PH203 Chapter 23 solutions Tactics Box 231 Using Kirchhoff's Loop Law Description: Knight/Jones/Field Tactics Box 231 Using Kirchhoff s loop law is illustrated Learning Goal: To practice Tactics Box 231 Using Kirchhoff 'loop law Circuit analysis is based on Kirchhoff's laws, which can be summarized as follows: Kirchhoff's junction law says that the total current into a junction must equal the total current leaving the junction Kirchhoff's loop law says that if we add all of the potential differences around the loop formed by a circuit, the sum of these potential differences must be zero While Kirchhoff's junction law is needed only when there are one or more junctions in a circuit, Kirchhoff's loop law is used for analyzing any type of circuit, as explained in the following Tactics Box TACTICS BOX 231 Using Kirchhoff s loop law 1 Draw a circuit diagram Label all known and unknown quantities 2 Assign a direction to the current Draw and label a current arrow to show your choice Choose the direction of the current based on how the batteries or sources of emf "want" the current to go If you choose the current direction opposite the actual direction, the final value for the current that you calculate will have the correct magnitude but will be negative, letting you know that the direction is opposite the direction you chose 3 "Travel" around the loop Start at any point in the circuit; then go all the way around the loop in the direction you assigned to the current in step 2 As you go through each circuit element, is interpreted to mean For a battery with current in the negative-to-positive direction: For a battery in the positive-to-negative direction (ie, the current is going into the
positive terminal of the battery): For a resistor: 4 Apply the loop law: Part A
The current in the circuit shown in the figure is 020 What is the potential difference across the battery traveling in the direction shown in the figure? A1 Find the potential difference across the resistor What is the potential difference across the resistor in the direction assigned to the current in the figure? Express your answer in volts
Express your answer in volts Now apply Kirchhoff's loop law to find the potential difference across the battery Express your answer in volts Part B Find the current in the circuit shown in the figure B1 Find the potential difference across the battery If we assign the counterclockwise direction to the current, what is the potential difference across the battery,? Express your answer in volts
B2 Find the potential difference across the 40-ohm resistor If we assign the counterclockwise direction to the current in the circuit, what is the potential difference across the 40-ohm resistor? Express your answer in terms of the variable B3 Find the potential difference across the 50-ohm resistor If we assign the counterclockwise direction to the current in the circuit, what is the potential difference across the 50-ohm resistor? Express your answer in terms of the variable Now apply Kirchhoff's loop law and solve for Express your answer in amperes Part C What is the potential difference across the unknown element in the circuit shown in the figure? Express your answer as if traveling across the element in the direction shown in the figure
C1 Apply Kirchhoff's loop law Complete the expression below and write an equation for Kirchhoff's loop law applied to the circuit in the figure Express your answer in terms of the variable Express your answer in volts Series And Parallel Connections Description: Several calculations of increasing complexity that help the students practice finding the equivalent resistance of the circuits combining series and parallel connections Learning Goal: To learn to calculate the equivalent resistance of the circuits combining series and parallel connections Resistors are often connected to each other in electric circuits Finding the equivalent resistance of combinations of resistors is a common and important task Equivalent resistance is defined as the single resistance that can replace the given combination of resistors in such a manner that the currents in the rest of the circuit do not change Finding the equivalent resistance is relatively straighforward if the circuit contains only series and
parallel connections of resistors An example of a series connection is shown in the diagram: For such a connection, the current is the same for all individual resistors and the total voltage is the sum of the voltages across the individual resistors Using Ohm's law ( ), one can show that, for a series connection, the equivalent resistance is the sum of the individual resistances Mathematically, these relationships can be written as: An example of a parallel connection is shown in the diagram:
For resistors connected in parallel the voltage is the same for all individual resistors because they are all connected to the same two points (A and B on the diagram) The total current is the sum of the currents through the individual resistors This should makes sense as the total current "splits" at points A and B Using Ohm's law, one can show that, for a parallel connection, the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances Mathematically, these relationships can be written as: NOTE: If you have already studied capacitors and the rules for finding the equivalent capacitance, you should notice that the rules for the capacitors are similar - but not quite the same as the ones discussed here In this problem, you will use the the equivalent resistance formulas to determine combinations of resistors Part A for various For the combination of resistors shown, find the equivalent resistance between points A and B
Express your answer in Ohms These resistors are connected in series; the current through each is the same Part B For the set-up shown, find the equivalent resistance between points A and B
Express your answer in Ohms This is a parallel connection since the voltage across each resistor is the same Part C For the combination of resistors shown, find the equivalent resistance between points A and B
C1 How to approach the question You cannot say that all three resistors are connected either in series or in parallel: this circuit has to be viewed as a combination of different connections Find the equivalent resistance of the "4-Ohm-12 Ohm" combination first C2 What kind of connection is this? The 2-Ohm resistor is connected: in series with the 4-Ohm resistor in series with the 12-Ohm resistor in series with the combination of the 4-Ohm and the 12-Ohm resistors in parallel with the 4-Ohm resistor in parallel with the 12-Ohm resistor in parallel with the combination of the 4-Ohm and the 12-Ohm resistors Express your answer in Ohms
In this case, you cannot say that all three resistors are connected either in series or in parallel You have a combination of a series and a parallel connection Some circuits may contain a large number of resistors connected in various ways To determine the equivalent resistance of such circuits, you have to take several steps, carefully selecting the "subcombinations" of resistors connected in relatively obvious ways Good record-keeping is essential here The next question helps you practice this skill Part D For the combination of resistors shown, find the equivalent resistance between points A and B D1 How to approach the question Find separately the equivalent resistances of the top and the bottom branches of the circuit; then combine them D2 Find for the "4-6-12" combination What is the equivalent resistance for the "4 ohm - 6 ohm - 12 Ohm" combination? Express your answer in ohms
D3 Find for the top branch What is the equivalent resistance for the top branch of the circuit (between C and D)? Express your answer in ohms D4 Find for the bottom branch What is the equivalent resistance for the bottom branch of the circuit (between E and F)? Express your answer in ohms Express your answer in Ohms The next level of analyzing a circuit is to determine the voltages across and the currents through the various branches of the circuit You will practice that skill in the future Of course, there are circuits that cannot possibly be represented as combinations of series and parallel connections However, there are ways to analyze those, too Brightness of Light Bulbs Ranking Task Description: Asks students to rank brightness of light bulbs in a mixed series and parallel circuit Part A Consider a circuit containing five identical light bulbs and an ideal battery Assume that the resistance of each light bulb remains constant Rank the bulbs (A through E) based on their brightness A1 How to approach the problem The greater the current in a bulb, the brighter the bulb is Compare the currents in the light bulbs A2 Comparing bulb A to bulb B When a wire "splits," the current splits according to the resistance in each pathway How does the resistance in the pathway through bulb A compare to the resistance in the pathway through bulb B? A3 Comparing bulb D to bulb E
Bulb D and bulb E are connected in series How much of the charge that goes through D also goes through E? A4 Comparing bulb C to bulb D or E When a wire "splits," the current splits according to the resistance in each pathway How does the resistance in the pathway of bulb C compare to the resistance in the pathway of bulbs D and E? Recall that two bulbs connected in series have more resistance than a single bulb A5 Comparing bulb C to bulb A or B What fraction of the total current (through the battery) goes through the pathway containing bulb A? Recall that the pathways through bulbs A and B have the same resistance, since the bulbs are identical Compare that number to the fraction of the current that goes through the pathway containing bulb C Rank from brightest to dimmest To rank items as equivalent, overlap them View Now consider what happens when a switch in the circuit is opened Part B What happens to the brightness of bulb A?
B1 How to approach this part Light bulbs D and E are, in effect, disconnected from the circuit The current exists only in light bulbs A, B, and C How would the change affect the currents in A, B, and C? B2 Consider changes in resistance How does the resistance of bulb C alone compare with the resistance of bulb C in parallel with bulbs D and E? It gets dimmer It gets brighter There is no change Part C What happens to bulb C? C1 How to approach this part This question can be answered by calculating the "before" and "after" currents through C in terms of the emf provided by the battery and the resistance of each bulb C2 Find the current in bulb C earlier Before the switch was open, the total resistance of the earlier circuit was, where is the resistance of one bulb What was the current in bulb C? Let the emf produced by the battery be Express your answer in terms of and C3 Find the current in bulb C now What is the current through bulb C after opening the switch? Let the emf produced by the battery be
Express your answer in terms of and It gets dimmer It gets brighter There is no change This is why appliances in your home are always connected in parallel Otherwise, turning some of them on or off would cause the current in others to change, which could damage them Equivalent Resistance Description: Find the equivalent resistance of a network of resistors with series and parallel connections The network geometry gets progressively more complicated by adding more resistors Consider the network of four resistors shown in the diagram, where 200, 500, 100, and 700 The resistors are connected to a constant voltage of magnitude Part A
Find the equivalent resistance A1 of the resistor network How to reduce the network of resistors The network of resistors shown in the diagram is a combination of series and parallel connections To determine its equivalent resistance, it is most convenient to reduce the network in successive stages First compute the equivalent resistance of the parallel connection between the resistors and, and imagine replacing the connection with a resistor with such resistance The resulting network will consist of three resistors in series Then find their equivalent resistance, which will also be the equivalent resistance of the original network A2 Find the resistance equivalent to and Find the equivalent resistance of the parallel connection between the resistors and A21 Two resistors in parallel Consider two resistors of resistance and that are connected in parallel They are equivalent to a resistor with resistance, which satisfies the following relation: Express your answer in ohms If you replace the resistors and with an equivalent resistor with resistance, the resulting network will consist of three resistors,, and connected in series Their equivalent resistance is also the equivalent resistance of the original network A3 Three resistors in series
Consider three resistors of resistance,, and that are connected in series They are equivalent to a resistor with resistance, which is given by Express your answer in ohms Part B Two resistors of resistance 300 and 300 are added to the network, and an additional resistor of resistance 300 is connected by a switch, as shown in the diagram Find the equivalent resistance B1 of the new resistor network when the switch is open How to reduce the extended network of resistors Since the switch is open, no current passes through the resistor, which can be ignored then As you did in Part A, reduce the network in successive stages Note that the new resistor is in
series with the resistors and, while the new resistor is in series with B2 Find the resistance equivalent to,, and Find the resistance equivalent to the resistor connection with,, and B21 Find the resistance equivalent to and Find the resistance equivalent to the connection between and B211 Two resistors in series Consider two resistors of resistance, and that are connected in series They are equivalent to a resistor with resistance Express your answer in ohms, which is given by If you replace the resistors and with their equivalent resistor (of resistance ), the resistor will result in parallel with B22 Two resistors in parallel Consider two resistors of resistance and that are connected in parallel They are equivalent to a resistor with resistance Express your answer in ohms, which satisfies the following relation:
If you replace the resistors,, and with an equivalent resistor with resistance, the resulting network will consist of four resistors,,, and all connected in series Their equivalent resistance is also the equivalent resistance of the original network B3 Four resistors in series Consider four resistors of resistance,,, and that are connected in series They are equivalent to a resistor with resistance Express your answer in ohms, which is given by Part C Find the equivalent resistance closed C1 of the resistor network described in Part B when the switch is How to reduce the network of resistors when the switch is closed When the switch is closed, current passes through the resistor ; therefore the resistor must be included in the calculation of the equivalent resistance Also when the switch is closed, the resistor is no longer connected in series with the resistors and, as was the case when the switch was open Instead, now is in parallel with and their equivalent resistor will be in series with and C2 Find the resistance equivalent to and Find the equivalent resistance of the parallel connection between the resistors and C21 Two resistors in parallel
Consider two resistors of resistance and that are connected in parallel They are equivalent to a resistor with resistance, which satisfies the following relation: Express your answer in ohms If you replace the resistors and with their equivalent resistor (of resistance ), and the resistors, and with their equivalent resistor (of resistance ), calculated in Part B, the resulting network will consist of four resistors,,, and all connected in series Their equivalent resistance is also the equivalent resistance of the original network C3 Four resistors in series Consider four resistors of resistance,,, and that are connected in series They are equivalent to a resistor with resistance, which is given by Express your answer in ohms Resistance and Wire Length Description: Cut a long resistor of known resistance into n identical shorter resistors Find the resistance of the short resistors reconnected in parallel You have been given a long length of wire You measure the resistance of the wire, and find it to be You then cut the wire into identical pieces
Part A If you connect the pieces in parallel as shown, what is the total resistance of the wires connected in parallel? A1 Find the resistance of the wire segments
A wire's resistance is proportional to its length Find the resistance of each of the short segments of wire A11 Formula for the resistance of a wire The formula for the resistance of a wire in terms of its length and cross-sectional area is where is the resistivity of the material of the wire, A2 Resistors in parallel For resistors in parallel of resistance for, the total resistance can be determined from the equality If all resistors have the same resistance, what is the total resistance of the parallel resistors? Express your answer in terms of and Express your answer in terms of and Measuring the EMF and Internal Resistance of a Battery
Description: Find the emf and internal resistance of a battery using a simple circuit Could be used as a pre-lab When switch S in the figure is open, the voltmeter V of the battery reads 312 When the switch is closed, the voltmeter reading drops to 295, and the ammeter A reads 168 Assume that the two meters are ideal, so they do not affect the circuit Part A Find the emf Express your answer in volts to three significant digits Part B Find the internal resistance B1 of the battery How to approach the problem The voltmeter reading is equal to minus the voltage drop across the internal resistor If there is no current flowing, there is no voltage drop across the internal resistor, but once current starts to flow, there will be a voltage drop across it Express your answer in ohms to four significant digits
Part C Find the circuit resistance C1 Find the voltage drop across the circuit resistor When the switch is closed, what is the voltage drop across the circuit resistor (of resistance )? Express your answer in volts to three significant digits You now know the voltage drop across the resistor and the current through it Use Ohm's law to find Express your answer in ohms to three significant digits This is the kind of circuit you would use in real life to measure the emf and internal resistance of a battery You need the second resistor to increase the resistance in the circuit so that the current flowing through the ammeter is not too large In fact, you would need to figure out roughly how big a resistance to use once you had determined the emf of the battery, depending on the range of the ammeter you were using Kirchhoff's Current Rule Ranking Task Description: Short conceptual problem about currents through resistors in various circuits (ranking task) The placement of resistors in a circuit is one factor that can determine the current passing through the resistor You will be given three circuits, and for each circuit you will be asked to compare the current through the various resistors In each of the circuits in Parts A to C, all resistors are identical Part A
Rank the resistors in the figure below (A to C) on the basis of the current that flows through them A1 Kirchhoff's current rule for circuit junctions Kirchhoff s current rule states that the current flowing into a junction (a point at which the number of paths available for current flow changes) must equal the current flowing out of the junction The portion of the current that flows through each available path depends on the resistance of each path Paths with less resistance will receive a larger share of the current Rank from largest to smallest To rank items as equivalent, overlap them View Part B Rank the resistors in the figure below (A to C) on the basis of the current that flows through them
B1 Kirchhoff's current rule for circuit junctions Kirchhoff s current rule states that the current flowing into a junction (a point at which the number of paths available for current flow changes) must equal the current flowing out of the junction The portion of the current that flows through each available path depends on the resistance of each path Paths with less resistance will receive a larger share of the current Rank from largest to smallest To rank items as equivalent, overlap them View Part C Rank the resistors in the figure below (A to D) on the basis of the current that flows through them
C1 Kirchhoff's current rule for circuit junctions Kirchhoff s current rule states that the current flowing into a junction (a point at which the number of paths available for current flow changes) must equal the current flowing out of the junction The portion of the current that flows through each available path depends on the resistance of each path Paths with less resistance will receive a larger share of the current Rank from largest to smallest To rank items as equivalent, overlap them View Kirchhoff's Loop Rule Conceptual Question Description: Short conceptual problem related to Kirchhoff's Loop (Voltage) Rule The circuit shown below
consists of four different resistors and a battery You don't know the strength of the battery or the value any of the four resistances Part A Select the expressions that will be equal to the voltage of the battery in the circuit, where example, is the potential drop across resistor A A1 Kirchhoff's voltage rule for closed circuit loops, for Kirchhoff s loop rule states that in any closed circuit loop, the voltage supplied by a battery must be used by the devices in the loop Therefore, the voltage drop across all of the resistors in a single closed circuit loop must add up to the voltage of the battery Carefully identify all of the closed loops in this circuit Check all that apply
Capacitors in Parallel Description: A series of questions on basic calculations for capacitors connected in parallel; there is a similar skill-builder ("Capacitors in Series") about capacitors connected in series (Knight potential difference notation) Learning Goal: To understand how to calculate capacitance, voltage, and charge for a parallel combination of capacitors Frequently, several capacitors are connected together to form a collection of capacitors We may be interested in determining the overall capacitance of such a collection The simplest configuration to analyze involves capacitors connected in series or in parallel More complicated setups can often (though not always!) be treated by combining the rules for these two cases Consider the example of a parallel combination of capacitors: Three capacitors are connected to each other and to a battery as shown in the figure The individual capacitances are,, and, and the battery's voltage is Part A If the potential of plate 1 is, then, in equilibrium, what are the potentials of plates 3 and 6? Assume that the negative terminal of the battery is at zero potential A1 Electrostatic equilibrium When electrostatic equilibrium is reached, all objects connected by a conductor (by wires, for example) must have the same potential Which plates on this diagram are at the same potential? and and
and and Part B If the charge of the first capacitor (the one with capacitance ) is, then what are the charges of the second and third capacitors? B1 Definition of capacitance Capacitance is given by, where is the charge of the capacitor and is the voltage across it B2 Voltages across the capacitors As established earlier, the voltage across each capacitor is for capacitors connected in parallel The voltage is always the same and and and and Part C Suppose we consider the system of the three capacitors as a single "equivalent" capacitor Given the charges of the three individual capacitors calculated in the previous part, find the total charge for this equivalent capacitor Express your answer in terms of and
Part D Using the value of, find the equivalent capacitance for this combination of capacitors D1 Using the definition of capacitance Use the general formula to find The charge on the "equivalent" capacitor is, and the voltage across this capacitor is the voltage across the battery, Express your answer in terms of The formula for combining three capacitors in parallel is How do you think this formula may be generalized to capacitors? Equivalent Capacitance Description: Find the equivalent capacitance of a network of capacitors with series and parallel connections Consider the combination of capacitors shown in the diagram, where 300, 110, 300, and 500
Part A Find the equivalent capacitance A1 of the network of capacitors How to reduce the network of capacitors To find the equivalent capacitance of the given network of capacitors, it is most convenient to reduce the network in successive stages First, replace the capacitors,, and, which are in parallel, with a single capacitor with an equivalent capacitance By doing so, you will reduce the network to a series connection of two capacitors At this point, you only need to find their equivalent capacitance A2 Find the capacitance equivalent to,, and Find the capacitance equivalent to the parallel connection of the capacitors,, and A21 Find the capacitance equivalent to and Find the capacitance equivalent to the parallel connection of the capacitors and A211 Two capacitors in parallel
Consider two capacitors of capacitance and connected in parallel They are equivalent to a capacitor with capacitance given by Express your answer in microfarads If you replace the capacitors and with a capacitor of capacitance, the resulting network would be a parallel connection between and Express your answer in microfarads If you replace the capacitors,, and with a capacitor of capacitance, the resulting network would be a series connection between and Its equivalent capacitance is also the equivalent capacitance of the original network A3 Two capacitors in series Consider two capacitors of capacitance and connected in series They are equivalent to a capacitor of capacitance that satisfies the following relation: Express your answer in microfarads Part B Two capacitors of capacitance 600 and 300 are added to the network, as shown in the diagram
Find the equivalent capacitance B1 of the new network of capacitors How to reduce the extended network of capacitors To determine the equivalent capacitance of the extended network of capacitors, it is again convenient to reduce the network in successive stages First, determine the equivalent capacitance of the series connection of the capacitors and Then, combine it with the equivalent capacitance of the parallel connection of,, and, and replace the five capacitors with their equivalent capacitor The resulting network will consist of two capacitors in series At this point, you only need to find their equivalent capacitance B2 Find the equivalent capacitance of,,,, and Find the equivalent capacitance of the combination of capacitors,,,, and B21 Find the equivalent capacitance of and Find the equivalent capacitance of the series connection of and B211 Two capacitors in series
Consider two capacitors of capacitance and connected in series They are equivalent to a capacitor of capacitance that satisfies the following relation: Express your answer in microfarads B22 Find the equivalent capacitance of,, and Find the equivalent capacitance of the parallel connection of,, and B221 Three capacitors in parallel Consider three capacitors of capacitance,, and connected in parallel They are equivalent to a capacitor with capacitance given by Express your answer in microfarads Express your answer in microfarads If you replace the capacitors,,,, and with a capacitor of capacitance, the resulting network would be a series connection between and Its equivalent capacitance is also the equivalent capacitance of the original network B3 Two capacitors in series
Consider two capacitors of capacitance and connected in series They are equivalent to a capacitor of capacitance that satisfies the following relation: Express your answer in microfarads RC Circuit and Current Conceptual Question Description: Short conceptual problem about current through a charging and discharging RC circuit In the diagram below, the two resistors, and, are identical and the capacitor is initially uncharged with the switch open Part A How does the current through compare with the current through immediately after the switch is first closed?
A1 Using Kirchhoff's junction rule for currents At all times, the current through (entering the junction) must be equal to the sum of the currents through through branches and the capacitor s branch (exiting the junction) The relative sizes of the currents and the capacitor s branch are determined by the effective resistances of these The current through the current through Part B How does the current through compare with the current through a very long time after the switch has been closed? B1 Using Kirchhoff's junction rule for currents At all times, the current through (entering the junction) must be equal to the sum of the currents through through branches B2 and the capacitor s branch (exiting the junction) The relative sizes of the currents and the capacitor s branch are determined by the effective resistances of these Current associated with a fully charged capacitor When a capacitor is fully charged, the current in the branch of the circuit containing the capacitor is zero The current through the current through Part C How does the current through compare with the current through immediately after the switch is opened (after being closed a very long time)? C1 Effect of a discharging capacitor
The charge is "held" on the plates of the capacitor by the potential difference of the battery When the battery is removed from the system, the negative charge on one plate of the capacitor will flow toward the positive charge on the other plate if a conducting path (circuit loop) exists During this discharging phase, the capacitor behaves in a way very analogous to a battery The current through the current through Ionic Potentials across Cell Membranes Conceptual Question Description: Short conceptual problem dealing with ionic potentials across cell membranes In its resting state, the membrane surrounding a neuron is permeable to potassium ions but only slightly permeable to sodium ions Thus, positive K ions can flow through the membrane in an attempt to equalize K concentration, but Na ions cannot as quickly This leads to an excess of Na ions outside of the cell If the space outside the cell is defined as zero electric potential, then the electric potential of the interior of the cell is negative This resting potential is typically about 80 A schematic of this situation is shown in the figure In response to an electrical stimulus, certain channels in the membrane can become permeable to Na ions Due to the concentration gradient, Na ions rush into the cell and the interior of the cell reaches an electric potential of about 40 This process is termed depolarization In response to depolarization, the membrane again becomes less permeable to Na ions, and the K ions flow out of the interior of the cell through channels established by the positive electric potential inside of the cell This then reestablishing the resting potential This is termed repolarization Only a small percentage of the available Na and K ions participate in each depolarization/repolarization cycle, so the cell can respond to many stimuli in succession without depleting its "stock" of available Na and K ions A graph of an electric potential inside a cell vs time is shown in the next figure
depolarization/repolarization cycle Part A for a single During the resting phase, what is the electric potential energy of a typical Na ion outside of the cell? A1 The electron volt Electric potential energy is defined as The electric charge on individual particles is always a multiple of the fundamental charge charge on a single proton) Rather than substituting a numerical value for, it is often more convenient to use the constant as a unit Thus, a proton located at a potential of 100 has energy (the which can be written as, or Thus, the proton has 100 electron volts of energy (Electron volts can be converted to the more traditional unit of energy, the joule, by multiplying by the conversion factor and recalling that Thus, )
40 +40 80 +80 0 Part B During the resting phase, what is the electrical potential energy of a typical K ion inside of the cell? B1 The electron volt Electric potential energy is defined as The electric charge on individual particles is always a multiple of the fundamental charge charge on a single proton) Rather than substituting a numerical value for, it is often more convenient to use the constant as a unit Thus, a proton located at a potential of 100 has energy (the which can be written as, or Thus, the proton has 100 electron volts of energy (Electron volts can be converted to the more traditional unit of energy, the joule, by multiplying by the conversion factor and recalling that Thus, )
40 +40 80 +80 0 Part C During depolarization, what is the work done (by the electric field) on the first few Na ions that enter the cell? C1 The electron volt Electric potential energy is defined as The electric charge on individual particles is always a multiple of the fundamental charge charge on a single proton) Rather than substituting a numerical value for, it is often more convenient to use the constant as a unit Thus, a proton located at a potential of 100 has energy (the which can be written as, or Thus, the proton has 100 electron volts of energy (Electron volts can be converted to the more traditional unit of energy, the joule, by multiplying by the conversion factor and recalling that Thus, ) C2 Algebraic sign of the work In general, work is defined as the product of the force applied parallel (or antiparallel) to the displacement of an object Thus, The work done by a force is positive if the force and the displacement are parallel; it is negative if
The work done by a force is positive if the force and the displacement are parallel; it is negative if the force and displacement are opposite in direction C3 Magnitude of the work Work transfers energy into or out of a system Therefore, in the absence of other energy transfers, the magnitude of the work done on an object is equal to the magnitude of the object s change in energy Since the primary form of energy present in this example is electric potential energy, the magnitude of the work done is equal to the change in the ion s electric potential energy 40 +40 80 +80 120 +120 0 Part D During repolarization, what is the work done (by the electric field) on the first few K ions that exit the cell? D1 The electron volt Electric potential energy is defined as The electric charge on individual particles is always a multiple of the fundamental charge charge on a single proton) Rather than substituting a numerical value for, it is often more convenient to use the constant as a unit Thus, a proton located at a potential of 100 has energy (the which can be written as, or Thus, the proton has 100 electron volts of energy (Electron volts can be converted to the more traditional unit of energy, the joule, by multiplying by the conversion factor
traditional unit of energy, the joule, by multiplying by the conversion factor and recalling that Thus, ) D2 Algebraic sign of the work In general, work is defined as the product of the force applied parallel (or antiparallel) to the displacement of an object Thus, The work done by a force is positive if the force and the displacement are parallel; the work done is negative if the force and displacement are opposite in direction D3 Magnitude of the work Work transfers energy into or out of a system Therefore, in the absence of other energy transfers, the magnitude of the work done on an object is equal to the magnitude of the object s change in energy Since the primary form of energy present in this example is electric potential energy, the magnitude of the work done is equal to the change in the ion s electric potential energy 40 +40 80 +80 120 +120 0 Problems P2311 Prepare: Please refer to Figure P2311 The three resistances in (a), (b), and (c) are parallel resistors We will thus use Equation 2312 to find the equivalent resistance Solve: (a) The equivalent resistance is
(b) The equivalent resistance is (c) The equivalent resistance is Assess: We must learn how to combine series and parallel resistors P2313 Prepare: For resistors in parallel, Adding four of the resistors in parallel gives Solve: We can put four of the resistors in parallel The total resistance is now Assess: There are other ways to arrive at the same using more from our collection of resistors P2315 Prepare: The resistance of the three resistors in parallel is Solve: That parallel combination in series with the resistor adds to an equivalent resistance of So that s the answer: the three resistors in parallel with each other, and then that combination in series with the resistor Assess: There are typical resistances that one can buy; generally one can t (cheaply) buy resistors with every possible value of resistance So it is common to combine the resistors you have in combinations of series and parallel to arrive at a different value of equivalent resistance called for in the circuit P2316 Prepare: When resistors are connected in parallel the combination has less resistance than any of the individual resistors Two resistors in parallel have an equivalent resistance of Solve: We can connect three such parallel pairs in series so that the total resistance is Assess: It is often necessary to combine standard value resistors in creative ways to arrive at a non-standard required resistance P2317 Prepare: The connecting wires are ideal with zero resistance We have to reduce the circuit to a single equivalent resistor by continuing to identify resistors that are in series or parallel combinations
Solve: For the first step, the resistors 30 Ω and 45 Ω are in parallel Their equivalent resistance is R eq 1 18 Ω For the second step, resistors 42 Ω and R eq 1 18 Ω are in series Therefore, R eq 2 R eq 1 + 42 Ω 18 Ω + 42 Ω 60 Ω For the third step, the resistors 40 Ω and R eq 2 60 Ω are in parallel So, R eq 3 24 Ω The equivalent resistance of the circuit is 24 Ω Assess: Have a good understanding of how series and parallel resistors combine to obtain equivalent resistors P2324 Prepare: Please refer to Figure P2324 The batteries are ideal, the connecting wires are ideal, and the ammeter has a negligibly small resistance Solve: Kirchhoff s junction law tells us that the current flowing through the 20 Ω resistance in the middle branch is I 1 + I 2 30 A We can therefore determine I 1 by applying Kirchhoff s loop law to the left loop Starting clockwise from the lower left corner, + 90 V I 1 (30 Ω) (30 A)(20 Ω) 0 V I 1 10 A I 2 (30 A I 1 ) (30 A 10 A) 20 A Finally, to determine the emf we apply Kirchhoff s loop law to the right loop and start counterclockwise from the lower right corner of the loop: I 2 (45 Ω) (30 A)(20 Ω) 0 V (20 A)(45 Ω) 60 V 0 V 150 V Assess: The currents and the emf look reasonable P2327 Prepare: The battery and the connecting wires are ideal The figure shows how to simplify the circuit in Figure P2327 using the laws of series and parallel resistances We have labeled the resistors as R 1 60 Ω, R 2 15 Ω, R 3 60 Ω, and R 4 40 Ω Having reduced the circuit to a single equivalent resistance R eq, we will reverse the procedure and build up the circuit using the loop law and the junction law to find the current and potential difference of each resistor
Solve: R 3 and R 4 are combined to get R 34 10 Ω, and then R 34 and R 2 are combined to obtain R 234 : R 234 6 Ω Next, R 234 and R 1 are combined to obtain R eq R 234 + R 1 60 Ω + 60 Ω 12 Ω From the final circuit, Thus, the current through the battery and R 1 is I R1 20 A and the potential difference across R 1 is I(R 1 ) (20 A) (60 Ω) 12 V As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference ΔV In Step 1 of the previous figure, R eq 12 Ω is returned to R 1 60 Ω and R 234 60 Ω in series Both resistors must have the same 20 A current as R eq We then use Ohm s law to find ΔV R1 (2 0A)(60 Ω) 12 V ΔV R234 (20 A)(60 Ω) 12 V As a check, 12 V + 12 V 24 V, which was ΔV of the R eq resistor In Step 2, the resistance R 234 is returned to R 2 and R 34 in parallel Both resistors must have the same ΔV 12 V as the resistor R 234 Then from Ohm s law, As a check, I R2 + I R34 20 A, which was the current I of the R 234 resistor In Step 3, R 34 is returned to R 3 and R 4 in series Both resistors must have the same 12 A as the R 34 resistor We then use Ohm s law to find (ΔV) R3 (12 A)(60 Ω) 72 V (ΔV) R4 (12 A)(40 Ω) 48 V As a check, 72 V + 48 V 12 V, which was ΔV of the resistor R 34
Resistor Potential difference (V) R 1 12 R 2 12 R 3 72 R 4 48 The three steps as we rebuild our circuit are shown Current (A) 20 08 12 12 Assess: This problem requires a good understanding of how to first reduce a circuit to a single equivalent resistance and then to build up a circuit P2328 Prepare: The circuit reduction process shows that So the current through the battery (and the first resistor) is Solve: The potential difference across the first resistor is Use the loop law around the left-most loop to deduce that Moving on, the equivalent resistance of the four right-most resistors (those not in the leftmost loop) is, so the junction law tells us the current splits evenly at point 1, giving through the middle horizontal resistor Then the potential difference across that middle horizontal resistor is Use the loop law around the loop containing the battery and point 2 (but not point 3) to deduce that Because both downstream branches from point 3 have the same resistance ( ) the current splits evenly there too, giving the current through the right-most horizontal resistor of The potential difference across that resistor is then The loop law around the outer loop then gives Assess: Checking with other calculations (such as different applications of the loop law) give the same results P2336 Prepare: Please refer to figure P2336 The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure Solve: Because C 1 and C 2 are in series, their equivalent capacitance C eq 12 is
C eq 12 12 µf Then, C eq 12 and C 3 are in parallel So, C eq C eq 12 + C 3 12 µf + 25 µf 37 µf Assess: We must understand well how to combine series and parallel capacitance P2338 Prepare: Assume ideal battery, wires, and capacitors Solve: (a) The two on the right are in parallel, so we add them to get gives That in series with (b) The amount of charge that flows while charging is given by Assess: The values are typical P2339 Prepare: For capacitors in series we know the equivalent capacitance is less than any of the individual capacitances The charge on capacitors in series is the same Solve: (a) All three are in series, so (b) The charge on each capacitor is the same as we would calculate on an equivalent capacitor Assess: Indeed, the equivalent capacitance is less than the smallest capacitor P2344 Prepare: The capacitor discharges through a resistor The switch in the circuit in Figure P2344 is in position a When the switch is in position b the circuit consists of a capacitor and a resistor Current and voltage during a capacitor discharge are given by Equations 2322 Because the charge on a capacitor is Q CΔV, the decay of the capacitor charge is given by Q Q 0 e t/ τ Solve: (a) The switch has been in position a for a long time That means the capacitor is fully charged to a charge Immediately after the switch is moved to the b position, the charge on the capacitor is Q 0 18 µc The current through the resistor is Note that as soon as the switch is closed, the potential difference across the capacitor ΔV C appears across the 50 Ω resistor (b) The charge Q 0 decays as Q Q 0 e t / τ, where
Thus, the charge is 11 µc The resistor current is (c) Likewise, the charge is Q 24 µc and the current is I 24 ma Assess: All of these values seem reasonable P2345 Prepare: Before the action potential the membrane potential is approximately and at the peak of depolarization it is We ll use positive values since we only need the strength of the electric field Solve: Before the action potential, At the peak of depolarization, Assess: The high values are due to the thinness of the membrane P2354 Prepare: The internal resistance of the battery is in series with the external resistor So the total resistance of the circuit is As a preliminary calculation, use Ohm s law to find the current in the circuit Solve: (a) The potential difference between the terminals of the battery is (b) The total power dissipated is The power dissipated internally in the battery is or 1/3 of the total Assess: While ideal batteries are sources of constant emf, real batteries with internal resistance have a potential difference between the terminals that depends on the external resistance (the load) P2358 Prepare: Please refer to Fig P2358 The connecting wires are ideal, but the battery is not We will designate the current in the 50 Ω resistor I 5 and the voltage drop ΔV 5 Similar designations will be used for the other resistors Solve: Since the 10 Ω resistor is dissipating 40 W, ΔV 10 I 10 R 10 (20 A)(10 Ω) 200 V The 20 Ω resistor is in parallel with the 10 Ω resistor, so they have the same potential difference: ΔV 20 ΔV 10 200 V From Ohm s law,
The combined current through the 10 Ω and 20 Ω resistors first passes through the 50 Ω resistor Applying Kirchhoff s junction law at the junction between the three resistors, I 5 I 10 + I 20 10 A + 20 A 30 A ΔV 5 I 5 R 5 (30 A)(50 Ω) 15 V Knowing the currents and potential differences, we can now find the power dissipated: P 5 I 5 ΔV 5 (30 A)(150 V) 45 P 20 I 20 ΔV 20 (10 A)(200 V) 20 W P2364 Prepare: Please refer to Figure P2364 Assume the batteries and the connecting wires are ideal Solve: (a) The two batteries in this circuit are oriented to oppose each other The direction of the current is counterclockwise because the 12 V battery wins (b) There are no junctions, so the same current I flows through all circuit elements Applying Kirchhoff s loop law in the counterclockwise direction and starting at the lower right corner, ΔV i 12 V I(12 Ω) I(60 Ω) - 60 V IR 0 Note that the IR terms are all negative because we re applying the loop law in the direction of current flow, and the potential decreases as current flows through a resistor We can easily solve to find the unknown resistance R: 60 V I(18 Ω) IR 0 (c) The power is P I 2 R (025 A) 2 (6 Ω) 038 W (d) The potential difference across a resistor is ΔV IR, giving ΔV 6 15 V, and ΔV 12 30 V Starting from the lower left corner, the graph goes around the circuit clockwise, opposite from the direction in which we applied the loop law In this direction, we speak of potential as lost in the batteries and gained in the resistors P2374 Prepare: Please refer to Figure P2374 In an RC circuit, the capacitor voltage discharge is given by Equation 2322, ie, ΔV ΔV 0 e t/ τ Solve: From Figure P2374, we note that ΔV 0 30 V and ΔV 10 V at t 2 ms So,
P2377 Prepare: The chapter says that when the sodium channels open, the potential inside the cell changes from to relative to outside the cell This will be accompanied by a movement of charge as the ions flow into the cell The number of channels is the total charge that moves divided by the amount of charge per channel (which is given as ions per channel) Solve: Assess: Example 2315 says there are a great many channels in a cell, and our answer bears this out P2380 Prepare: When the defibrillator is attached to a person, a capacitor is discharged and the potential difference across the capacitor and current through the resistor (the heart) should look like that for a capacitor discharging in an RC circuit For this case, the potential difference across a capacitor is given by and the current through the resistor (the heart) is given by Solve: The set of graphs labeled A match the above two functions Assess: Knowing the general shape of a function is a valuable skill P2381 Prepare: In an circuit the time constant is Solve: The correct choice is C Assess: We were told that the capacitor discharges quickly, and our answer agrees with that 32 ms is quick, but reasonable P2382 Prepare: The initial current may be determined by and the time constant may be determined by Solve: Examining the above expressions for the initial current and time constant, we see that the initial current will decrease and the time constant will increase if the resistance increases The correct choice is B Assess: The ability to look at an expression and answer a what if question is a valuable skill P2383 Prepare: The time constant does not depend on the voltage; it is still So lowering the voltage leaves the time constant unchanged Solve: The correct choice is B Assess: The time constant gives us an idea of how fast the capacitor will discharge, ie, we know the potential difference will decay to 37% of its initial value after one time constant, and that is true no matter what the value of the initial voltage is