21.1 Calculating Resistance, Current, Voltage Drop, and Power Dissipation: Analysis of
a Series Circuit Suppose the voltage output of the battery in Figure 21.3 is 12.0 V, and the resistances are 1 = 1.00 Ω, 2 = 6.00 Ω, and 3 = 13.0 Ω. (a) What is the total resistance? (b) Find the current. (c) Calculate the voltage drop in each resistor, and show these add to equal the voltage output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors. Strategy and Solution for (a) The total resistance is simply the sum of the individual resistances, as given by this equation.00 Ω + 6.00 Ω + 13.0 Ω = 20.0 Ω. (21.6) Strategy and Solution for (b) The current is found using Ohm’s law, =. Entering the value of the applied voltage and the total resistance yields the current for the circuit: = s = 12.0 V 20.0 Ω = 0.600 A. (21.7) Strategy and Solution for (c) The voltage—or drop—in a resistor is given by Ohm’s law. Entering the current and the value of the first resistance yields Similarly, and 1 = 1 = (0.600 A)(1.0 Ω ) = 0.600 V. 2 = 2 = (0.600 A)(6.0 Ω ) = 3.60 V 3 = 3 = (0.600 A)(13.0 Ω ) = 7.80 V. Discussion for (c) The three drops add to 12.0 V, as predicted: 1 + 2 + 3 = (0.600 + 3.60 + 7.80) V = 12.0 V. (21.8) (21.9) (21.10) (21.11) Strategy and Solution for (d) The easiest way to calculate power in watts (W) dissipated by a resistor in a DC circuit is to use Joule’s law, =, where is electric power. In this case, each resistor has the same full current flowing through it. By substituting Ohm’s law = into Joule’s law, we get the power dissipated by the first resistor as Similarly, and 1 = 21
= (0.600 A)2(1.00 Ω ) = 0.360 W. 2 = 22 = (0.600 A)2(6.00 Ω ) = 2.16 W 3 = 23 = (0.600 A)2(13.0 Ω ) = 4.68 W. (21.12) (21.13) (21.14) Discussion for (d) Power can also be calculated using either = or = 2 full voltage of the source). The same values will be obtained. Strategy and Solution for (e), where is the voltage drop across the resistor (not the This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 917 The easiest way to calculate power output of the source is to use =, where is the source voltage. This gives = (0.600 A)(12.0 V) = 7.20 W. (21.15) Discussion for (e) Note, coincidentally, that the total power dissipated by the resistors is also 7.20 W, the same as the power put out by the source. That is, 1 + 2 + 3 = (0.360 + 2.16 + 4.68) W = 7.20 W. (21.16) Power is energy per unit time (watts), and so conservation of energy requires the power output of the source to be equal to the total power dissipated by the resistors. Major Features of Resistors in Series 1. Series resistances add: s = 1 + 2 + 3 +.... 2. The same current flows through each resistor in series. 3. Individual resistors in series do not get the total source voltage, but divide it. Resistors in Parallel Figure 21.4 shows resistors in parallel, wired to a voltage source. Resistors are in parallel when each resistor is connected directly to the voltage source by connecting wires having negligible resistance. Each resistor thus has the full voltage of the source applied to it. Each resistor draws the same current it would if it alone were connected to the voltage source (provided the voltage source is not overloaded). For example, an automobile’s headlights, radio, and so on, are wired in parallel, so that they utilize the full voltage of the source and can operate completely independently. The same is true in your house, or any building. (See Figure
21.4(b).) 918 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Figure 21.4 (a) Three resistors connected in parallel to a battery and the equivalent single or parallel resistance. (b) Electrical power setup in a house. (credit: Dmitry G, Wikimedia Commons) To find an expression for the equivalent parallel resistance p, let us consider the currents that flow and how they are related to resistance. Since each resistor in the circuit has the full voltage, the currents flowing through the individual resistors are 1 = 1 of these currents:. Conservation of charge implies that the total current produced by the source is the sum, and 3 = 3, 2 = 2 Substituting the expressions for the individual currents gives = 1 + 2 + 3. Note that Ohm’s law for the equivalent single resistance gives = p = 1 p. (21.17) (21.18) (21.19) The terms inside the parentheses in the last two equations must be equal. Generalizing to any number of resistors, the total resistance p of a parallel connection is related to the individual resistances by 3 +.... (21.20) This relationship results in a total resistance p that is less than the smallest of the individual resistances. (This is seen in the next example.) When resistors are connected in parallel, more current flows from the source than would flow for any of them individually, and so the total resistance is lower. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 919 Example 21.2 Calculating Resistance, Current, Power Dissipation, and Power Output: Analysis of a Parallel Circuit Let the voltage output of the battery and resistances in the parallel connection in Figure 21.4 be the same as the previously considered series connection: = 12.0 V, 1 = 1.00 Ω, 2 = 6.00 Ω, and 3 = 13.0 Ω. (a) What is the total resistance? (b) Find the total current. (c) Calculate the currents in each resistor, and show these add to equal the total current output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors. Strategy and Solution for (a
) The total resistance for a parallel combination of resistors is found using the equation below. Entering known values gives.00 Ω + 1 6.00 Ω + 1 13.0 Ω. Thus, 1 p = 1.00 Ω + 0.1667 Ω + 0.07692 Ω = 1.2436 Ω. (Note that in these calculations, each intermediate answer is shown with an extra digit.) We must invert this to find the total resistance p. This yields The total resistance with the correct number of significant digits is p = 0.804 Ω. p = 1 1.2436 Ω = 0.8041 Ω. Discussion for (a) p is, as predicted, less than the smallest individual resistance. Strategy and Solution for (b) The total current can be found from Ohm’s law, substituting p for the total resistance. This gives = p = 12.0 V 0.8041 Ω = 14.92 A. (21.21) (21.22) (21.23) (21.24) Discussion for (b) Current for each device is much larger than for the same devices connected in series (see the previous example). A circuit with parallel connections has a smaller total resistance than the resistors connected in series. Strategy and Solution for (c) The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage. Thus, Similarly, and 1 = 1 = 12.0 V 1.00 Ω = 12.0 A. 2 = 2 = 12.0 V 6.00 Ω = 2.00 A 3 = 3 = 12.0 V 13.0 Ω = 0.92 A. Discussion for (c) The total current is the sum of the individual currents: 1 + 2 + 3 = 14.92 A. This is consistent with conservation of charge. Strategy and Solution for (d) The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three are known. Let us use = 2, since each resistor gets full voltage. Thus, (21.25) (21.26) (21.27) (21.28) 920 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 1 = 2 1 = (12.0 V)2 1.00 Ω = 144 W. 2 = 2 2 =
(12.0 V)2 6.00 Ω = 24.0 W 3 = 2 3 = (12.0 V)2 13.0 Ω = 11.1 W. Similarly, and Discussion for (d) (21.29) (21.30) (21.31) The power dissipated by each resistor is considerably higher in parallel than when connected in series to the same voltage source. Strategy and Solution for (e) The total power can also be calculated in several ways. Choosing =, and entering the total current, yields = = (14.92 A)(12.0 V) = 179 W. Discussion for (e) Total power dissipated by the resistors is also 179 W: 1 + 2 + 3 = 144 W + 24.0 W + 11.1 W = 179 W. This is consistent with the law of conservation of energy. Overall Discussion Note that both the currents and powers in parallel connections are greater than for the same devices in series. (21.32) (21.33) Major Features of Resistors in Parallel 1. Parallel resistance is found from combination. +..., and it is smaller than any individual resistance in the 2. Each resistor in parallel has the same full voltage of the source applied to it. (Power distribution systems most often use parallel connections to supply the myriad devices served with the same voltage and to allow them to operate independently.) 3. Parallel resistors do not each get the total current; they divide it. Combinations of Series and Parallel More complex connections of resistors are sometimes just combinations of series and parallel. These are commonly encountered, especially when wire resistance is considered. In that case, wire resistance is in series with other resistances that are in parallel. Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in Figure 21.5. Various parts are identified as either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left. The process is more time consuming than difficult. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 921 Figure 21.5 This combination of seven resistors has both series and parallel parts. Each is identified and reduced to an equivalent resistance, and these are further reduced until a single equivalent resistance is reached. The simplest combination of series and parallel resistance, shown in Figure 21.6, is also
the most instructive, since it is found in many applications. For example, 1 could be the resistance of wires from a car battery to its electrical devices, which are in parallel. 2 and 3 could be the starter motor and a passenger compartment light. We have previously assumed that wire resistance is negligible, but, when it is not, it has important effects, as the next example indicates. Example 21.3 Calculating Resistance, IR Drop, Current, and Power Dissipation: Combining Series and Parallel Circuits Figure 21.6 shows the resistors from the previous two examples wired in a different way—a combination of series and parallel. We can consider 1 to be the resistance of wires leading to 2 and 3. (a) Find the total resistance. (b) What is the drop in 1? (c) Find the current 2 through 2. (d) What power is dissipated by 2? Figure 21.6 These three resistors are connected to a voltage source so that 2 and 3 are in parallel with one another and that combination is in series with 1. Strategy and Solution for (a) To find the total resistance, we note that 2 and 3 are in parallel and their combination p is in series with 1. Thus the total (equivalent) resistance of this combination is tot = 1 + p. First, we find p using the equation for resistors in parallel and entering known values.00 Ω + 1 13.0 Ω = 0.2436 Ω. (21.34) (21.35) 922 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Inverting gives So the total resistance is Discussion for (a) p = 1 0.2436 Ω = 4.11 Ω. tot = 1 + p = 1.00 Ω + 4.11 Ω = 5.11 Ω. (21.36) (21.37) The total resistance of this combination is intermediate between the pure series and pure parallel values ( 20.0 Ω and 0.804 Ω, respectively) found for the same resistors in the two previous examples. Strategy and Solution for (b) To find the drop in 1, we note that the full current flows through 1. Thus its drop is We must find before we can calculate 1. The total current is found using Ohm’s law for the circuit. That is, 1 =. = tot = 12.0 V 5.11 Ω = 2.35
A. Entering this into the expression above, we get 1 = = (2.35 A)(1.00 Ω ) = 2.35 V. Discussion for (b) (21.38) (21.39) (21.40) The voltage applied to 2 and 3 is less than the total voltage by an amount 1. When wire resistance is large, it can significantly affect the operation of the devices represented by 2 and 3. Strategy and Solution for (c) To find the current through 2, we must first find the voltage applied to it. We call this voltage p, because it is applied to a parallel combination of resistors. The voltage applied to both 2 and 3 is reduced by the amount 1, and so it is Now the current 2 through resistance 2 is found using Ohm’s law: p = − 1 = 12.0 V − 2.35 V = 9.65 V. 2 = p 2 = 9.65 V 6.00 Ω = 1.61 A. (21.41) (21.42) Discussion for (c) The current is less than the 2.00 A that flowed through 2 when it was connected in parallel to the battery in the previous parallel circuit example. Strategy and Solution for (d) The power dissipated by 2 is given by 2 = (2 )22 = (1.61 A)2(6.00 Ω ) = 15.5 W. (21.43) Discussion for (d) The power is less than the 24.0 W this resistor dissipated when connected in parallel to the 12.0-V source. Applying the Science Practices: Circuit Construction Kit (DC only) Plan an experiment to analyze the effect on currents and potential differences due to rearrangement of resistors and variations in voltage sources. Your experimental investigation should include data collection for at least five different scenarios of rearranged resistors (i.e., several combinations of series and parallel) and three scenarios of different voltage sources. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 923 Practical Implications One implication of this last example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relatively large, as in a worn (or a very long) extension cord, then this loss can be significant. If a large current is drawn, the drop
in the wires can also be significant. For example, when you are rummaging in the refrigerator and the motor comes on, the refrigerator light dims momentarily. Similarly, you can see the passenger compartment light dim when you start the engine of your car (although this may be due to resistance inside the battery itself). What is happening in these high-current situations is illustrated in Figure 21.7. The device represented by 3 has a very low resistance, and so when it is switched on, a large current flows. This increased current causes a larger drop in the wires represented by 1, reducing the voltage across the light bulb (which is 2 ), which then dims noticeably. Figure 21.7 Why do lights dim when a large appliance is switched on? The answer is that the large current the appliance motor draws causes a significant drop in the wires and reduces the voltage across the light. Check Your Understanding Can any arbitrary combination of resistors be broken down into series and parallel combinations? See if you can draw a circuit diagram of resistors that cannot be broken down into combinations of series and parallel. Solution No, there are many ways to connect resistors that are not combinations of series and parallel, including loops and junctions. In such cases Kirchhoff’s rules, to be introduced in Kirchhoff’s Rules, will allow you to analyze the circuit. Problem-Solving Strategies for Series and Parallel Resistors 1. Draw a clear circuit diagram, labeling all resistors and voltage sources. This step includes a list of the knowns for the problem, since they are labeled in your circuit diagram. 2. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. 3. Determine whether resistors are in series, parallel, or a combination of both series and parallel. Examine the circuit diagram to make this assessment. Resistors are in series if the same current must pass sequentially through them. 4. Use the appropriate list of major features for series or parallel connections to solve for the unknowns. There is one list for series and another for parallel. If your problem has a combination of series and parallel, reduce it in steps by considering individual groups of series or parallel connections, as done in this module and the examples. Special note: When finding, the reciprocal must be taken with care. 5. Check to see whether the answers are reasonable and consistent. Units and numerical results must be reasonable. Total series resistance should be greater, whereas
total parallel resistance should be smaller, for example. Power should be greater for the same devices in parallel compared with series, and so on. 21.2 Electromotive Force: Terminal Voltage By the end of this section, you will be able to: Learning Objectives 924 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments • Compare and contrast the voltage and the electromagnetic force of an electric power source. • Describe what happens to the terminal voltage, current, and power delivered to a load as internal resistance of the voltage source increases. • Explain why it is beneficial to use more than one voltage source connected in parallel. The information presented in this section supports the following AP® learning objectives and science practices: • 5.B.9.7 The student is able to refine and analyze a scientific question for an experiment using Kirchhoff’s loop rule for circuits that includes determination of internal resistance of the battery and analysis of a nonohmic resistor. (S.P. 4.1, 4.2, 5.1, 5.3) When you forget to turn off your car lights, they slowly dim as the battery runs down. Why don’t they simply blink off when the battery’s energy is gone? Their gradual dimming implies that battery output voltage decreases as the battery is depleted. Furthermore, if you connect an excessive number of 12-V lights in parallel to a car battery, they will be dim even when the battery is fresh and even if the wires to the lights have very low resistance. This implies that the battery’s output voltage is reduced by the overload. The reason for the decrease in output voltage for depleted or overloaded batteries is that all voltage sources have two fundamental parts—a source of electrical energy and an internal resistance. Let us examine both. Electromotive Force You can think of many different types of voltage sources. Batteries themselves come in many varieties. There are many types of mechanical/electrical generators, driven by many different energy sources, ranging from nuclear to wind. Solar cells create voltages directly from light, while thermoelectric devices create voltage from temperature differences. A few voltage sources are shown in Figure 21.8. All such devices create a potential difference and can supply current if connected to a resistance. On the small scale, the potential difference creates an electric field that exerts force on charges, causing current. We thus use the name electromotive force, abbreviated emf. Emf is not a force at all
; it is a special type of potential difference. To be precise, the electromotive force (emf) is the potential difference of a source when no current is flowing. Units of emf are volts. Figure 21.8 A variety of voltage sources (clockwise from top left): the Brazos Wind Farm in Fluvanna, Texas (credit: Leaflet, Wikimedia Commons); the Krasnoyarsk Dam in Russia (credit: Alex Polezhaev); a solar farm (credit: U.S. Department of Energy); and a group of nickel metal hydride batteries (credit: Tiaa Monto). The voltage output of each depends on its construction and load, and equals emf only if there is no load. Electromotive force is directly related to the source of potential difference, such as the particular combination of chemicals in a battery. However, emf differs from the voltage output of the device when current flows. The voltage across the terminals of a battery, for example, is less than the emf when the battery supplies current, and it declines further as the battery is depleted or loaded down. However, if the device’s output voltage can be measured without drawing current, then output voltage will equal emf (even for a very depleted battery). Internal Resistance As noted before, a 12-V truck battery is physically larger, contains more charge and energy, and can deliver a larger current than a 12-V motorcycle battery. Both are lead-acid batteries with identical emf, but, because of its size, the truck battery has a smaller internal resistance. Internal resistance is the inherent resistance to the flow of current within the source itself. Figure 21.9 is a schematic representation of the two fundamental parts of any voltage source. The emf (represented by a script E in the figure) and internal resistance are in series. The smaller the internal resistance for a given emf, the more current and the more power the source can supply. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 925 Figure 21.9 Any voltage source (in this case, a carbon-zinc dry cell) has an emf related to its source of potential difference, and an internal resistance related to its construction. (Note that the script E stands for emf.). Also shown are the output terminals across which the terminal voltage is measured. Since = emf
−, terminal voltage equals emf only if there is no current flowing. The internal resistance can behave in complex ways. As noted, increases as a battery is depleted. But internal resistance may also depend on the magnitude and direction of the current through a voltage source, its temperature, and even its history. The internal resistance of rechargeable nickel-cadmium cells, for example, depends on how many times and how deeply they have been depleted. Things Great and Small: The Submicroscopic Origin of Battery Potential Various types of batteries are available, with emfs determined by the combination of chemicals involved. We can view this as a molecular reaction (what much of chemistry is about) that separates charge. The lead-acid battery used in cars and other vehicles is one of the most common types. A single cell (one of six) of this battery is seen in Figure 21.10. The cathode (positive) terminal of the cell is connected to a lead oxide plate, while the anode (negative) terminal is connected to a lead plate. Both plates are immersed in sulfuric acid, the electrolyte for the system. Figure 21.10 Artist’s conception of a lead-acid cell. Chemical reactions in a lead-acid cell separate charge, sending negative charge to the anode, which is connected to the lead plates. The lead oxide plates are connected to the positive or cathode terminal of the cell. Sulfuric acid conducts the charge as well as participating in the chemical reaction. The details of the chemical reaction are left to the reader to pursue in a chemistry text, but their results at the molecular level help explain the potential created by the battery. Figure 21.11 shows the result of a single chemical reaction. Two electrons are placed on the anode, making it negative, provided that the cathode supplied two electrons. This leaves the cathode positively charged, because it has lost two electrons. In short, a separation of charge has been driven by a chemical reaction. Note that the reaction will not take place unless there is a complete circuit to allow two electrons to be supplied to the cathode. Under many circumstances, these electrons come from the anode, flow through a resistance, and return to the cathode. Note also that since the chemical reactions involve substances with resistance, it is not possible to create the emf without an internal resistance. 926 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Figure 21.11 Artist’s conception of two electrons being forced
onto the anode of a cell and two electrons being removed from the cathode of the cell. The chemical reaction in a lead-acid battery places two electrons on the anode and removes two from the cathode. It requires a closed circuit to proceed, since the two electrons must be supplied to the cathode. Why are the chemicals able to produce a unique potential difference? Quantum mechanical descriptions of molecules, which take into account the types of atoms and numbers of electrons in them, are able to predict the energy states they can have and the energies of reactions between them. In the case of a lead-acid battery, an energy of 2 eV is given to each electron sent to the anode. Voltage is defined as the electrical potential energy divided by charge: = E. An electron volt is the energy given to a single electron by a voltage of 1 V. So the voltage here is 2 V, since 2 eV is given to each electron. It is the energy produced in each molecular reaction that produces the voltage. A different reaction produces a different energy and, hence, a different voltage. Terminal Voltage The voltage output of a device is measured across its terminals and, thus, is called its terminal voltage. Terminal voltage is given by = emf −, (21.44) where is the internal resistance and is the current flowing at the time of the measurement. is positive if current flows away from the positive terminal, as shown in Figure 21.9. You can see that the larger the current, the smaller the terminal voltage. And it is likewise true that the larger the internal resistance, the smaller the terminal voltage. Suppose a load resistance load is connected to a voltage source, as in Figure 21.12. Since the resistances are in series, the total resistance in the circuit is load +. Thus the current is given by Ohm’s law to be = emf load +. (21.45) Figure 21.12 Schematic of a voltage source and its load load. Since the internal resistance is in series with the load, it can significantly affect the terminal voltage and current delivered to the load. (Note that the script E stands for emf.) We see from this expression that the smaller the internal resistance, the greater the current the voltage source supplies to its load load. As batteries are depleted, increases. If becomes a significant fraction of the load resistance, then the current is significantly reduced, as the following example illustrates. This content is available for free at http://cnx.
org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 927 Example 21.4 Calculating Terminal Voltage, Power Dissipation, Current, and Resistance: Terminal Voltage and Load A certain battery has a 12.0-V emf and an internal resistance of 0.100 Ω. (a) Calculate its terminal voltage when connected to a 10.0- Ω load. (b) What is the terminal voltage when connected to a 0.500- Ω load? (c) What power does the 0.500- Ω load dissipate? (d) If the internal resistance grows to 0.500 Ω, find the current, terminal voltage, and power dissipated by a 0.500- Ω load. Strategy The analysis above gave an expression for current when internal resistance is taken into account. Once the current is found, the terminal voltage can be calculated using the equation = emf −. Once current is found, the power dissipated by a resistor can also be found. Solution for (a) Entering the given values for the emf, load resistance, and internal resistance into the expression above yields = emf load + = 12.0 V 10.1 Ω = 1.188 A. Enter the known values into the equation = emf − to get the terminal voltage: = emf − = 12.0 V − (1.188 A)(0.100 Ω) = 11.9 V. Discussion for (a) (21.46) (21.47) The terminal voltage here is only slightly lower than the emf, implying that 10.0 Ω is a light load for this particular battery. Solution for (b) Similarly, with load = 0.500 Ω, the current is = emf load + = 12.0 V 0.600 Ω = 20.0 A. = emf − = 12.0 V − (20.0 A)(0.100 Ω) = 10.0 V. The terminal voltage is now Discussion for (b) (21.48) (21.49) This terminal voltage exhibits a more significant reduction compared with emf, implying 0.500 Ω is a heavy load for this battery. Solution for (c) The power dissipated by the 0.500 - Ω load can be found using the formula = 2. Entering the known values gives load = 2load = (20.0
A)2(0.500 Ω) = 2.00×102 W. (21.50) Discussion for (c) Note that this power can also be obtained using the expressions 2 this case). Solution for (d) or, where is the terminal voltage (10.0 V in Here the internal resistance has increased, perhaps due to the depletion of the battery, to the point where it is as great as the load resistance. As before, we first find the current by entering the known values into the expression, yielding Now the terminal voltage is = emf load + = 12.0 V 1.00 Ω = 12.0 A. = emf − = 12.0 V − (12.0 A)(0.500 Ω) = 6.00 V, and the power dissipated by the load is (21.51) (21.52) 928 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments load = 2load = (12.0 A)2(0.500 Ω ) = 72.0 W. (21.53) Discussion for (d) We see that the increased internal resistance has significantly decreased terminal voltage, current, and power delivered to a load. Applying the Science Practices: Internal Resistance The internal resistance of a battery can be estimated using a simple activity. The circuit shown in the figure below includes a resistor R in series with a battery along with an ammeter and voltmeter to measure the current and voltage respectively. Figure 21.13 The currents and voltages measured for several R values are shown in the table below. Using the data given in the table and applying graphical analysis, determine the emf and internal resistance of the battery. Your response should clearly explain the method used to obtain the result. Table 21.1 Resistance Current (A) Voltage (V) R1 R2 R3 R4 3.53 2.07 1.46 1.13 4.24 4.97 5.27 5.43 Answer Plot the measured currents and voltages on a graph. The terminal voltage of a battery is equal to the emf of the battery minus the voltage drop across the internal resistance of the battery or V = emf – Ir. Using this linear relationship and the plotted graph, the internal resistance and emf of the battery can be found. The graph for this case is shown below. The equation is V = -0.50I + 6.0 and hence the internal resistance will be equal to 0.
5 Ω and emf will be equal to 6 V. Figure 21.14 Battery testers, such as those in Figure 21.15, use small load resistors to intentionally draw current to determine whether the terminal voltage drops below an acceptable level. They really test the internal resistance of the battery. If internal resistance is high, the battery is weak, as evidenced by its low terminal voltage. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 929 Figure 21.15 These two battery testers measure terminal voltage under a load to determine the condition of a battery. The large device is being used by a U.S. Navy electronics technician to test large batteries aboard the aircraft carrier USS Nimitz and has a small resistance that can dissipate large amounts of power. (credit: U.S. Navy photo by Photographer’s Mate Airman Jason A. Johnston) The small device is used on small batteries and has a digital display to indicate the acceptability of their terminal voltage. (credit: Keith Williamson) Some batteries can be recharged by passing a current through them in the direction opposite to the current they supply to a resistance. This is done routinely in cars and batteries for small electrical appliances and electronic devices, and is represented pictorially in Figure 21.16. The voltage output of the battery charger must be greater than the emf of the battery to reverse current through it. This will cause the terminal voltage of the battery to be greater than the emf, since = emf −, and is now negative. Figure 21.16 A car battery charger reverses the normal direction of current through a battery, reversing its chemical reaction and replenishing its chemical potential. Multiple Voltage Sources There are two voltage sources when a battery charger is used. Voltage sources connected in series are relatively simple. When voltage sources are in series, their internal resistances add and their emfs add algebraically. (See Figure 21.17.) Series connections of voltage sources are common—for example, in flashlights, toys, and other appliances. Usually, the cells are in series in order to produce a larger total emf. But if the cells oppose one another, such as when one is put into an appliance backward, the total emf is less, since it is the algebraic sum of the individual emfs. A battery is a multiple connection of voltaic cells, as shown in Figure 21.18
. The disadvantage of series connections of cells is that their internal resistances add. One of the authors once owned a 1957 MGA that had two 6-V batteries in series, rather than a single 12-V battery. This arrangement produced a large internal resistance that caused him many problems in starting the engine. Figure 21.17 A series connection of two voltage sources. The emfs (each labeled with a script E) and internal resistances add, giving a total emf of emf1 + emf2 and a total internal resistance of 1 + 2. 930 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Figure 21.18 Batteries are multiple connections of individual cells, as shown in this modern rendition of an old print. Single cells, such as AA or C cells, are commonly called batteries, although this is technically incorrect. If the series connection of two voltage sources is made into a complete circuit with the emfs in opposition, then a current of magnitude = emf1 – emf2 1 + 2 flows. See Figure 21.19, for example, which shows a circuit exactly analogous to the battery charger discussed above. If two voltage sources in series with emfs in the same sense are connected to a load load, as in Figure 21.20, then = emf1 + emf2 1 + 2 + load flows. Figure 21.19 These two voltage sources are connected in series with their emfs in opposition. Current flows in the direction of the greater emf and is limited to = emf1 − emf2 1 + 2 by the sum of the internal resistances. (Note that each emf is represented by script E in the figure.) A battery charger connected to a battery is an example of such a connection. The charger must have a larger emf than the battery to reverse current through it. Figure 21.20 This schematic represents a flashlight with two cells (voltage sources) and a single bulb (load resistance) in series. The current that flows is = emf1 + emf2 1 + 2 + load. (Note that each emf is represented by script E in the figure.) Take-Home Experiment: Flashlight Batteries Find a flashlight that uses several batteries and find new and old batteries. Based on the discussions in this module, predict the brightness of the flashlight when different combinations of batteries are used. Do your predictions match what you observe? Now place new batteries in the flashlight and leave the flashlight switched on for several hours
. Is the flashlight still quite bright? Do the same with the old batteries. Is the flashlight as bright when left on for the same length of time with old and new batteries? What does this say for the case when you are limited in the number of available new batteries? Figure 21.21 shows two voltage sources with identical emfs in parallel and connected to a load resistance. In this simple case, the total emf is the same as the individual emfs. But the total internal resistance is reduced, since the internal resistances are in parallel. The parallel connection thus can produce a larger current. Here, = emf tot + load flows through the load, and tot is less than those of the individual batteries. For example, some diesel-powered cars use two 12-V batteries in parallel; they produce a total emf of 12 V but can deliver the larger current needed to start a diesel engine. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 931 Figure 21.21 Two voltage sources with identical emfs (each labeled by script E) connected in parallel produce the same emf but have a smaller total internal resistance than the individual sources. Parallel combinations are often used to deliver more current. Here = flows emf tot + load through the load. Animals as Electrical Detectors A number of animals both produce and detect electrical signals. Fish, sharks, platypuses, and echidnas (spiny anteaters) all detect electric fields generated by nerve activity in prey. Electric eels produce their own emf through biological cells (electric organs) called electroplaques, which are arranged in both series and parallel as a set of batteries. Electroplaques are flat, disk-like cells; those of the electric eel have a voltage of 0.15 V across each one. These cells are usually located toward the head or tail of the animal, although in the case of the electric eel, they are found along the entire body. The electroplaques in the South American eel are arranged in 140 rows, with each row stretching horizontally along the body and containing 5,000 electroplaques. This can yield an emf of approximately 600 V, and a current of 1 A—deadly. The mechanism for detection of external electric fields is similar to that for producing nerve signals in the cell through depolarization and repolarization—the movement of ions across
the cell membrane. Within the fish, weak electric fields in the water produce a current in a gel-filled canal that runs from the skin to sensing cells, producing a nerve signal. The Australian platypus, one of the very few mammals that lay eggs, can detect fields of 30 mV m, while sharks have been found to be able to sense a field in their snouts as small as 100 mV m (Figure 21.22). Electric eels use their own electric fields produced by the electroplaques to stun their prey or enemies. Figure 21.22 Sand tiger sharks (Carcharias taurus), like this one at the Minnesota Zoo, use electroreceptors in their snouts to locate prey. (credit: Jim Winstead, Flickr) Solar Cell Arrays Another example dealing with multiple voltage sources is that of combinations of solar cells—wired in both series and parallel combinations to yield a desired voltage and current. Photovoltaic generation (PV), the conversion of sunlight directly into 932 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments electricity, is based upon the photoelectric effect, in which photons hitting the surface of a solar cell create an electric current in the cell. Most solar cells are made from pure silicon—either as single-crystal silicon, or as a thin film of silicon deposited upon a glass or metal backing. Most single cells have a voltage output of about 0.5 V, while the current output is a function of the amount of sunlight upon the cell (the incident solar radiation—the insolation). Under bright noon sunlight, a current of about 100 mA/cm2 of cell surface area is produced by typical single-crystal cells. Individual solar cells are connected electrically in modules to meet electrical-energy needs. They can be wired together in series or in parallel—connected like the batteries discussed earlier. A solar-cell array or module usually consists of between 36 and 72 cells, with a power output of 50 W to 140 W. The output of the solar cells is direct current. For most uses in a home, AC is required, so a device called an inverter must be used to convert the DC to AC. Any extra output can then be passed on to the outside electrical grid for sale to the utility. Take-Home Experiment: Virtual Solar Cells One can assemble a “virtual” solar cell array by using playing cards, or business or index cards, to represent a solar cell. Combinations of these cards in series
and/or parallel can model the required array output. Assume each card has an output of 0.5 V and a current (under bright light) of 2 A. Using your cards, how would you arrange them to produce an output of 6 A at 3 V (18 W)? Suppose you were told that you needed only 18 W (but no required voltage). Would you need more cards to make this arrangement? 21.3 Kirchhoff’s Rules By the end of this section, you will be able to: Learning Objectives • Analyze a complex circuit using Kirchhoff’s rules, applying the conventions for determining the correct signs of various terms. The information presented in this section supports the following AP® learning objectives and science practices: • 5.B.9.1 The student is able to construct or interpret a graph of the energy changes within an electrical circuit with only a single battery and resistors in series and/or in, at most, one parallel branch as an application of the conservation of energy (Kirchhoff’s loop rule). (S.P. 1.1, 1.4) • 5.B.9.2 The student is able to apply conservation of energy concepts to the design of an experiment that will demonstrate the validity of Kirchhoff’s loop rule in a circuit with only a battery and resistors either in series or in, at most, one pair of parallel branches. (S.P. 4.2, 6.4, 7.2) • 5.B.9.3 The student is able to apply conservation of energy (Kirchhoff’s loop rule) in calculations involving the total electric potential difference for complete circuit loops with only a single battery and resistors in series and/or in, at most, one parallel branch. (S.P. 2.2, 6.4, 7.2) • 5.B.9.4 The student is able to analyze experimental data including an analysis of experimental uncertainty that will demonstrate the validity of Kirchhoff’s loop rule. (S.P. 5.1) • 5.B.9.5 The student is able to use conservation of energy principles (Kirchhoff’s loop rule) to describe and make predictions regarding electrical potential difference, charge, and current in steady-state circuits composed of various combinations of resistors and capacitors. (S.P. 6.4) • 5.C.
3.1 The student is able to apply conservation of electric charge (Kirchhoff’s junction rule) to the comparison of electric current in various segments of an electrical circuit with a single battery and resistors in series and in, at most, one parallel branch and predict how those values would change if configurations of the circuit are changed. (S.P. 6.4, 7.2) • 5.C.3.2 The student is able to design an investigation of an electrical circuit with one or more resistors in which evidence of conservation of electric charge can be collected and analyzed. (S.P. 4.1, 4.2, 5.1) • 5.C.3.3 The student is able to use a description or schematic diagram of an electrical circuit to calculate unknown values of current in various segments or branches of the circuit. (S.P. 1.4, 2.2) • 5.C.3.4 The student is able to predict or explain current values in series and parallel arrangements of resistors and other branching circuits using Kirchhoff’s junction rule and relate the rule to the law of charge conservation. (S.P. 6.4, 7.2) • 5.C.3.5 The student is able to determine missing values and direction of electric current in branches of a circuit with resistors and NO capacitors from values and directions of current in other branches of the circuit through appropriate selection of nodes and application of the junction rule. (S.P. 1.4, 2.2) Many complex circuits, such as the one in Figure 21.23, cannot be analyzed with the series-parallel techniques developed in Resistors in Series and Parallel and Electromotive Force: Terminal Voltage. There are, however, two circuit analysis rules that can be used to analyze any circuit, simple or complex. These rules are special cases of the laws of conservation of charge and conservation of energy. The rules are known as Kirchhoff’s rules, after their inventor Gustav Kirchhoff (1824–1887). This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 933 Figure 21.23 This circuit cannot be reduced to a combination of series and parallel connections. Kirchhoff’s rules, special applications of the laws of conservation of charge and energy, can be used to
analyze it. (Note: The script E in the figure represents electromotive force, emf.) Kirchhoff’s Rules • Kirchhoff’s first rule—the junction rule. The sum of all currents entering a junction must equal the sum of all currents leaving the junction. • Kirchhoff’s second rule—the loop rule. The algebraic sum of changes in potential around any closed circuit path (loop) must be zero. Explanations of the two rules will now be given, followed by problem-solving hints for applying Kirchhoff’s rules, and a worked example that uses them. Kirchhoff’s First Rule Kirchhoff’s first rule (the junction rule) is an application of the conservation of charge to a junction; it is illustrated in Figure 21.24. Current is the flow of charge, and charge is conserved; thus, whatever charge flows into the junction must flow out. Kirchhoff’s first rule requires that 1 = 2 + 3 (see figure). Equations like this can and will be used to analyze circuits and to solve circuit problems. Making Connections: Conservation Laws Kirchhoff’s rules for circuit analysis are applications of conservation laws to circuits. The first rule is the application of conservation of charge, while the second rule is the application of conservation of energy. Conservation laws, even used in a specific application, such as circuit analysis, are so basic as to form the foundation of that application. Figure 21.24 The junction rule. The diagram shows an example of Kirchhoff’s first rule where the sum of the currents into a junction equals the sum of the currents out of a junction. In this case, the current going into the junction splits and comes out as two currents, so that 1 = 2 + 3. Here 1 must be 11 A, since 2 is 7 A and 3 is 4 A. 934 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Kirchhoff’s Second Rule Kirchhoff’s second rule (the loop rule) is an application of conservation of energy. The loop rule is stated in terms of potential,, rather than potential energy, but the two are related since PEelec =. Recall that emf is the potential difference of a source when no current is flowing. In a closed loop, whatever energy is supplied by emf must be transferred into other forms by devices in the loop, since there are no other
ways in which energy can be transferred into or out of the circuit. Figure 21.25 illustrates the changes in potential in a simple series circuit loop. Kirchhoff’s second rule requires emf − − 1 − 2 = 0. Rearranged, this is emf = + 1 + 2, which means the emf equals the sum of the (voltage) drops in the loop. Figure 21.25 The loop rule. An example of Kirchhoff’s second rule where the sum of the changes in potential around a closed loop must be zero. (a) In this standard schematic of a simple series circuit, the emf supplies 18 V, which is reduced to zero by the resistances, with 1 V across the internal resistance, and 12 V and 5 V across the two load resistances, for a total of 18 V. (b) This perspective view represents the potential as something like a roller coaster, where charge is raised in potential by the emf and lowered by the resistances. (Note that the script E stands for emf.) Applying Kirchhoff’s Rules By applying Kirchhoff’s rules, we generate equations that allow us to find the unknowns in circuits. The unknowns may be currents, emfs, or resistances. Each time a rule is applied, an equation is produced. If there are as many independent equations as unknowns, then the problem can be solved. There are two decisions you must make when applying Kirchhoff’s rules. These decisions determine the signs of various quantities in the equations you obtain from applying the rules. 1. When applying Kirchhoff’s first rule, the junction rule, you must label the current in each branch and decide in what direction it is going. For example, in Figure 21.23, Figure 21.24, and Figure 21.25, currents are labeled 1, 2, 3, and, and arrows indicate their directions. There is no risk here, for if you choose the wrong direction, the current will be of the correct magnitude but negative. 2. When applying Kirchhoff’s second rule, the loop rule, you must identify a closed loop and decide in which direction to go around it, clockwise or counterclockwise. For example, in Figure 21.25 the loop was traversed in the same direction as the current (clockwise). Again, there is no risk; going around the circuit in the opposite direction reverses the sign of every
term in the equation, which is like multiplying both sides of the equation by –1. Figure 21.26 and the following points will help you get the plus or minus signs right when applying the loop rule. Note that the resistors and emfs are traversed by going from a to b. In many circuits, it will be necessary to construct more than one loop. In traversing each loop, one needs to be consistent for the sign of the change in potential. (See Example 21.5.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 935 Figure 21.26 Each of these resistors and voltage sources is traversed from a to b. The potential changes are shown beneath each element and are explained in the text. (Note that the script E stands for emf.) • When a resistor is traversed in the same direction as the current, the change in potential is −. (See Figure 21.26.) • When a resistor is traversed in the direction opposite to the current, the change in potential is +. (See Figure 21.26.) • When an emf is traversed from – to + (the same direction it moves positive charge), the change in potential is +emf. (See Figure 21.26.) • When an emf is traversed from + to – (opposite to the direction it moves positive charge), the change in potential is − emf. (See Figure 21.26.) Example 21.5 Calculating Current: Using Kirchhoff’s Rules Find the currents flowing in the circuit in Figure 21.27. Figure 21.27 This circuit is similar to that in Figure 21.23, but the resistances and emfs are specified. (Each emf is denoted by script E.) The currents in each branch are labeled and assumed to move in the directions shown. This example uses Kirchhoff’s rules to find the currents. Strategy This circuit is sufficiently complex that the currents cannot be found using Ohm’s law and the series-parallel techniques—it is necessary to use Kirchhoff’s rules. Currents have been labeled 1, 2, and 3 in the figure and assumptions have been made about their directions. Locations on the diagram have been labeled with letters a through h. In the solution we will apply the junction and loop rules, seeking three independent equations to allow
us to solve for the three unknown currents. Solution We begin by applying Kirchhoff’s first or junction rule at point a. This gives 1 = 2 + 3, (21.54) since 1 flows into the junction, while 2 and 3 flow out. Applying the junction rule at e produces exactly the same equation, so that no new information is obtained. This is a single equation with three unknowns—three independent equations are needed, and so the loop rule must be applied. 936 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Now we consider the loop abcdea. Going from a to b, we traverse 2 in the same (assumed) direction of the current 2, and so the change in potential is −22. Then going from b to c, we go from – to +, so that the change in potential is +emf1. Traversing the internal resistance 1 from c to d gives −21. Completing the loop by going from d to a again traverses a resistor in the same direction as its current, giving a change in potential of −11. The loop rule states that the changes in potential sum to zero. Thus, −22 + emf1 − 21 − 11 = −2(2 + 1) + emf1 − 11 = 0. Substituting values from the circuit diagram for the resistances and emf, and canceling the ampere unit gives Now applying the loop rule to aefgha (we could have chosen abcdefgha as well) similarly gives + 11 + 33 + 32 − emf2= +1 1 + 3 3 + 2 − emf2 = 0. −3 2 + 18 − 6 1 = 0. (21.55) (21.56) (21.57) Note that the signs are reversed compared with the other loop, because elements are traversed in the opposite direction. With values entered, this becomes These three equations are sufficient to solve for the three unknown currents. First, solve the second equation for 2 : + 6 1 + 2 3 − 45 = 0. Now solve the third equation for 3 : 2 = 6 − 2 1. 3 = 22.5 − 3 1. Substituting these two new equations into the first one allows us to find a value for 1 : 1 = 2 + 3 = (6 − 2 1) + (22.5 − 3 1) = 28.5 − 5 1. Combining terms gives
6 1 = 28.5, and 1 = 4.75 A. Substituting this value for 1 back into the fourth equation gives.50 2 = −3.50 A. The minus sign means 2 flows in the direction opposite to that assumed in Figure 21.27. Finally, substituting the value for 1 into the fifth equation gives 3 = 22.5−3 1 = 22.5 − 14.25 3 = 8.25 A. (21.58) (21.59) (21.60) (21.61) (21.62) (21.63) (21.64) (21.65) (21.66) (21.67) Discussion Just as a check, we note that indeed 1 = 2 + 3. The results could also have been checked by entering all of the values into the equation for the abcdefgha loop. Problem-Solving Strategies for Kirchhoff’s Rules 1. Make certain there is a clear circuit diagram on which you can label all known and unknown resistances, emfs, and currents. If a current is unknown, you must assign it a direction. This is necessary for determining the signs of potential changes. If you assign the direction incorrectly, the current will be found to have a negative value—no harm done. 2. Apply the junction rule to any junction in the circuit. Each time the junction rule is applied, you should get an equation with a current that does not appear in a previous application—if not, then the equation is redundant. 3. Apply the loop rule to as many loops as needed to solve for the unknowns in the problem. (There must be as many independent equations as unknowns.) To apply the loop rule, you must choose a direction to go around the loop. Then This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 937 carefully and consistently determine the signs of the potential changes for each element using the four bulleted points discussed above in conjunction with Figure 21.26. 4. Solve the simultaneous equations for the unknowns. This may involve many algebraic steps, requiring careful checking and rechecking. 5. Check to see whether the answers are reasonable and consistent. The numbers should be of the correct order of magnitude, neither exceedingly large nor vanishingly small. The signs should be reasonable—for example, no resistance should be negative.
Check to see that the values obtained satisfy the various equations obtained from applying the rules. The currents should satisfy the junction rule, for example. The material in this section is correct in theory. We should be able to verify it by making measurements of current and voltage. In fact, some of the devices used to make such measurements are straightforward applications of the principles covered so far and are explored in the next modules. As we shall see, a very basic, even profound, fact results—making a measurement alters the quantity being measured. Check Your Understanding Can Kirchhoff’s rules be applied to simple series and parallel circuits or are they restricted for use in more complicated circuits that are not combinations of series and parallel? Solution Kirchhoff's rules can be applied to any circuit since they are applications to circuits of two conservation laws. Conservation laws are the most broadly applicable principles in physics. It is usually mathematically simpler to use the rules for series and parallel in simpler circuits so we emphasize Kirchhoff’s rules for use in more complicated situations. But the rules for series and parallel can be derived from Kirchhoff’s rules. Moreover, Kirchhoff’s rules can be expanded to devices other than resistors and emfs, such as capacitors, and are one of the basic analysis devices in circuit analysis. Making Connections: Parallel Resistors A simple circuit shown below – with two parallel resistors and a voltage source – is implemented in a laboratory experiment with ɛ= 6.00 ± 0.02 V and R1 = 4.8 ± 0.1 Ω and R2 = 9.6 ± 0.1 Ω. The values include an allowance for experimental uncertainties as they cannot be measured with perfect certainty. For example if you measure the value for a resistor a few times, you may get slightly different results. Hence values are expressed with some level of uncertainty. Figure 21.28 In the laboratory experiment the currents measured in the two resistors are I1 = 1.27 A and I2 = 0.62 A respectively. Let us examine these values using Kirchhoff’s laws. For the two loops, E - I1R1 = 0 or I1 = E/R1 E - I2R2 = 0 or I2 = E/R2 Converting the given uncertainties for voltage and resistances into percentages, we get E = 6.00 V ± 0.33% R1 = 4.8 Ω ± 2.08%
R2 = 9.6 Ω ± 1.04% We now find the currents for the two loops. While the voltage is divided by the resistance to find the current, uncertainties in voltage and resistance are directly added to find the uncertainty in the current value. I1 = (6.00/4.8) ± (0.33%+2.08%) = 1.25 ± 2.4% = 1.25 ± 0.03 A I2 = (6.00/9.6) ± (0.33%+1.04%) = 0.63 ± 1.4% = 0.63 ± 0.01 A Finally you can check that the two measured values in this case are within the uncertainty ranges found for the currents. However there can also be additional experimental uncertainty in the measurements of currents. 938 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 21.4 DC Voltmeters and Ammeters By the end of this section, you will be able to: Learning Objectives • Explain why a voltmeter must be connected in parallel with the circuit. • Draw a diagram showing an ammeter correctly connected in a circuit. • Describe how a galvanometer can be used as either a voltmeter or an ammeter. • Find the resistance that must be placed in series with a galvanometer to allow it to be used as a voltmeter with a given reading. • Explain why measuring the voltage or current in a circuit can never be exact. Voltmeters measure voltage, whereas ammeters measure current. Some of the meters in automobile dashboards, digital cameras, cell phones, and tuner-amplifiers are voltmeters or ammeters. (See Figure 21.29.) The internal construction of the simplest of these meters and how they are connected to the system they monitor give further insight into applications of series and parallel connections. Figure 21.29 The fuel and temperature gauges (far right and far left, respectively) in this 1996 Volkswagen are voltmeters that register the voltage output of “sender” units, which are hopefully proportional to the amount of gasoline in the tank and the engine temperature. (credit: Christian Giersing) Voltmeters are connected in parallel with whatever device’s voltage is to be measured. A parallel connection is used because objects in parallel experience the same potential difference. (See Figure 21.30, where the voltmeter is represented by the symbol V.) Ammeters are connected
in series with whatever device’s current is to be measured. A series connection is used because objects in series have the same current passing through them. (See Figure 21.31, where the ammeter is represented by the symbol A.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 939 Figure 21.30 (a) To measure potential differences in this series circuit, the voltmeter (V) is placed in parallel with the voltage source or either of the resistors. Note that terminal voltage is measured between points a and b. It is not possible to connect the voltmeter directly across the emf without including its internal resistance,. (b) A digital voltmeter in use. (credit: Messtechniker, Wikimedia Commons) Figure 21.31 An ammeter (A) is placed in series to measure current. All of the current in this circuit flows through the meter. The ammeter would have the same reading if located between points d and e or between points f and a as it does in the position shown. (Note that the script capital E stands for emf, and stands for the internal resistance of the source of potential difference.) Analog Meters: Galvanometers Analog meters have a needle that swivels to point at numbers on a scale, as opposed to digital meters, which have numerical readouts similar to a hand-held calculator. The heart of most analog meters is a device called a galvanometer, denoted by G. Current flow through a galvanometer, G, produces a proportional needle deflection. (This deflection is due to the force of a magnetic field upon a current-carrying wire.) The two crucial characteristics of a given galvanometer are its resistance and current sensitivity. Current sensitivity is the current that gives a full-scale deflection of the galvanometer’s needle, the maximum current that the instrument can measure. For example, a galvanometer with a current sensitivity of 50 μA has a maximum deflection of its needle when 50 μA flows through it, reads half-scale when 25 μA flows through it, and so on. 940 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments If such a galvanometer has a 25- Ω resistance, then a voltage of only = = full-scale reading. By connecting resistors to this galvanometer in different ways, you can use it as
either a voltmeter or ammeter that can measure a broad range of voltages or currents. (25 Ω) = 1.25 mV produces a 50 μA Galvanometer as Voltmeter Figure 21.32 shows how a galvanometer can be used as a voltmeter by connecting it in series with a large resistance,. The value of the resistance is determined by the maximum voltage to be measured. Suppose you want 10 V to produce a fullscale deflection of a voltmeter containing a 25-Ω galvanometer with a 50-μA sensitivity. Then 10 V applied to the meter must produce a current of 50 μA. The total resistance must be tot = + = = 10 V 50 μA = 200 kΩ, or = tot − = 200 kΩ − 25 Ω ≈ 200 k Ω. (21.68) (21.69) ( is so large that the galvanometer resistance,, is nearly negligible.) Note that 5 V applied to this voltmeter produces a halfscale deflection by producing a 25-μA current through the meter, and so the voltmeter’s reading is proportional to voltage as desired. This voltmeter would not be useful for voltages less than about half a volt, because the meter deflection would be small and difficult to read accurately. For other voltage ranges, other resistances are placed in series with the galvanometer. Many meters have a choice of scales. That choice involves switching an appropriate resistance into series with the galvanometer. Figure 21.32 A large resistance placed in series with a galvanometer G produces a voltmeter, the full-scale deflection of which depends on the choice of. The larger the voltage to be measured, the larger must be. (Note that represents the internal resistance of the galvanometer.) Galvanometer as Ammeter The same galvanometer can also be made into an ammeter by placing it in parallel with a small resistance, often called the shunt resistance, as shown in Figure 21.33. Since the shunt resistance is small, most of the current passes through it, allowing an ammeter to measure currents much greater than those producing a full-scale deflection of the galvanometer. Suppose, for example, an ammeter is needed that gives a full-scale deflection for 1.0 A, and contains the same 25- Ω galvanometer with its 50-μA sensitivity. Since and are in parallel, the voltage across them is the same. These drops are
= G so that = G =. Solving for, and noting that G is 50 μA and is 0.999950 A, we have = G = (25 Ω ) 50 μA 0.999950 A = 1.2510−3 Ω. (21.70) Figure 21.33 A small shunt resistance placed in parallel with a galvanometer G produces an ammeter, the full-scale deflection of which depends on the choice of. The larger the current to be measured, the smaller must be. Most of the current ( ) flowing through the meter is shunted through to protect the galvanometer. (Note that represents the internal resistance of the galvanometer.) Ammeters may also have multiple scales for greater flexibility in application. The various scales are achieved by switching various shunt resistances in parallel with the galvanometer—the greater the maximum current to be measured, the smaller the shunt resistance must be. Taking Measurements Alters the Circuit When you use a voltmeter or ammeter, you are connecting another resistor to an existing circuit and, thus, altering the circuit. Ideally, voltmeters and ammeters do not appreciably affect the circuit, but it is instructive to examine the circumstances under which they do or do not interfere. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 941 First, consider the voltmeter, which is always placed in parallel with the device being measured. Very little current flows through the voltmeter if its resistance is a few orders of magnitude greater than the device, and so the circuit is not appreciably affected. (See Figure 21.34(a).) (A large resistance in parallel with a small one has a combined resistance essentially equal to the small one.) If, however, the voltmeter’s resistance is comparable to that of the device being measured, then the two in parallel have a smaller resistance, appreciably affecting the circuit. (See Figure 21.34(b).) The voltage across the device is not the same as when the voltmeter is out of the circuit. Figure 21.34 (a) A voltmeter having a resistance much larger than the device ( Voltmeter >> ) with which it is in parallel produces a parallel resistance essentially the same as the device and does not appreciably affect the circuit being measured. (b) Here the voltmeter has the same resistance
as the device ( Voltmeter ≅ ), so that the parallel resistance is half of what it is when the voltmeter is not connected. This is an example of a significant alteration of the circuit and is to be avoided. An ammeter is placed in series in the branch of the circuit being measured, so that its resistance adds to that branch. Normally, the ammeter’s resistance is very small compared with the resistances of the devices in the circuit, and so the extra resistance is negligible. (See Figure 21.35(a).) However, if very small load resistances are involved, or if the ammeter is not as low in resistance as it should be, then the total series resistance is significantly greater, and the current in the branch being measured is reduced. (See Figure 21.35(b).) A practical problem can occur if the ammeter is connected incorrectly. If it was put in parallel with the resistor to measure the current in it, you could possibly damage the meter; the low resistance of the ammeter would allow most of the current in the circuit to go through the galvanometer, and this current would be larger since the effective resistance is smaller. Figure 21.35 (a) An ammeter normally has such a small resistance that the total series resistance in the branch being measured is not appreciably increased. The circuit is essentially unaltered compared with when the ammeter is absent. (b) Here the ammeter’s resistance is the same as that of the branch, so that the total resistance is doubled and the current is half what it is without the ammeter. This significant alteration of the circuit is to be avoided. One solution to the problem of voltmeters and ammeters interfering with the circuits being measured is to use galvanometers with greater sensitivity. This allows construction of voltmeters with greater resistance and ammeters with smaller resistance than when less sensitive galvanometers are used. There are practical limits to galvanometer sensitivity, but it is possible to get analog meters that make measurements accurate to a few percent. Note that the inaccuracy comes from altering the circuit, not from a fault in the meter. Connections: Limits to Knowledge Making a measurement alters the system being measured in a manner that produces uncertainty in the measurement. For macroscopic systems, such as the circuits discussed in this module, the alteration can usually be made negligibly small, but it cannot be eliminated entirely. For submicroscopic systems, such as atoms, nuclei, and
smaller particles, measurement alters the system in a manner that cannot be made arbitrarily small. This actually limits knowledge of the system—even limiting what nature can know about itself. We shall see profound implications of this when the Heisenberg uncertainty principle is discussed in the modules on quantum mechanics. There is another measurement technique based on drawing no current at all and, hence, not altering the circuit at all. These are called null measurements and are the topic of Null Measurements. Digital meters that employ solid-state electronics and null measurements can attain accuracies of one part in 106. 950 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments heart defibrillator is slightly more complex than the one in Figure 21.42, to compensate for magnetic and AC effects that will be covered in Magnetism. Check Your Understanding When is the potential difference across a capacitor an emf? Solution Only when the current being drawn from or put into the capacitor is zero. Capacitors, like batteries, have internal resistance, so their output voltage is not an emf unless current is zero. This is difficult to measure in practice so we refer to a capacitor’s voltage rather than its emf. But the source of potential difference in a capacitor is fundamental and it is an emf. PhET Explorations: Circuit Construction Kit (DC only) An electronics kit in your computer! Build circuits with resistors, light bulbs, batteries, and switches. Take measurements with the realistic ammeter and voltmeter. View the circuit as a schematic diagram, or switch to a life-like view. Figure 21.45 Circuit Construction Kit (DC only) (http://cnx.org/content/m55370/1.3/circuit-construction-kit-dc_en.jar) Glossary ammeter: an instrument that measures current analog meter: a measuring instrument that gives a readout in the form of a needle movement over a marked gauge bridge device: a device that forms a bridge between two branches of a circuit; some bridge devices are used to make null measurements in circuits capacitance: the maximum amount of electric potential energy that can be stored (or separated) for a given electric potential capacitor: an electrical component used to store energy by separating electric charge on two opposing plates conservation laws: require that energy and charge be conserved in a system current: the flow of charge through an electric circuit past a given point of measurement current sensitivity: the maximum current that a galvanometer can read digital meter: a measuring instrument
that gives a readout in a digital form electromotive force (emf): the potential difference of a source of electricity when no current is flowing; measured in volts full-scale deflection: the maximum deflection of a galvanometer needle, also known as current sensitivity; a galvanometer with a full-scale deflection of 50 μA has a maximum deflection of its needle when 50 μA flows through it galvanometer: an analog measuring device, denoted by G, that measures current flow using a needle deflection caused by a magnetic field force acting upon a current-carrying wire internal resistance: the amount of resistance within the voltage source Joule’s law: the relationship between potential electrical power, voltage, and resistance in an electrical circuit, given by: = junction rule: Kirchhoff’s first rule, which applies the conservation of charge to a junction; current is the flow of charge; thus, whatever charge flows into the junction must flow out; the rule can be stated 1 = 2 + 3 Kirchhoff’s rules: a set of two rules, based on conservation of charge and energy, governing current and changes in potential in an electric circuit loop rule: Kirchhoff’s second rule, which states that in a closed loop, whatever energy is supplied by emf must be transferred into other forms by devices in the loop, since there are no other ways in which energy can be transferred into or out of This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 951 the circuit. Thus, the emf equals the sum of the (voltage) drops in the loop and can be stated: emf = + 1 + 2 null measurements: methods of measuring current and voltage more accurately by balancing the circuit so that no current flows through the measurement device ohmmeter: an instrument that applies a voltage to a resistance, measures the current, calculates the resistance using Ohm’s law, and provides a readout of this calculated resistance Ohm’s law: the relationship between current, voltage, and resistance within an electrical circuit: = parallel: the wiring of resistors or other components in an electrical circuit such that each component receives an equal voltage from the power source; often pictured in a ladder-shaped diagram, with each component on a rung of the ladder potential difference: the difference in electric potential between two points in an electric
circuit, measured in volts potentiometer: a null measurement device for measuring potentials (voltages) RC circuit: a circuit that contains both a resistor and a capacitor resistance: causing a loss of electrical power in a circuit resistor: a component that provides resistance to the current flowing through an electrical circuit series: a sequence of resistors or other components wired into a circuit one after the other shunt resistance: a small resistance placed in parallel with a galvanometer G to produce an ammeter; the larger the current to be measured, the smaller must be; most of the current flowing through the meter is shunted through to protect the galvanometer terminal voltage: the voltage measured across the terminals of a source of potential difference voltage: the electrical potential energy per unit charge; electric pressure created by a power source, such as a battery voltage drop: the loss of electrical power as a current travels through a resistor, wire or other component voltmeter: an instrument that measures voltage Wheatstone bridge: a null measurement device for calculating resistance by balancing potential drops in a circuit Section Summary 21.1 Resistors in Series and Parallel • The total resistance of an electrical circuit with resistors wired in a series is the sum of the individual resistances: s = 1 + 2 + 3 +.... • Each resistor in a series circuit has the same amount of current flowing through it. • The voltage drop, or power dissipation, across each individual resistor in a series is different, and their combined total adds up to the power source input. • The total resistance of an electrical circuit with resistors wired in parallel is less than the lowest resistance of any of the components and can be determined using the formula +.... • Each resistor in a parallel circuit has the same full voltage of the source applied to it. • The current flowing through each resistor in a parallel circuit is different, depending on the resistance. • If a more complex connection of resistors is a combination of series and parallel, it can be reduced to a single equivalent resistance by identifying its various parts as series or parallel, reducing each to its equivalent, and continuing until a single resistance is eventually reached. 21.2 Electromotive Force: Terminal Voltage • All voltage sources have two fundamental parts—a source of electrical energy that has a characteristic electromotive force (emf), and an internal resistance. • The emf is the potential difference of a source when no current is flowing. • The numerical value of the emf depends on the source of potential difference. • The internal resistance of a voltage source
affects the output voltage when a current flows. • The voltage output of a device is called its terminal voltage and is given by = emf −, where is the electric current and is positive when flowing away from the positive terminal of the voltage source. 952 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments • When multiple voltage sources are in series, their internal resistances add and their emfs add algebraically. • Solar cells can be wired in series or parallel to provide increased voltage or current, respectively. 21.3 Kirchhoff’s Rules • Kirchhoff’s rules can be used to analyze any circuit, simple or complex. • Kirchhoff’s first rule—the junction rule: The sum of all currents entering a junction must equal the sum of all currents leaving the junction. • Kirchhoff’s second rule—the loop rule: The algebraic sum of changes in potential around any closed circuit path (loop) must be zero. • The two rules are based, respectively, on the laws of conservation of charge and energy. • When calculating potential and current using Kirchhoff’s rules, a set of conventions must be followed for determining the correct signs of various terms. • The simpler series and parallel rules are special cases of Kirchhoff’s rules. 21.4 DC Voltmeters and Ammeters • Voltmeters measure voltage, and ammeters measure current. • A voltmeter is placed in parallel with the voltage source to receive full voltage and must have a large resistance to limit its effect on the circuit. • An ammeter is placed in series to get the full current flowing through a branch and must have a small resistance to limit its effect on the circuit. • Both can be based on the combination of a resistor and a galvanometer, a device that gives an analog reading of current. • Standard voltmeters and ammeters alter the circuit being measured and are thus limited in accuracy. 21.5 Null Measurements • Null measurement techniques achieve greater accuracy by balancing a circuit so that no current flows through the measuring device. • One such device, for determining voltage, is a potentiometer. • Another null measurement device, for determining resistance, is the Wheatstone bridge. • Other physical quantities can also be measured with null measurement techniques. 21.6 DC Circuits Containing Resistors and Capacitors • An circuit is one that has both a resistor and a capacitor. • The time constant for an circuit
is =. • When an initially uncharged ( 0 = 0 at = 0 ) capacitor in series with a resistor is charged by a DC voltage source, the voltage rises, asymptotically approaching the emf of the voltage source; as a function of time, = emf(1 − − / )(charging). • Within the span of each time constant, the voltage rises by 0.632 of the remaining value, approaching the final voltage asymptotically. If a capacitor with an initial voltage 0 is discharged through a resistor starting at = 0, then its voltage decreases exponentially as given by In each time constant, the voltage falls by 0.368 of its remaining initial value, approaching zero asymptotically. = 0− / (discharging). • • Conceptual Questions 21.1 Resistors in Series and Parallel 1. A switch has a variable resistance that is nearly zero when closed and extremely large when open, and it is placed in series with the device it controls. Explain the effect the switch in Figure 21.46 has on current when open and when closed. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 953 Figure 21.46 A switch is ordinarily in series with a resistance and voltage source. Ideally, the switch has nearly zero resistance when closed but has an extremely large resistance when open. (Note that in this diagram, the script E represents the voltage (or electromotive force) of the battery.) 2. What is the voltage across the open switch in Figure 21.46? 3. There is a voltage across an open switch, such as in Figure 21.46. Why, then, is the power dissipated by the open switch small? 4. Why is the power dissipated by a closed switch, such as in Figure 21.46, small? 5. A student in a physics lab mistakenly wired a light bulb, battery, and switch as shown in Figure 21.47. Explain why the bulb is on when the switch is open, and off when the switch is closed. (Do not try this—it is hard on the battery!) Figure 21.47 A wiring mistake put this switch in parallel with the device represented by. (Note that in this diagram, the script E represents the voltage (or electromotive force) of the battery.) 6. Knowing that the severity of a shock depends on the magnitude of the current through
your body, would you prefer to be in series or parallel with a resistance, such as the heating element of a toaster, if shocked by it? Explain. 7. Would your headlights dim when you start your car’s engine if the wires in your automobile were superconductors? (Do not neglect the battery’s internal resistance.) Explain. 8. Some strings of holiday lights are wired in series to save wiring costs. An old version utilized bulbs that break the electrical connection, like an open switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on 120 V and has 40 identical bulbs, what is the normal operating voltage of each? Newer versions use bulbs that short circuit, like a closed switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on 120 V and has 39 remaining identical bulbs, what is then the operating voltage of each? 9. If two household lightbulbs rated 60 W and 100 W are connected in series to household power, which will be brighter? Explain. 10. Suppose you are doing a physics lab that asks you to put a resistor into a circuit, but all the resistors supplied have a larger resistance than the requested value. How would you connect the available resistances to attempt to get the smaller value asked for? 11. Before World War II, some radios got power through a “resistance cord” that had a significant resistance. Such a resistance cord reduces the voltage to a desired level for the radio’s tubes and the like, and it saves the expense of a transformer. Explain why resistance cords become warm and waste energy when the radio is on. 12. Some light bulbs have three power settings (not including zero), obtained from multiple filaments that are individually switched and wired in parallel. What is the minimum number of filaments needed for three power settings? 21.2 Electromotive Force: Terminal Voltage 13. Is every emf a potential difference? Is every potential difference an emf? Explain. 14. Explain which battery is doing the charging and which is being charged in Figure 21.48. 954 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Figure 21.48 15. Given a battery, an assortment of resistors, and a variety of voltage and current measuring devices, describe how you would determine the internal resistance of the battery. 16. Two different 12-V automobile batteries on
a store shelf are rated at 600 and 850 “cold cranking amps.” Which has the smallest internal resistance? 17. What are the advantages and disadvantages of connecting batteries in series? In parallel? 18. Semitractor trucks use four large 12-V batteries. The starter system requires 24 V, while normal operation of the truck’s other electrical components utilizes 12 V. How could the four batteries be connected to produce 24 V? To produce 12 V? Why is 24 V better than 12 V for starting the truck’s engine (a very heavy load)? 21.3 Kirchhoff’s Rules 19. Can all of the currents going into the junction in Figure 21.49 be positive? Explain. Figure 21.49 20. Apply the junction rule to junction b in Figure 21.50. Is any new information gained by applying the junction rule at e? (In the figure, each emf is represented by script E.) Figure 21.50 21. (a) What is the potential difference going from point a to point b in Figure 21.50? (b) What is the potential difference going from c to b? (c) From e to g? (d) From e to d? 22. Apply the loop rule to loop afedcba in Figure 21.50. 23. Apply the loop rule to loops abgefa and cbgedc in Figure 21.50. 21.4 DC Voltmeters and Ammeters This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 955 24. Why should you not connect an ammeter directly across a voltage source as shown in Figure 21.51? (Note that script E in the figure stands for emf.) Figure 21.51 25. Suppose you are using a multimeter (one designed to measure a range of voltages, currents, and resistances) to measure current in a circuit and you inadvertently leave it in a voltmeter mode. What effect will the meter have on the circuit? What would happen if you were measuring voltage but accidentally put the meter in the ammeter mode? 26. Specify the points to which you could connect a voltmeter to measure the following potential differences in Figure 21.52: (a) the potential difference of the voltage source; (b) the potential difference across 1 ; (c) across 2 ; (d) across 3
; (e) across 2 and 3. Note that there may be more than one answer to each part. Figure 21.52 27. To measure currents in Figure 21.52, you would replace a wire between two points with an ammeter. Specify the points between which you would place an ammeter to measure the following: (a) the total current; (b) the current flowing through 1 ; (c) through 2 ; (d) through 3. Note that there may be more than one answer to each part. 21.5 Null Measurements 28. Why can a null measurement be more accurate than one using standard voltmeters and ammeters? What factors limit the accuracy of null measurements? 29. If a potentiometer is used to measure cell emfs on the order of a few volts, why is it most accurate for the standard emfs to be the same order of magnitude and the resistances to be in the range of a few ohms? 21.6 DC Circuits Containing Resistors and Capacitors 30. Regarding the units involved in the relationship =, verify that the units of resistance times capacitance are time, that is, Ω ⋅ F = s. 31. The time constant in heart defibrillation is crucial to limiting the time the current flows. If the capacitance in the defibrillation unit is fixed, how would you manipulate resistance in the circuit to adjust the constant? Would an adjustment of the applied voltage also be needed to ensure that the current delivered has an appropriate value? 32. When making an ECG measurement, it is important to measure voltage variations over small time intervals. The time is limited by the constant of the circuit—it is not possible to measure time variations shorter than. How would you manipulate and in the circuit to allow the necessary measurements? 33. Draw two graphs of charge versus time on a capacitor. Draw one for charging an initially uncharged capacitor in series with a resistor, as in the circuit in Figure 21.41, starting from t = 0. Draw the other for discharging a capacitor through a resistor, as in the circuit in Figure 21.42, starting at t = 0, with an initial charge 0. Show at least two intervals of. 34. When charging a capacitor, as discussed in conjunction with Figure 21.41, how long does it take for the voltage on the capacitor to reach emf? Is this a problem? 956 Chapter 21 \| Circuits, Bioelectricity, and DC
Instruments 35. When discharging a capacitor, as discussed in conjunction with Figure 21.42, how long does it take for the voltage on the capacitor to reach zero? Is this a problem? 36. Referring to Figure 21.41, draw a graph of potential difference across the resistor versus time, showing at least two intervals of. Also draw a graph of current versus time for this situation. 37. A long, inexpensive extension cord is connected from inside the house to a refrigerator outside. The refrigerator doesn’t run as it should. What might be the problem? 38. In Figure 21.44, does the graph indicate the time constant is shorter for discharging than for charging? Would you expect ionized gas to have low resistance? How would you adjust to get a longer time between flashes? Would adjusting affect the discharge time? 39. An electronic apparatus may have large capacitors at high voltage in the power supply section, presenting a shock hazard even when the apparatus is switched off. A “bleeder resistor” is therefore placed across such a capacitor, as shown schematically in Figure 21.53, to bleed the charge from it after the apparatus is off. Why must the bleeder resistance be much greater than the effective resistance of the rest of the circuit? How does this affect the time constant for discharging the capacitor? Figure 21.53 A bleeder resistor bl discharges the capacitor in this electronic device once it is switched off. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 957 Problems & Exercises 21.1 Resistors in Series and Parallel Note: Data taken from figures can be assumed to be accurate to three significant digits. 1. (a) What is the resistance of ten 275-Ω resistors connected in series? (b) In parallel? 2. (a) What is the resistance of a 1.00×102 -Ω, a 2.50-kΩ, and a 4.00-k Ω resistor connected in series? (b) In parallel? 3. What are the largest and smallest resistances you can obtain by connecting a 36.0-Ω, a 50.0-Ω, and a 700-Ω resistor together? 4. An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a
15-A, 120-V circuit. (The three devices are in parallel when plugged into the same socket.). (a) What current is drawn by each device? (b) Will this combination blow the 15-A fuse? 5. Your car’s 30.0-W headlight and 2.40-kW starter are ordinarily connected in parallel in a 12.0-V system. What power would one headlight and the starter consume if connected in series to a 12.0-V battery? (Neglect any other resistance in the circuit and any change in resistance in the two devices.) 6. (a) Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel. 7. Referring to the example combining series and parallel circuits and Figure 21.6, calculate 3 in the following two different ways: (a) from the known values of and 2 ; (b) using Ohm’s law for 3. In both parts explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors. 8. Referring to Figure 21.6: (a) Calculate 3 and note how it compares with 3 found in the first two example problems in this module. (b) Find the total power supplied by the source and compare it with the sum of the powers dissipated by the resistors. 9. Refer to Figure 21.7 and the discussion of lights dimming when a heavy appliance comes on. (a) Given the voltage source is 120 V, the wire resistance is 0.400 Ω, and the bulb is nominally 75.0 W, what power will the bulb dissipate if a total of 15.0 A passes through the wires when the motor comes on? Assume negligible change in bulb resistance. (b) What power is consumed by the motor? 10. A 240-kV power transmission line carrying 5.00×102 A is hung from grounded metal towers by ceramic insulators, each having a 1.00109 -Ω resistance. Figure 21.54. (a) What is the resistance to ground of 100 of these insulators? (b) Calculate the power dissipated by 100 of them. (c) What fraction of the power carried by the line is this? Explicitly show how you follow the steps
in the Problem-Solving Strategies for Series and Parallel Resistors. Figure 21.54 High-voltage (240-kV) transmission line carrying 5.00×102 A is hung from a grounded metal transmission tower. The row of ceramic insulators provide 1.00×109 Ω of resistance each. 11. Show that if two resistors 1 and 2 are combined and one is much greater than the other ( 1 >>2 ): (a) Their series resistance is very nearly equal to the greater resistance 1. (b) Their parallel resistance is very nearly equal to smaller resistance 2. 12. Unreasonable Results Two resistors, one having a resistance of 145 Ω, are connected in parallel to produce a total resistance of 150 Ω. (a) What is the value of the second resistance? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? 13. Unreasonable Results Two resistors, one having a resistance of 900 kΩ, are connected in series to produce a total resistance of 0.500 MΩ. (a) What is the value of the second resistance? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? 21.2 Electromotive Force: Terminal Voltage 14. Standard automobile batteries have six lead-acid cells in series, creating a total emf of 12.0 V. What is the emf of an individual lead-acid cell? 15. Carbon-zinc dry cells (sometimes referred to as nonalkaline cells) have an emf of 1.54 V, and they are produced as single cells or in various combinations to form other voltages. (a) How many 1.54-V cells are needed to make the common 9-V battery used in many small electronic devices? (b) What is the actual emf of the approximately 9-V battery? (c) Discuss how internal resistance in the series connection of cells will affect the terminal voltage of this approximately 9-V battery. 16. What is the output voltage of a 3.0000-V lithium cell in a digital wristwatch that draws 0.300 mA, if the cell’s internal resistance is 2.00 Ω? 17. (a) What is the terminal voltage of a large 1.54-V carbonzinc dry cell used in a physics lab to supply 2.00 A to a circuit, if the cell’s internal resistance is 0.100
Ω? (b) How much 958 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments represent the situation. (b) If the internal resistance of the power supply is 2000 Ω, what is the current through his body? (c) What is the power dissipated in his body? (d) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in this situation to be 1.00 mA or less? (e) Will this modification compromise the effectiveness of the power supply for driving low-resistance devices? Explain your reasoning. 27. Electric fish generate current with biological cells called electroplaques, which are physiological emf devices. The electroplaques in the South American eel are arranged in 140 rows, each row stretching horizontally along the body and each containing 5000 electroplaques. Each electroplaque has an emf of 0.15 V and internal resistance of 0.25 Ω. If the water surrounding the fish has resistance of 800 Ω, how much current can the eel produce in water from near its head to near its tail? 28. Integrated Concepts A 12.0-V emf automobile battery has a terminal voltage of 16.0 V when being charged by a current of 10.0 A. (a) What is the battery’s internal resistance? (b) What power is dissipated inside the battery? (c) At what rate (in ºC/min ) will its temperature increase if its mass is 20.0 kg and it has a specific heat of 0.300 kcal/kg ⋅ ºC, assuming no heat escapes? 29. Unreasonable Results A 1.58-V alkaline cell with a 0.200-Ω internal resistance is supplying 8.50 A to a load. (a) What is its terminal voltage? (b) What is the value of the load resistance? (c) What is unreasonable about these results? (d) Which assumptions are unreasonable or inconsistent? 30. Unreasonable Results (a) What is the internal resistance of a 1.54-V dry cell that supplies 1.00 W of power to a 15.0-Ω bulb? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? 21.3 Kirchhoff’s Rules 31. Apply the loop rule to loop abcdefgha in Figure 21.27
. 32. Apply the loop rule to loop aedcba in Figure 21.27. 33. Verify the second equation in Example 21.5 by substituting the values found for the currents 1 and 2. 34. Verify the third equation in Example 21.5 by substituting the values found for the currents 1 and 3. 35. Apply the junction rule at point a in Figure 21.55. electrical power does the cell produce? (c) What power goes to its load? 18. What is the internal resistance of an automobile battery that has an emf of 12.0 V and a terminal voltage of 15.0 V while a current of 8.00 A is charging it? 19. (a) Find the terminal voltage of a 12.0-V motorcycle battery having a 0.600-Ω internal resistance, if it is being charged by a current of 10.0 A. (b) What is the output voltage of the battery charger? 20. A car battery with a 12-V emf and an internal resistance of 0.050 Ω is being charged with a current of 60 A. Note that in this process the battery is being charged. (a) What is the potential difference across its terminals? (b) At what rate is thermal energy being dissipated in the battery? (c) At what rate is electric energy being converted to chemical energy? (d) What are the answers to (a) and (b) when the battery is used to supply 60 A to the starter motor? 21. The hot resistance of a flashlight bulb is 2.30 Ω, and it is run by a 1.58-V alkaline cell having a 0.100-Ω internal resistance. (a) What current flows? (b) Calculate the power supplied to the bulb using 2 bulb. (c) Is this power the same as calculated using 2 bulb? 22. The label on a portable radio recommends the use of rechargeable nickel-cadmium cells (nicads), although they have a 1.25-V emf while alkaline cells have a 1.58-V emf. The radio has a 3.20-Ω resistance. (a) Draw a circuit diagram of the radio and its batteries. Now, calculate the power delivered to the radio. (b) When using Nicad cells each having an internal resistance of 0.0400 Ω. (c) When using alkaline cells each having an internal resistance of 0
.200 Ω. (d) Does this difference seem significant, considering that the radio’s effective resistance is lowered when its volume is turned up? 23. An automobile starter motor has an equivalent resistance of 0.0500 Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal resistance. (a) What is the current to the motor? (b) What voltage is applied to it? (c) What power is supplied to the motor? (d) Repeat these calculations for when the battery connections are corroded and add 0.0900 Ω to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.) 24. A child’s electronic toy is supplied by three 1.58-V alkaline cells having internal resistances of 0.0200 Ω in series with a 1.53-V carbon-zinc dry cell having a 0.100-Ω internal resistance. The load resistance is 10.0 Ω. (a) Draw a circuit diagram of the toy and its batteries. (b) What current flows? (c) How much power is supplied to the load? (d) What is the internal resistance of the dry cell if it goes bad, resulting in only 0.500 W being supplied to the load? 25. (a) What is the internal resistance of a voltage source if its terminal voltage drops by 2.00 V when the current supplied increases by 5.00 A? (b) Can the emf of the voltage source be found with the information supplied? 26. A person with body resistance between his hands of 10.0 k Ω accidentally grasps the terminals of a 20.0-kV power supply. (Do NOT do this!) (a) Draw a circuit diagram to This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 959 44. Find the resistance that must be placed in series with a 25.0-Ω galvanometer having a 50.0-μA sensitivity (the same as the one discussed in the text) to allow it to be used as a voltmeter with a 0.100-V full-scale reading. 45. Find the resistance that must be placed in series with a 25.0-Ω galvanometer having a 50.0
-μA sensitivity (the same as the one discussed in the text) to allow it to be used as a voltmeter with a 3000-V full-scale reading. Include a circuit diagram with your solution. 46. Find the resistance that must be placed in parallel with a 25.0-Ω galvanometer having a 50.0-μA sensitivity (the same as the one discussed in the text) to allow it to be used as an ammeter with a 10.0-A full-scale reading. Include a circuit diagram with your solution. 47. Find the resistance that must be placed in parallel with a 25.0-Ω galvanometer having a 50.0-μA sensitivity (the same as the one discussed in the text) to allow it to be used as an ammeter with a 300-mA full-scale reading. 48. Find the resistance that must be placed in series with a 10.0-Ω galvanometer having a 100-μA sensitivity to allow it to be used as a voltmeter with: (a) a 300-V full-scale reading, and (b) a 0.300-V full-scale reading. 49. Find the resistance that must be placed in parallel with a 10.0-Ω galvanometer having a 100-μA sensitivity to allow it to be used as an ammeter with: (a) a 20.0-A full-scale reading, and (b) a 100-mA full-scale reading. 50. Suppose you measure the terminal voltage of a 1.585-V alkaline cell having an internal resistance of 0.100 Ω by placing a 1.00-k Ω voltmeter across its terminals. (See Figure 21.57.) (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio. Figure 21.57 51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00 Ω by placing a 1.00-k Ω voltmeter across its terminals. (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio. 52. A certain ammeter has a resistance of 5.0010−5 Ω on its 3.00-A scale and contains a 10.0-
Ω galvanometer. What is the sensitivity of the galvanometer? 53. A 1.00-MΩ voltmeter is placed in parallel with a 75.0-k Ω resistor in a circuit. (a) Draw a circuit diagram of the connection. (b) What is the resistance of the combination? (c) If the voltage across the combination is kept the same as it was across the 75.0-k Ω resistor alone, what is the percent increase in current? (d) If the current through the combination is kept the same as it was through the 75.0-k Ω resistor alone, what is the percentage decrease in voltage? (e) Are the changes found in parts (c) and (d) significant? Discuss. Figure 21.55 36. Apply the loop rule to loop abcdefghija in Figure 21.55. 37. Apply the loop rule to loop akledcba in Figure 21.55. 38. Find the currents flowing in the circuit in Figure 21.55. Explicitly show how you follow the steps in the ProblemSolving Strategies for Series and Parallel Resistors. 39. Solve Example 21.5, but use loop abcdefgha instead of loop akledcba. Explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors. 40. Find the currents flowing in the circuit in Figure 21.50. 41. Unreasonable Results Consider the circuit in Figure 21.56, and suppose that the emfs are unknown and the currents are given to be 1 = 5.00 A, 2 = 3.0 A, and 3 = –2.00 A. (a) Could you find the emfs? (b) What is wrong with the assumptions? Figure 21.56 21.4 DC Voltmeters and Ammeters 42. What is the sensitivity of the galvanometer (that is, what current gives a full-scale deflection) inside a voltmeter that has a 1.00-M Ω resistance on its 30.0-V scale? 43. What is the sensitivity of the galvanometer (that is, what current gives a full-scale deflection) inside a voltmeter that has a 25.0-k Ω resistance on its 100-V scale? 960 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 54. A 0.0200-Ω ammeter is placed in series with a 10.00
-Ω resistor in a circuit. (a) Draw a circuit diagram of the connection. (b) Calculate the resistance of the combination. (c) If the voltage is kept the same across the combination as it was through the 10.00-Ω resistor alone, what is the percent decrease in current? (d) If the current is kept the same through the combination as it was through the 10.00-Ω resistor alone, what is the percent increase in voltage? (e) Are the changes found in parts (c) and (d) significant? Discuss. 55. Unreasonable Results Suppose you have a 40.0-Ω galvanometer with a 25.0-μA sensitivity. (a) What resistance would you put in series with it to allow it to be used as a voltmeter that has a full-scale deflection for 0.500 mV? (b) What is unreasonable about this result? (c) Which assumptions are responsible? 56. Unreasonable Results (a) What resistance would you put in parallel with a 40.0-Ω galvanometer having a 25.0-μA sensitivity to allow it to be used as an ammeter that has a full-scale deflection for 10.0-μA? (b) What is unreasonable about this result? (c) Which assumptions are responsible? 21.5 Null Measurements 57. What is the emfx of a cell being measured in a potentiometer, if the standard cell’s emf is 12.0 V and the potentiometer balances for x = 5.000 Ω and s = 2.500 Ω? 58. Calculate the emfx of a dry cell for which a potentiometer is balanced when x = 1.200 Ω, while an alkaline standard cell with an emf of 1.600 V requires s = 1.247 Ω to balance the potentiometer. 59. When an unknown resistance x is placed in a Wheatstone bridge, it is possible to balance the bridge by adjusting 3 to be 2500 Ω. What is x if 2 1 = 0.625? 60. To what value must you adjust 3 to balance a Wheatstone bridge, if the unknown resistance x is 100 Ω, 1 is 50.0 Ω, and 2 is 175 Ω? 61. (a) What is the unknown emfx in a potentiometer that balances when x is 10.0 Ω
, and balances when s is 15.0 Ω for a standard 3.000-V emf? (b) The same emfx is placed in the same potentiometer, which now balances when s is 15.0 Ω for a standard emf of 3.100 V. At what resistance x will the potentiometer balance? 62. Suppose you want to measure resistances in the range from 10.0 Ω to 10.0 kΩ using a Wheatstone bridge that = 2.000. Over what range should 3 be has 2 1 adjustable? This content is available for free at http://cnx.org/content/col11844/1.13 21.6 DC Circuits Containing Resistors and Capacitors 63. The timing device in an automobile’s intermittent wiper system is based on an time constant and utilizes a 0.500-μF capacitor and a variable resistor. Over what range must be made to vary to achieve time constants from 2.00 to 15.0 s? 64. A heart pacemaker fires 72 times a minute, each time a 25.0-nF capacitor is charged (by a battery in series with a resistor) to 0.632 of its full voltage. What is the value of the resistance? 65. The duration of a photographic flash is related to an time constant, which is 0.100 μs for a certain camera. (a) If the resistance of the flash lamp is 0.0400 Ω during discharge, what is the size of the capacitor supplying its energy? (b) What is the time constant for charging the capacitor, if the charging resistance is 800 kΩ? 66. A 2.00- and a 7.50-μF capacitor can be connected in series or parallel, as can a 25.0- and a 100-kΩ resistor. Calculate the four time constants possible from connecting the resulting capacitance and resistance in series. 67. After two time constants, what percentage of the final voltage, emf, is on an initially uncharged capacitor, charged through a resistance? 68. A 500-Ω resistor, an uncharged 1.50-μF capacitor, and a 6.16-V emf are connected in series. (a) What is the initial current? (b) What is the time constant? (c) What is the current after one time constant? (d) What is the voltage on the capacitor after one time
constant? 69. A heart defibrillator being used on a patient has an time constant of 10.0 ms due to the resistance of the patient and the capacitance of the defibrillator. (a) If the defibrillator has an 8.00-μF capacitance, what is the resistance of the path through the patient? (You may neglect the capacitance of the patient and the resistance of the defibrillator.) (b) If the initial voltage is 12.0 kV, how long does it take to decline to 6.00×102 V? 70. An ECG monitor must have an time constant less than 1.00×102 μs to be able to measure variations in voltage over small time intervals. (a) If the resistance of the circuit (due mostly to that of the patient’s chest) is 1.00 kΩ, what is the maximum capacitance of the circuit? (b) Would it be difficult in practice to limit the capacitance to less than the value found in (a)? 71. Figure 21.58 shows how a bleeder resistor is used to discharge a capacitor after an electronic device is shut off, allowing a person to work on the electronics with less risk of shock. (a) What is the time constant? (b) How long will it take to reduce the voltage on the capacitor to 0.250% (5% of 5%) of its full value once discharge begins? (c) If the capacitor is charged to a voltage 0 through a 100-Ω resistance, calculate the time it takes to rise to 0.8650 (This is about two time constants.) Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 961 Consider a rechargeable lithium cell that is to be used to power a camcorder. Construct a problem in which you calculate the internal resistance of the cell during normal operation. Also, calculate the minimum voltage output of a battery charger to be used to recharge your lithium cell. Among the things to be considered are the emf and useful terminal voltage of a lithium cell and the current it should be able to supply to a camcorder. Figure 21.58 72. Using the exact exponential treatment, find how much time is required to discharge a 250-μF capacitor through a 500-Ω resistor down to 1.00% of its original voltage. 73. Using the exact exponential treatment, find how much time is required to charge an initially uncharged 100-
pF capacitor through a 75.0-M Ω resistor to 90.0% of its final voltage. 74. Integrated Concepts If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an discharge through a flash tube can limit blurring. Assuming 1.00 mm of motion during one constant is acceptable, and given that the flash is driven by a 600-μF capacitor, what is the resistance in the flash tube? 75. Integrated Concepts A flashing lamp in a Christmas earring is based on an discharge of a capacitor through its resistance. The effective duration of the flash is 0.250 s, during which it produces an average 0.500 W from an average 3.00 V. (a) What energy does it dissipate? (b) How much charge moves through the lamp? (c) Find the capacitance. (d) What is the resistance of the lamp? 76. Integrated Concepts A 160-μF capacitor charged to 450 V is discharged through a 31.2-k Ω resistor. (a) Find the time constant. (b) Calculate the temperature increase of the resistor, given that its mass is 2.50 g and its specific heat is 1.67 kJ kg ⋅ ºC, noting that most of the thermal energy is retained in the short time of the discharge. (c) Calculate the new resistance, assuming it is pure carbon. (d) Does this change in resistance seem significant? 77. Unreasonable Results (a) Calculate the capacitance needed to get an time constant of 1.00×103 s with a 0.100-Ω resistor. (b) What is unreasonable about this result? (c) Which assumptions are responsible? 78. Construct Your Own Problem Consider a camera’s flash unit. Construct a problem in which you calculate the size of the capacitor that stores energy for the flash lamp. Among the things to be considered are the voltage applied to the capacitor, the energy needed in the flash and the associated charge needed on the capacitor, the resistance of the flash lamp during discharge, and the desired time constant. 79. Construct Your Own Problem 962 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Test Prep for AP® Courses 21.1 Resistors in Series and Parallel 1. Figure 21.59 The figure above shows a circuit containing two batteries and three identical resistors with resistance R. Which of the following changes
to the circuit will result in an increase in the current at point P? Select two answers. a. Reversing the connections to the 14 V battery. b. Removing the 2 V battery and connecting the wires to close the left loop. c. Rearranging the resistors so all three are in series. d. Removing the branch containing resistor Z. 2. In a circuit, a parallel combination of six 1.6-kΩ resistors is connected in series with a parallel combination of four 2.4-kΩ resistors. If the source voltage is 24 V, what will be the percentage of total current in one of the 2.4-kΩ resistors? a. 10% b. 12% c. 20% d. 25% 3. If the circuit in the previous question is modified by removing some of the 1.6 kΩ resistors, the total current in the circuit is 24 mA. How many resistors were removed? a. 1 b. 2 c. 3 d. 4 4. Figure 21.60 Two resistors, with resistances R and 2R are connected to a voltage source as shown in this figure. If the power dissipated in R is 10 W, what is the power dissipated in 2R? a. 1 W b. 2.5 W c. 5 W d. 10 W 5. In a circuit, a parallel combination of two 20-Ω and one 10-Ω resistors is connected in series with a 4-Ω resistor. The source voltage is 36 V. a. Find the resistor(s) with the maximum current. b. Find the resistor(s) with the maximum voltage drop. c. Find the power dissipated in each resistor and hence the total power dissipated in all the resistors. Also find the power output of the source. Are they equal or not? Justify your answer. d. Will the answers for questions (a) and (b) differ if a 3 Ω resistor is added in series to the 4 Ω resistor? If yes, repeat the question(s) for the new resistor combination. This content is available for free at http://cnx.org/content/col11844/1.13 e. If the values of all the resistors and the source voltage are doubled, what will be the effect on the current? 21.2 Electromotive Force: Terminal Voltage 6. Suppose there are two voltage sources – Sources
A and B – with the same emfs but different internal resistances, i.e., the internal resistance of Source A is lower than Source B. If they both supply the same current in their circuits, which of the following statements is true? a. External resistance in Source A’s circuit is more than Source B’s circuit. b. External resistance in Source A’s circuit is less than Source B’s circuit. c. External resistance in Source A’s circuit is the same as Source B’s circuit. d. The relationship between external resistances in the two circuits can’t be determined. 7. Calculate the internal resistance of a voltage source if the terminal voltage of the source increases by 1 V when the current supplied decreases by 4 A? Suppose this source is connected in series (in the same direction) to another source with a different voltage but same internal resistance. What will be the total internal resistance? How will the total internal resistance change if the sources are connected in the opposite direction? 21.3 Kirchhoff’s Rules 8. An experiment was set up with the circuit diagram shown. Assume R1 = 10 Ω, R2 = R3 = 5 Ω, r = 0 Ω and E = 6 V. Figure 21.61 a. One of the steps to examine the set-up is to test points with the same potential. Which of the following points can be tested? a. Points b, c and d. b. Points d, e and f. c. Points f, h and j. d. Points a, h and i. b. At which three points should the currents be measured so that Kirchhoff’s junction rule can be directly confirmed? a. Points b, c and d. b. Points d, e and f. c. Points f, h and j. d. Points a, h and i. c. If the current in the branch with the voltage source is upward and currents in the other two branches are downward, i.e. Ia = Ii + Ic, identify which of the following can be true? Select two answers. a. b. c. d. Ii = Ij - If Ie = Ih - Ii Ic = Ij - Ia Id = Ih - Ij Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 963 d. The measurements reveal that the current through
R1 is 0.5 A and R3 is 0.6 A. Based on your knowledge of Kirchoff’s laws, confirm which of the following statements are true. a. The measured current for R1 is correct but for R3 is incorrect. b. The measured current for R3 is correct but for R1 is incorrect. c. Both the measured currents are correct. d. Both the measured currents are incorrect. e. The graph shown in the following figure is the energy dissipated at R1 as a function of time. Figure 21.62 Which of the following shows the graph for energy dissipated at R2 as a function of time? c. Figure 21.65 d. Figure 21.66 9. For this question, consider the circuit shown in the following figure. Figure 21.67 a. Assuming that none of the three currents (I1, I2, and I3) are equal to zero, which of the following statements is false? a. b. c. The current through R3 is equal to the current I3 = I1 + I2 at point a. I2 = I3 - I1 at point e. through R5. d. The current through R1 is equal to the current through R5. b. Which of the following statements is true? a. E1 + E2 + I1R1 - I2R2 + I1r1 - I2r2 + I1R5 = 0 b. - E1 + E2 + I1R1 - I2R2 + I1r1 - I2r2 - I1R5 = 0 c. E1 - E2 - I1R1 + I2R2 - I1r1 + I2r2 - I1R5 = 0 d. E1 + E2 - I1R1 + I2R2 - I1r1 + I2r2 + I1R5 = 0 c. If I1 = 5 A and I3 = -2 A, which of the following statements is false? a. The current through R1 will flow from a to b and will be equal to 5 A. b. The current through R3 will flow from a to j and will be equal to 2 A. c. The current through R5 will flow from d to e and will be equal to 5 A. d. None of the above. d. If I1 = 5 A and
I3 = -2 A, I2 will be equal to a. 3 A -3 A b. c. 7 A -7 A d. 10. a. Figure 21.63 b. Figure 21.64 964 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Figure 21.68 In an experiment this circuit is set up. Three ammeters are used to record the currents in the three vertical branches (with R1, R2, and E). The readings of the ammeters in the resistor branches (i.e. currents in R1 and R2) are 2 A and 3 A respectively. a. Find the equation obtained by applying Kirchhoff’s loop rule in the loop involving R1 and R2. b. What will be the reading of the third ammeter (i.e. the branch with E)? If E were replaced by 3E, how would this reading change? If the original circuit is modified by adding another voltage source (as shown in the following circuit), find the readings of the three ammeters. c. Figure 21.69 11. 0.19 A. According to the results calculated in part (d) identify the resistor(s). Justify any difference in measured and calculated value. 21.6 DC Circuits Containing Resistors and Capacitors 12. A battery is connected to a resistor and an uncharged capacitor. The switch for the circuit is closed at t = 0 s. a. While the capacitor is being charged, which of the following is true? a. Current through and voltage across the resistor increase. b. Current through and voltage across the resistor decrease. c. Current through and voltage across the resistor first increase and then decrease. d. Current through and voltage across the resistor first decrease and then increase. b. When the capacitor is fully charged, which of the following is NOT zero? a. Current in the resistor. b. Voltage across the resistor. c. Current in the capacitor. d. None of the above. 13. An uncharged capacitor C is connected in series (with a switch) to a resistor R1 and a voltage source E. Assume E = 24 V, R1 = 1.2 kΩ and C = 1 mF. a. What will be the current through the circuit as the switch is closed? Draw a circuit diagram and show the direction of current after the switch is closed. How long will it take for the capacitor to be 99% charged
? b. After full charging, this capacitor is connected in series to another resistor, R2 = 1 kΩ. What will be the current in the circuit as soon as it’s connected? Draw a circuit diagram and show the direction of current. How long will it take for the capacitor voltage to reach 3.24 V? Figure 21.70 In this circuit, assume the currents through R1, R2 and R3 are I1, I2 and I3 respectively and all are flowing in the clockwise direction. a. Find the equation obtained by applying Kirchhoff’s junction rule at point A. b. Find the equations obtained by applying Kirchhoff’s loop rule in the upper and lower loops. c. Assume R1 = R2 = 6 Ω, R3 = 12 Ω, r1 = r2 = 0 Ω, E1 = 6 V and E2 = 4 V. Calculate I1, I2 and I3. d. For the situation in which E2 is replaced by a closed switch, repeat parts (a) and (b). Using the values for R1, R2, R3, r1 and E1 from part (c) calculate the currents through the three resistors. e. For the circuit in part (d) calculate the output power of the voltage source and across all the resistors. Examine if energy is conserved in the circuit. f. A student implemented the circuit of part (d) in the lab and measured the current though one of the resistors as This content is available for free at http://cnx.org/content/col11844/1.13 Appendix A 1531 A ATOMIC MASSES Table A1 Atomic Masses Atomic Number 10 11 Name neutron Hydrogen Deuterium Tritium Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Atomic Mass Number, A Symbol 10 11 11 12 13 14 13 14 15 15 16 18 18 19 20 22 22 1H 2H or D 3H or T 3He 4He 6Li 7Li 7Be 9Be 10B 11B 11C 12C 13C 14C 13N 14N 15N 15O 16O 18O 18F 19F 20Ne 22Ne 22Na 3.016 050 3.016 030 4.002 603 6.015 121 7.016 003 7.016 928 9.012 182
10.012 937 11.009 305 11.011 432 12.000 000 13.003 355 14.003 241 13.005 738 14.003 074 15.000 108 15.003 065 15.994 915 17.999 160 18.000 937 Atomic Mass (u) 1.008 665 Percent Abundance or Decay Mode − Half-life, t1/2 10.37 min 1.007 825 99.985% 2.014 102 0.015% − 12.33 y 1.38×10−4% ≈100% 7.5% 92.5% EC 100% 19.9% 80.1% EC, + 98.90% 1.10% − + 99.63% 0.37% EC, + 99.76% 0.200% EC, + 53.29 d 5730 y 9.96 min 122 s 1.83 h 2.602 y 18.998 403 100% 19.992 435 90.51% 21.991 383 21.994 434 9.22% + 1532 Appendix A Atomic Number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 12 13 14 Magnesium Aluminum Silicon 15 Phosphorus 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Sulfur Chlorine Argon Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium 23 24 24 27 28 31 31 32 32 35 35 37 40 39 40 40 45 48 51 52 55 56 59 60 58 60 63 65 64 66 69 23Na 24Na 24Mg 27Al 28Si 31Si 31P 32P 32S 35S 35Cl 37Cl 40Ar 39K 40K 40Ca 45Sc 48Ti 51V 52Cr 22.989 767 23.990 961 100% − 23.985 042 78.99% 26.981 539 100% 14.96 h 27.976 927 92.23% 2.62h 30.975 362 30.973 762 31.973 907 − 100% − 31.972 070 95.02% 14.28 d 87.4 d 34.969 031 34.968 852 36.965 903 39.962 384 38.963 707 39.963 999 44.
955 910 47.947 947 50.943 962 51.940 509 − 75.77% 24.23% 99.60% 93.26% 100% 73.8% 99.75% 83.79% 100% 0.0117%, EC, − 1.28×109 y 39.962 591 96.94% 55Mn 54.938 047 5.271 y 56Fe 59Co 60Co 58Ni 60Ni 63Cu 65Cu 64Zn 66Zn 69Ga 55.934 939 91.72% 58.933 198 59.933 819 57.935 346 59.930 788 62.939 598 64.927 793 63.929 145 65.926 034 68.925 580 100% − 68.27% 26.10% 69.17% 30.83% 48.6% 27.9% 60.1% This content is available for free at http://cnx.org/content/col11844/1.13 Appendix A 1533 Atomic Number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 71.922 079 73.921 177 74.921 594 79.916 520 78.918 336 83.911 507 84.911 794 85.909 267 87.905 619 89.907 738 88.905 849 89.907 152 27.4% 36.5% 100% 49.7% 50.69% 57.0% 72.17% 9.86% 82.58% − 100% − 89.904 703 51.45% 92.906 377 100% 97.905 406 24.13% 97.907 215 101.904 348 102.905 500 105.903 478 106.905 092 108.904 757 113.903 357 114.903 880 − 31.6% 100% 27.33% 51.84% 48.16% 28.73% 32 Germanium 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Arsenic Selenium Bromine Krypton Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium
Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon 72 74 75 80 79 84 85 86 88 90 89 90 90 93 98 98 102 103 106 107 109 114 115 120 121 130 127 131 132 136 72Ge 74Ge 75As 80Se 79Br 84Kr 85Rb 86Sr 88Sr 90Sr 89Y 90Y 90Zr 93Nb 98Mo 98Tc 102Ru 103Rh 106Pd 107Ag 109Ag 114Cd 115In 120Sn 121Sb 130Te 127I 131I 132Xe 136Xe 95.7%, − 4.4×1014y 119.902 200 32.59% 120.903 821 57.3% 129.906 229 126.904 473 130.906 114 131.904 144 135.907 214 33.8%, − 2.5×1021y 100% − 26.9% 8.9% 8.040 d 28.8 y 64.1 h 4.2×106y 1534 Atomic Number, Z 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Name Cesium Barium Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutecium Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 Appendix A 133 134 137 138 139 140 141 142 145 152 153 158 159 164 165 166 169 174 175 180 181 184 187 191 192 191 193 195 197 198 199 133Cs 134Cs 137Ba 138Ba 139La 140Ce 141Pr 142Nd 145Pm 152Sm 153Eu 158Gd 159Tb 164Dy 165Ho 166Er 132.905 429 133.906 696 136.905 812 137.905 232 138.906 346 139.905 433 140.907 647 100% EC, − 11.23% 71.70% 99.91% 88.48% 100% 141.907 719 27.13% 144.912 743 151.919 729 152.921 225 EC,
26.7% 52.2% 157.924 099 24.84% 2.06 y 17.7 y 158.925 342 163.929 171 164.930 319 165.930 290 169Tm 168.934 212 174Yb 175Lu 180Hf 181Ta 173.938 859 174.940 770 179.946 545 180.947 992 184W 183.950 928 100% 28.2% 100% 33.6% 100% 31.8% 97.41% 35.10% 99.98% 30.67% 187Re 191Os 192Os 191Ir 193Ir 195Pt 197Au 198Au 199Hg 186.955 744 190.960 920 191.961 467 190.960 584 192.962 917 194.964 766 196.966 543 197.968 217 62.6%, − 4.6×1010y − 41.0% 37.3% 62.7% 33.8% 100% − 15.4 d 2.696 d 198.968 253 16.87% This content is available for free at http://cnx.org/content/col11844/1.13 Appendix A 1535 Atomic Number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 202Hg 201.970 617 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 Thallium Lead Bismuth Polonium Astatine Radon Francium Radium Actinium Thorium Protactinium Uranium Neptunium Plutonium Americium Curium Berkelium 202 205 206 207 208 210 211 212 209 211 210 218 222 223 226 227 228 232 231 233 235 236 238 239 239 239 243 245 247 205Tl 206Pb 207Pb 208Pb 210Pb 211Pb 212Pb 209Bi 211Bi 210Po 218At 222Rn 223Fr 226Ra 227Ac 228Th 232Th 231Pa 233U 235U 236U 238U 239U 239Np 239Pu 204.974 401 205.974 440 206.975 872 207.976 627 209.984 163 210.988 735 211.991 871 208.980 374 210.987 255 209.982 848 218.008 684 222.017
570 223.019 733 226.025 402 227.027 750 228.028 715 29.86% 70.48% 24.1% 22.1% 52.4% − − − 100% − − − − 22.3 y 36.1 min 10.64 h 2.14 min 138.38 d 1.6 s 3.82 d 21.8 min 1.60×103y 21.8 y 1.91 y 232.038 054 100%, 1.41×1010y 231.035 880 233.039 628 235.043 924 0.720%, 236.045 562 238.050 784 99.2745%, 239.054 289 239.052 933 239.052 157 − − 243Am 243.061 375 fission 245Cm 245.065 483 247Bk 247.070 300 3.28×104y 1.59×103y 7.04×108y 2.34×107y 4.47×109y 23.5 min 2.355 d 2.41×104y 7.37×103y 8.50×103y 1.38×103y 1536 Appendix A Atomic Number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 98 99 100 101 102 103 104 105 106 107 108 109 Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium Rutherfordium Dubnium Seaborgium Bohrium Hassium Meitnerium 249 254 253 255 255 257 261 262 263 262 264 266 249Cf 254Es 253Fm 255Md 255No 257Lr 261Rf 262Db 263Sg 262Bh 264Hs 266Mt 249.074 844 254.088 019 253.085 173 255.091 081 255.093 260 257.099 480 261.108 690 262.113 760 263.11 86 262.123 1 264.128 5 266.137 8 − EC, EC, EC, EC, fission fission 351 y 276 d 3.00 d 27 min 3.1 min 0.646 s 1.08 min 34 s 0.8 s 0.102 s 0.08 ms 3.4 ms This content is available for free at http://cnx.org/content/col11844/1.13 Appendix B 1537 B SELECT
ED RADIOACTIVE ISOTOPES Decay modes are, −, + would + are roughly one-half the maxima. decay. IT is a transition from a metastable excited state. Energies for ±, electron capture (EC) and isomeric transition (IT). EC results in the same daughter nucleus as decays are the maxima; average energies Table B1 Selected Radioactive Isotopes Isotope t 1/2 DecayMode(s) Energy(MeV) Percent γ -Ray Energy(MeV) Percent 100% 100% 100% 90% 1.27 100% 0.0186 0.156 1.20 0.55 1.71 0.167 0.710 1.31 0.827 100% 100% 100% 89% 87% 0.257 100% 3.69 28% 1.80 43% 0.273 0.466 0.318 45% 55% 100% 3H 14C 13N 22Na 32P 35S 36Cl 40K 43K 45Ca 51Cr 52Mn 52Fe 59Fe 60Co 65Zn 67Ga 12.33 y 5730 y 9.96 min 2.602 y 14.28 d 87.4 d 3.00×105y 1.28×109y 22.3 h 165 d 27.70 d 5.59d 8.27 h 44.6 d 5.271 y 244.1 d 78.3 h − − + + − − − − − − EC + + − s − EC EC 75Se 118.5 d EC s s s s s s 0.373 0.618 0.320 1.33 1.43 0.169 0.378 1.10 1.29 1.17 1.33 1.12 0.0933 0.185 0.300 others 0.121 0.136 0.265 87% 87% 10% 28% 28% 43% 43% 57% 43% 100% 100% 51% 70% 35% 19% 20% 65% 68% 1538 Isotope t 1/2 DecayMode(s) Energy(MeV) Percent γ -Ray Energy(MeV) Appendix B Percent 20% 9% 0.280 others 1.08 0.514 100% 0.142 0.392 0.159 0.364 others 0.0400 0.372 0.411 others 0.662 0.030 0.044 0.537 others 0.412
100% 100% ≈100% 85% 35% 32% 25% 95% 25% 65% 24% ≈100% 0.0733 100% 0.186 100% ≈100% s numerous <0.400% s s 23% 77% 11% 15% 73% 0.050 23% numerous <0.250% 7.5×10−5 0.013 0.052 73% 15% 10% 86Rb 18.8 d 85Sr 90Sr 90Y 99mTc 113mIn 123I 131I 64.8 d 28.8 y 64.1 h 6.02 h 99.5 min 13.0 h 8.040 d − s EC − − IT IT EC − s 129Cs 32.3 h EC 137Cs 30.17 y 140Ba 12.79 d 198Au 197Hg 210Po 226Ra 2.696 d 64.1 h 138.38 d 1.60×103y 235U 238U 7.038×108y 4.468×109y 237Np 2.14×106y 239Pu 2.41×104y − s − − EC s s s s 0.69 1.77 0.546 2.28 0.248 0.607 others 0.511 1.17 1.035 s s 9% 91% 100% 100% 7% 93% 95% 5% ≈100% s 1.161 ≈100% 100% 5% 95% 5.41 4.68 4.87 4.68 4.22 4.27 numerous 4.96 (max.) 5.19 5.23 5.24 This content is available for free at http://cnx.org/content/col11844/1.13 Appendix B 1539 Isotope t 1/2 DecayMode(s) Energy(MeV) Percent γ -Ray Energy(MeV) Percent 243Am 7.37×103y s Max. 5.44 s 5.37 5.32 others 88% 11% others 0.075 others 1540 Appendix B This content is available for free at http://cnx.org/content/col11844/1.13 Appendix C 1541 C USEFUL INFORMATION This appendix is broken into several tables. • Table C1, Important Constants • Table C2, Submicroscopic Masses • Table C3, Solar System Data • Table C4, Metric Prefixes for
Powers of Ten and Their Symbols • Table C5, The Greek Alphabet • Table C6, SI units • Table C7, Selected British Units • Table C8, Other Units • Table C9, Useful Formulae Table C1 Important Constants [1] Symbol Meaning Best Value Approximate Value σ ε0 μ0 Speed of light in vacuum Gravitational constant Avogadro’s number Boltzmann’s constant Gas constant StefanBoltzmann constant Coulomb force constant Charge on electron Permittivity of free space Permeability of free space Planck’s constant 2.99792458 × 108 m / s 3.00 × 108 m / s 6.67384(80) × 10−11 N ⋅ m2 / kg2 6.67 × 10−11 N ⋅ m2 / kg2 6.02214129(27) × 1023 6.02 × 1023 1.3806488(13) × 10−23 J / K 1.38 × 10−23 J / K 8.3144621(75) J / mol ⋅ K 8.31 J / mol ⋅ K = 1.99 cal / mol ⋅ K = 0.0821atm ⋅ L / mol ⋅ K 5.670373(21) × 10−8 W / m2 ⋅ K 5.67 × 10−8 W / m2 ⋅ K 8.987551788... × 109 N ⋅ m2 / C2 8.99 × 109 N ⋅ m2 / C2 −1.602176565(35) × 10−19 C −1.60 × 10−19 C 8.854187817... × 10−12 C2 / N ⋅ m2 8.85 × 10−12 C2 / N ⋅ m2 4π × 10−7 T ⋅ m / A 1.26 × 10−6 T ⋅ m / A 6.62606957(29) × 10−34 J ⋅ s 6.63 × 10−34 J ⋅ s Table C2 Submicroscopic Masses [2] Symbol Meaning Best Value Approximate Value Electron mass 9.10938291(40)×10−31kg 9.11×10−31kg Proton mass 1.6726
21777(74)×10−27kg 1.6726×10−27kg 1. Stated values are according to the National Institute of Standards and Technology Reference on Constants, Units, and Uncertainty, www.physics.nist.gov/cuu (http://www.physics.nist.gov/cuu) (accessed May 18, 2012). Values in parentheses are the uncertainties in the last digits. Numbers without uncertainties are exact as defined. 2. Stated values are according to the National Institute of Standards and Technology Reference on Constants, Units, and Uncertainty, www.physics.nist.gov/cuu (http://www.physics.nist.gov/cuu) (accessed May 18, 2012). Values in parentheses are the uncertainties in the last digits. Numbers without uncertainties are exact as defined. 1542 Appendix C Symbol Meaning Best Value Approximate Value u Neutron mass 1.674927351(74)×10−27kg 1.6749×10−27kg Atomic mass unit 1.660538921(73)×10−27kg 1.6605×10−27kg Table C3 Solar System Data Sun mass average radius Earth-sun distance (average) Earth mass average radius orbital period Moon mass average radius orbital period (average) 1.99×1030kg 6.96×108m 1.496×1011m 5.9736×1024kg 6.376×106m 3.16×107s 7.35×1022kg 1.74×106m 2.36×106s Earth-moon distance (average) 3.84×108m Table C4 Metric Prefixes for Powers of Ten and Their Symbols Prefix Symbol Value Prefix Symbol Value tera giga mega kilo hecto deka T G M k h da 1012 109 106 103 102 101 deci centi milli micro nano pico — — 100( = 1) femto d c m n p f 10−1 10−2 10−3 10−6 10−9 10−12 10−15 Table C5 The Greek Alphabet Alpha Α Eta Η Nu Ν Tau Τ Beta Β Theta Θ Xi Ξ Upsilon Υ Gamma Γ Iota Ι Omicron Ο Phi Φ Delta Δ Kappa Κ Pi Epsilon Ε Lambda Λ Rho Π
Chi Ρ Psi Χ Ψ Zeta Ζ Mu Μ Sigma Σ Omega Ω This content is available for free at http://cnx.org/content/col11844/1.13 Appendix C 1543 Table C6 SI Units Entity Abbreviation Name Fundamental units Length Mass Time Current Supplementary unit Angle Derived units Force Energy Power Pressure m kg s A rad meter kilogram second ampere radian N = kg ⋅ m / s2 newton J = kg ⋅ m2 / s2 joule W = J / s Pa = N / m2 watt pascal hertz volt farad coulomb Frequency Hz = 1 / s Electronic potential V = J / C Capacitance Charge Resistance Ω = V / A ohm Magnetic field T = N / (A ⋅ m) tesla Nuclear decay rate Bq = 1 / s becquerel Table C7 Selected British Units Length 1 inch (in.) = 2.54 cm (exactly) 1 foot (ft) = 0.3048 m 1 mile (mi) = 1.609 km Force 1 pound (lb) = 4.448 N Energy 1 British thermal unit (Btu) = 1.055×103 J Power 1 horsepower (hp) = 746 W Pressure 1 lb / in2 = 6.895×103 Pa Table C8 Other Units Length 1 light year (ly) = 9.46×1015 m 1 astronomical unit (au) = 1.50×1011 m 1 nautical mile = 1.852 km 1 angstrom(Å) = 10−10 m Area 1 acre (ac) = 4.05×103 m2 1 square foot (ft2) = 9.29×10−2 m2 1 barn () = 10−28 m2 Volume 1 liter () = 10−3 m3 1544 Appendix C 1 U.S. gallon (gal) = 3.785×10−3 m3 Mass 1 solar mass = 1.99×1030 kg Time Speed Angle 1 metric ton = 103 kg 1 atomic mass unit () = 1.6605×10−27 kg 1 year () = 3.16×107 s 1 day () = 86400 s 1 mile per hour (mph) = 1.609 km / h 1 nautical mile per hour (naut) = 1.852 km / h 1 degree () = 1.745×10−2
: 2×1031 if an average human lifetime is taken to be about 70 years. 33 Sample answer: 50 atoms 35 Sample answers: (a) 1012 cells/hummingbird (b) 1016 cells/human Chapter 2 Problems & Exercises 1 (a) 7 m (b) 7 m (c) +7 m 3 (a) 13 m (b) 9 m (c) +9 m 5 (a) 3.0×104 m/s (b) 0 m/s 7 2×107 years 9 34.689 m/s = 124.88 km/h 11 (a) 40.0 km/h (b) 34.3 km/h, 25º S of E. (c) average speed = 3.20 km/h, - = 0. 13 384,000 km 15 (a) 6.61×1015 rev/s (b) 0 m/s 16 4.29 m/s2 18 (a) 1.43 s (b) −2.50 m/s2 20 (a) 10.8 m/s This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) 1559 Figure 2.48. 21 38.9 m/s (about 87 miles per hour) 23 (a) 16.5 s (b) 13.5 s (c) −2.68 m/s2 25 (a) 20.0 m (b) −1.00 m/s (c) This result does not really make sense. If the runner starts at 9.00 m/s and decelerates at 2.00 m/s2, then she will have stopped after 4.50 s. If she continues to decelerate, she will be running backwards. 27 0.799 m 29 (a) 28.0 m/s (b) 50.9 s (c) 7.68 km to accelerate and 713 m to decelerate 31 (a) 51.4 m (b) 17.1 s 33 (a) −80.4 m/s2 (b) 9.33×10−2 s 35 (a) 7.7 m/s (b) −15×102 m/s2. This is about 3 times the deceleration of the pilots, who were falling from thousands of meters high! 37 (a) 32.6
m/s2 (b) 162 m/s (c) > max, because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be 1560 Answer Key greatest at the beginning, so it would not be accelerating at 32.6 m/s2 during the last few meters, but substantially less, and the final velocity would be less than 162 m/s. 39 104 s 40 (a) = 12.2 m/s ; = 4.07 m/s2 (b) = 11.2 m/s 41 (a) 1 = 6.28 m ; 1 = 10.1 m/s (b) 2 = 10.1 m ; 2 = 5.20 m/s (c) 3 = 11.5 m ; 3 = 0.300 m/s (d) 4 = 10.4 m ; 4 = −4.60 m/s 43 0 = 4.95 m/s 45 (a) = −9.80 m/s2 ; 0 = 13.0 m/s ; 0 = 0 m (b) = 0m/s. Unknown is distance to top of trajectory, where velocity is zero. Use equation 2 = 0 because it contains all known values except for, so we can solve for. Solving for gives 2 + 2( − 0) 2 = 2( − 00 m/s)2 − (13.0 m/s)2 −9.80 m/s2 2 = 8.62 m (2.100) Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result. (c) 2.65 s 47 Figure 2.57. (a) 8.26 m (b) 0.717 s 49 1.91 s 51 (a) 94.0 m (b) 3.13 s This content is available for free at http://cnx.org/content/col11844/1.13 1561 = (11.7 − 6.95)×103 m (40.0 – 20.0) s = 238 m/s (2.114) Answer Key 53 (a) -70.0 m/s (downward) (b) 6.10 s 55 (a) 19.6 m (b) 18.5 m
57 (a) 305 m (b) 262 m, -29.2 m/s (c) 8.91 s 59 (a) 115 m/s (b) 5.0 m/s2 61 63 Figure 2.63. 65 (a) 6 m/s (b) 12 m/s (c) 3 m/s2 (d) 10 s Test Prep for AP® Courses 1 (a) 3 a. Use tape to mark off two distances on the track — one for cart A before the collision and one for the combined carts after the collision. Push cart A to give it an initial speed. Use a stopwatch to measure the time it takes for the cart(s) to cross the marked distances. The speeds are the distances divided by the times. If the measurement errors are of the same magnitude, they will have a greater effect after the collision. The speed of the combined carts will be less than the initial speed of cart A. As a result, these errors will be a greater percentage of the actual velocity value after the collision occurs. (Note: Other arguments could properly be made for ‘more error before the collision' and error that ‘equally affects both sets of measurement.') b. 5 1562 Answer Key The position vs. time graph should be represented with a positively sloped line whose slope steadily decreases to zero. The y-intercept of the graph may be any value. The line on the velocity vs. time graph should have a positive yintercept and a negative slope. Because the final velocity of the book is zero, the line should finish on the x-axis. The position vs. time graph should be represented with a negatively sloped line whose slope steadily decreases to zero. The y-intercept of the graph may be any value. The line on the velocity vs. time graph should have a negative yintercept and a positive slope. Because the final velocity of the book is zero, the line should finish on the x-axis.] 7 (c) Chapter 3 Problems & Exercises 1 (a) 480 m (b) 379 m, 18.4° east of north 3 north component 3.21 km, east component 3.83 km 5 19.5 m, 4.65° south of west 7 (a) 26.6 m, 65.1° north of east (b) 26.6 m, 65.1° south of west 9 52.9 m, 90.1° with

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