HW19Due: 11:59pm on Monday, October 19, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View] "An Introduction to EMF and Circuits" is a nice review of circuits, but Part M is not perfectly clear. MP is asking for the amount of energy extracted from the battery's chemicals during the 60-second period. GBA An Introduction to EMF and Circuits Learning Goal: To understand the concept of electromotive force and internal resistance; to understand the processes in oneloop circuits; to become familiar with the use of the ammeter and voltmeter. In order for the current in a conductor to exist continuously, the conductor must be part of a loop, that is, a closed path through which the charged particles can move without creating a "build-up." Such build-up, if it occurs, creates its own electric field that cancels out the external electric field, ultimately causing the current to stop. However, having a loop, or a closed circuit, is not enough to maintain the current; there must also be a source of energy. Its necessity is fairly obvious: As charged particles move along the circuit, they lose potential energy. In fact, electrostatic forces always push the particles in the direction that leads to a decrease in potential energy. At some point, each charged particle would reach the location in the circuit where it has the lowest possible potential energy. How can such a particle move toward a point where it would have a higher potential energy? Such a move requires that nonelectrostatic forces act upon the charged particle, pushing it toward higher potential energy despite the presence of electrostatic forces. In circuits, such forces exist inside a device commonly known as a battery. In a circuit, the battery serves as the energy source that keeps the charged particles in continuous motion by increasing their potential energy through the action of some kind of nonelectrostatic force. The amount of work that the battery does on each coulomb of charge that it "pushes through" is called (inappropriately) the electromotive force (pronounced "ee-em-ef" and abbreviated emf or denoted by ). Batteries are often referred to as sources of emf (rather than sources of energy, even though they are, fundamentally, sources of energy). The emf of a battery can be calculated using the definition mentioned above: . The units of emf are joules per coulomb, that is, volts. The terminals of a battery are often labeled and for "higher potential" and "lower potential," respectively. The potential difference between the terminals is called the terminal voltage of the battery. If no current is running through a battery, the terminal voltage is equal to the emf of the battery: . However, if there is a current in the circuit, the terminal voltage is less than the emf because the battery has its own internal resistance (usually labeled ). When charge passes through the battery, the battery does the amount of work on the charge; however, the charge also "loses" the amount of energy equal to in potential energy is , and the terminal voltage is . In order to answer the questions that follow, you should first review the meaning of the symbols describing various elements of the circuit, including the ammeter and the voltmeter; you should also know the way the ammeter and the voltmeter must be connected to the rest of the circuit in order to function properly. Note that the internal resistance is usually indicated as a separate resistor drawn next to the "battery" symbol. It is important to keep in mind that this resistor with resistance is actually inside the battery. In all diagrams, stands for emf, for the internal resistance of the battery, and for the resistance of the external circuit. As ( is the current through the circuit); therefore, the increase usual, we'll assume that the connecting wires have negligible resistance. We will also assume that both the ammeter and the voltmeter are ideal: That is, the ammeter has negligible resistance, and the voltmeter has a very large resistance. Part A For the circuit shown in the diagram , which potential difference corresponds to the terminal voltage of the battery? ANSWER: between points K and L between points L and M between points K and M Correct Keep in mind that the "resistor" with resistance is actually inside the battery. The next several questions refer to the four diagrams shown here labeled A, B, C, and D. ANSWER: CD Correct Part C In which diagram is the current through the battery nearly zero? Hint C. For example if both A and C are correct enter AC.1 How a voltmeter works A voltmeter works by measuring the voltage of anything to which it is connected in parallel in the circuit. a very high one. Part D In which diagram or diagrams does the ammeter correctly measure the current through the resistor with resistance Hint D.1 How to approach the problem In order to determine the current through the battery we must follow the current loop through the circuit. the voltmeter readings would actually be quite close to the terminal voltage if the ammeter has a very low resistance. Whichever loop has the highest resistance will have the lowest current. Since the latter has a very large resistance. diagram C clearly shows the best way to connect the voltmeter in order to measure the terminal voltage. For example if both A and C are correct enter AC. ANSWER: D Correct .1 How to approach the problem ? Note that current is conserved through a wire. ANSWER: CD Correct Part E In which diagram does the voltmeter correctly measure the terminal voltage of the battery? Choose the best answer.Part B In which diagram(s) (labeled A . ANSWER: A B C D Correct In diagrams A and B. and in order for an ammeter to measure the correct current passing through an element. However. and the voltmeter. we would like the voltmeter to have a very high internal resistance so that not much current flows through it. Hint E. Keep in mind that the voltmeter has a very high internal resistance.D) does the ammeter correctly measure the current through the battery? Hint B. For example if both A and C are correct enter AC. it must be in series with that element.1 How an ammeter works Hint not displayed Enter the letter(s) of the correct diagram(s) in alphabetical order. this current is essentially zero. Part F In which diagram does the voltmeter read almost zero? Enter the letter(s) of the correct diagram(s) in alphabetical order. ANSWER: A B C D Correct Diagram A is the only one in which the current through the battery is the same as the current through the voltmeter. As a result. Enter the letter(s) of the correct diagram(s) in alphabetical order. the charge that passes through one must pass through another. Therefore. Use two significant figures.0 volts and internal resistance 3.0-ohm resistor perform in one minute? How to approach the problem Hint not displayed Hint M.0-ohm resistor is connected to the terminals of the battery.1 across the 21. For example if both A and C are correct enter AC.00 ohms. the voltmeter is connected in series with the battery. Part H A voltmeter is connected to the terminals of the battery. the voltage across the resistor Since the ends of the resistor with resistance is the same as that between the terminals of the battery. and the potential difference (voltage) is nearly zero. ANSWER: = 0.0-ohm resistor from Part I? How to approach the problem Hint not displayed Express your answer in volts. therefore. ANSWER: = 10.5 Correct Part L What is the terminal voltage Hint L. The last group of questions refers to a battery that has emf 12. the charge flowing through that wire will not lose an appreciable amount of potential energy. Use two significant figures. what is the current through the 21. Use three significant figures.50 Correct Part J In the situation described in Part I. the current through the ammeter is the same as the current through the voltmeter. Therefore.0 Correct Part I The voltmeter is now removed and a 21. the currents must be the same. the ammeter reads no current. Part M How much work Hint M. ANSWER: = 0. ANSWER: = 12. Since the resistance of the voltmeter is very large.The voltmeter in diagram D is connected to two points that are also connected by a wire that has. the battery is not connected to any other external circuit elements.0-ohm resistor? Express your answer in amperes. What is the current through the battery? Express your answer in amperes. presumably.0-ohm resistor from Part I? Kirchhoff's voltage law Hint not displayed Express your answer in volts. the current is nearly zero. ANSWER: = 10. Use three significant figures. ANSWER: AB Correct In diagram A. In diagram B. Part G In which diagram or diagrams does the ammeter read almost zero? Enter the letter(s) of the correct diagram(s) in alphabetical order. What is the reading of the voltmeter ? Express your answer in volts. very low resistance.1 of the battery connected to the 21.5 Correct are attached to the terminals of the battery. Use three significant figures. Part K What is the potential difference Hint K.2 Find the charge .50 Correct Since the battery and the external resistor form one loop. Since the voltmeter has a very large resistance there is no (or nearly zero) current in the whole circuit.1 does the battery connected to the 21. Unlike the idealized voltmeter. the potential drop across it is 0 volts). How to approach the problem A real-world voltmeter can be thought of as an ideal ammeter (i. With a little algebraic manipulation.1 What is meant by "within 1%" Look at the above expression for the potential difference measured by the voltmeter: From this expression. If the voltmeter's resistance is large enough. In the next part you will calculate how large a typical voltmeter resistance needs to be.4 Using Kirchhoff's loop rule Hint not displayed Express your answer in terms of ANSWER: = Correct . Hint B.1 is connected across the terminals of a battery of emf and internal resistance . Use three significant figures. Express your answer numerically (in ohms) to at least three significant digits. It is a good idea to check that the answer gives the correct result in the limit that . Specifically. The large resistance ensures that the voltmeter draws very little current from the circuit. you should see that the value of is different from what it would be if the voltmeter were absent. The goal of this problem is to illustrate the effect of placing a voltmeter in a circuit.Hint not displayed Express your answer in joules. its resistance is 0 ohms).0% of the emf of the battery.e. find the minimum value of the voltmeter resistance for which the voltmeter reading is within 1. Part B If = volts and ohms. . Find the measured by the voltmeter. The specifications require that this expression be less than 1%. ANSWER: = 360 Correct How a Real Voltmeter Works A nonideal voltmeter.3 An expression for Hint not displayed Hint A.2 How to find the potential between points a and b Hint not displayed Hint A.6 . you can see that the percentage difference between the potential difference measured by the voltmeter and the actual emf of the battery is . Hint A. In this form it is easier to see why the voltmeter reading differs from the actual emf it is supposed to measure by only a small amount if . it displays . in series with a large resistance (see the figure).. The ammeter is calibrated so that instead of displaying the current through the voltmeter. which is the potential difference across the voltmeter (since the ammeter is ideal. and . the answer can also be written as . Part A A voltmeter with resistance potential difference Hint A. then this change will be small. a real voltmeter has a resistance that is not infinitely large. ANSWER: = 44. . The current through resistance passes through resistance because there is no junction in between the resistor and the ammeter that could allow it to go also. Part C Apply the loop rule to loop 2 (the smaller loop on the right). together with the meter readings ANSWER: Correct If you apply the juncion rule to the junction above . Part A The junction rule describes the conservation of which quantity? Note that this rule applies only to circuits that are in a steady state. The battery supplies a constant voltage . This allows reasonably accurate measurements of much larger resistances to be made. If the actual current is in the opposite direction from your current arrow. The figure contains two junctions (where three or more wires meet)--they are at the ends of the resistor labeled . some of which are much larger than the value you just obtained (on the order of megaohms). Circuit elements connected in a string like this elsewhere. the given resistances. Hint C. your answer for that current will be negative.3 Voltage drop across ammeter Hint not displayed Express the voltage drops in terms of . ammeters are ideal meters that read The direction of each loop and the direction of each current arrow that you draw on your own circuits are arbitrary. This problem introduces Kirchhoff's two rules for circuits: Kirchhoff's loop rule: The sum of the voltage changes across the circuit elements forming any closed loop is zero. water would accumulate at the junction. and any other given quantities.= Correct Typical voltmeters have a range of possible resistances. you just obtained for the junction labeled 1. the current passing through the battery must be are said to be in series and the same current must pass through each element.1 Elements in series have same current Hint not displayed Hint C. Loop 1 is the loop around the entire circuit. Just assign voltage drops consistently and sum both voltage drops and currents algebraically and you will get correct equations.1 At the junction Think of the analogy with water flow. The figure shows a circuit that illustrates the concept of loops. Similarly. ANSWER: current voltage resistance Correct Part B Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance Hint B. . The and respectively. an inconsequential change in overall sign of the equation because it equals zero). whereas loop 2 is the smaller loop on the right. you should find that the ezpression you get is equivalent to what and and the current .1 Elements in series is not labeled. Obviously the conservation of charge or current flow enforces the same relationship among the currents when they separate as when they recombine. Sum the voltage changes across each circuit element around this loop going in the direction of the arrow.2 Sign of voltage across resistors Hint not displayed Hint C.. This fact greatly reduces the number of independent current values in any practical circuit. Answer in terms of given quantities. and the resistors are labeled with their resistances. Hint A. Remember that the current meter is ideal. Kirchhoff's Rules and Applying Them Learning Goal: To understand the origins of both of Kirchhoff's rules and how to use them to solve a circuit problem. You should recognize that the current passing through the ammeter also ). what can you say (mathematically) about the sum of the three water currents at this junction? If this were not true. If a certain current of water comes to a split in the pipe. Kirchhoff's junction rule: The algebraic sum of the currents into (or out of) any junction in the circuit is zero. .e. which are colored red and labeled loop 1 and loop 2. To apply the loop rule you would add the voltage changes of all circuit elements around the chosen loop. The direction of any loop is even less imporant: The equation obtained from a counterclockwise loop is the same as that from a clockwise loop except for a negative sign in front of every term (i. 00 Part A Find the equivalent resistance of the resistor network. and = 7. = 4.6 Correct Part B Two resistors of resistance = 5. Networks of Resistors Consider the network of four resistors shown in the diagram. . Sum the voltage changes across each circuit element around this loop going in the direction of the arrow. If you . and any other given quantities.1 How to reduce the network of resistors Hint not displayed Hint A. no current passes through the resistor . you will generate an extra equation that is redundant with the other two. the inner loop through the battery. Express the voltage drops in terms of ANSWER: Correct There is one more loop in this circuit. Hint A. = 1. you can get enough equations to solve a circuit by either 1.ANSWER: Correct Part D Now apply the loop rule to loop 1 (the larger loop spanning the entire circuit).00 and = 3. ANSWER: 10..50 .3 Three resistors in series Hint not displayed Express your answer in ohms. selecting all of the internal loops (loops with no circuit elements inside the loop) or 2. as shown in the diagram. using a number of loops (not necessarily internal) equal to the number of internal loops.00 . the given resistances. . with the extra proviso that at least one loop pass through each circuit element. Find the equivalent resistance of the new resistor network when the switch is open. which can be ignored then. apply Kirchhoff's loop rule to this additional loop. reduce Since the switch is open.50 are added to the network. where . = 7.00 . and an additional resistor of resistance is connected by a switch. As you did in Part A.1 How to reduce the extended network of resistors . both ammeters.00 = 3. and resistors and . In general. The resistors are connected to a constant voltage of magnitude .2 Find the resistance equivalent to and Hint not displayed Hint A. Hint B. and is in series with the resistors and .2.1 Two resistors in series Consider two resistors of resistance resistance . Hint A.8 Correct Part C Find the equivalent resistance of the resistor network described in Part B when the switch is closed. They are equivalent to a resistor with .2 Two resistors in parallel Consider two resistors of resistance resistance and that are connected in parallel. Note that the new resistor resistor Hint B. Hint B.1. ANSWER: 14.1 How to reduce the network of resistors when the switch is closed Hint not displayed Hint C. . and that are connected in series. Express your answer in ohms. They are equivalent to a resistor with If you replace the resistors parallel with .2 Comparing bulb A to bulb B Hint not displayed Hint A. .3 Comparing bulb D to bulb E .3 Four resistors in series Hint not displayed Express your answer in ohms. which satisfies the following relation: . ANSWER: = Answer not displayed Hint B. Hint B.50 Correct and with their equivalent resistor (of resistance ).2 Find the resistance equivalent to and Hint not displayed Hint C. and .1 Find the resistance equivalent to Find the resistance equivalent to the connection between . while the new Find the resistance equivalent to Find the resistance equivalent to the resistor connection with and and . . Hint B.2.3 Four resistors in series Hint not displayed Express your answer in ohms. ANSWER: 10.the network in successive stages. Rank the bulbs (A through E) based on their brightness.1 How to approach the problem Hint not displayed Hint A. Hint C. ANSWER: = 7.2 is in series with .7 Correct Which Bulb is Brightest? Part A Consider a circuit containing five identical light bulbs and an ideal battery. Express your answer in ohms. which is given by . the resistor will result in . Assume that the resistance of each light bulb remains constant.2. overlap them.2 Correct . 0. It gets brighter.5 Correct . There is no change.4 Comparing bulb C to bulb D or E Hint not displayed Hint A.1 How to approach this part Hint not displayed Hint B.5 Comparing bulb C to bulb A or B Hint not displayed Rank from brightest to dimmest.2 Consider changes in resistance Hint not displayed ANSWER: It gets dimmer. ANSWER: View Correct Now consider what happens when a switch in the circuit is opened. Correct Part C If the resistance of each light bulb is 25 opened? ANSWER: 29. Part B What happens to the brightness of bulb A? Hint B.114 A Correct . what was the resistance of the entire network of bulbs before the switch was Part D If the battery voltage is 5 each bulb is 25 ANSWER: .Hint not displayed Hint A. what was the resistance of the entire network of bulbs after the switch was opened? Part F . To rank items as equivalent. what was the current through bulb C before the switch was opened? Recall that the resistance of Part E If the resistance of each light bulb is 25 ANSWER: 37. which could damage them. .77 out of a possible total of 50 points. ANSWER: = 8. Otherwise. Correct This is why appliances in your home are always connected in parallel. .133 A Correct Part G What happens to bulb C? Hint G. You received 49.5%. what was the current through bulb C after the switch was opened? Recall that the resistance of 0.78 ? Each emf source has negligible internal resistance. There is no change. Problem 26. It gets brighter. turning some of them on or off would cause the current in others to change.1 How to approach this part Hint not displayed ANSWER: It gets dimmer.33 Correct Score Summary: Your score on this assignment is 99.62 Part A What must the emf through the in the figure be in order for the current resistor to be 1.If the battery voltage is 5 each bulb is 25 ANSWER: .