Study better. Learn faster. Get the grade you want.
Discover why millions of students use us to learn better.

- StudyBlue
- Arizona
- Arizona State University - Tempe
- Physics
- Physics 131
- Adams/menendez
- [SOLUTIONS] Mastering Physics HW19.pdf

Sign up now and start studying this note for FREE

HW19 Due: 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 one- loop 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 ( is the current through the circuit); therefore, the increase 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 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. Part B In which diagram(s) (labeled A - D) does the ammeter correctly measure the current through the battery? Hint B.1How an ammeter works Hint not displayed Enter the letter(s) of the correct diagram(s) in alphabetical order. For example if both A and C are correct enter AC. ANSWER: CD Correct Part C In which diagram is the current through the battery nearly zero? Hint C.1 How to approach the problem In order to determine the current through the battery we must follow the current loop through the circuit. Whichever loop has the highest resistance will have the lowest current. Keep in mind that the voltmeter has a very high internal resistance. 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. Since the latter has a very large resistance, this current is essentially zero. 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 Note that current is conserved through a wire, and in order for an ammeter to measure the correct current passing through an element, it must be in series with that element. Enter the letter(s) of the correct diagram(s) in alphabetical order. For example if both A and C are correct enter AC. ANSWER: CD Correct Part E In which diagram does the voltmeter correctly measure the terminal voltage of the battery? Choose the best answer. Hint E.1How a voltmeter works A voltmeter works by measuring the voltage of anything to which it is connected in parallel in the circuit. As a result, we would like the voltmeter to have a very high internal resistance so that not much current flows through it. ANSWER: A B C D Correct In diagrams A and B, the voltmeter readings would actually be quite close to the terminal voltage if the ammeter has a very low resistance, and the voltmeter, a very high one. However, diagram C clearly shows the best way to connect the voltmeter in order to measure the terminal voltage. Part F In which diagram does the voltmeter read almost zero? Enter the letter(s) of the correct diagram(s) in alphabetical order. For example if both A and C are correct enter AC. ANSWER: D Correct The voltmeter in diagram D is connected to two points that are also connected by a wire that has, presumably, very low resistance. Therefore, the charge flowing through that wire will not lose an appreciable amount of potential energy, and the potential difference (voltage) is nearly zero. 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. For example if both A and C are correct enter AC. ANSWER: AB Correct In diagram A, the voltmeter is connected in series with the battery. Since the voltmeter has a very large resistance there is no (or nearly zero) current in the whole circuit. Therefore, the ammeter reads no current. In diagram B, the current through the ammeter is the same as the current through the voltmeter. Since the resistance of the voltmeter is very large, the current is nearly zero. The last group of questions refers to a battery that has emf 12.0 volts and internal resistance 3.00 ohms. Part H A voltmeter is connected to the terminals of the battery; the battery is not connected to any other external circuit elements. What is the reading of the voltmeter ? Express your answer in volts. Use three significant figures. ANSWER: = 12.0 Correct Part I The voltmeter is now removed and a 21.0-ohm resistor is connected to the terminals of the battery. What is the current through the battery? Express your answer in amperes. Use two significant figures. ANSWER: = 0.50 Correct Part J In the situation described in Part I, what is the current through the 21.0-ohm resistor? Express your answer in amperes. Use two significant figures. ANSWER: = 0.50 Correct Since the battery and the external resistor form one loop, the charge that passes through one must pass through another; therefore, the currents must be the same. Part K What is the potential difference across the 21.0-ohm resistor from Part I? Hint K.1 How to approach the problem Hint not displayed Express your answer in volts. Use three significant figures. ANSWER: = 10.5 Correct Part L What is the terminal voltage of the battery connected to the 21.0-ohm resistor from Part I? Hint L.1Kirchhoff's voltage law Hint not displayed Express your answer in volts. Use three significant figures. ANSWER: = 10.5 Correct Since the ends of the resistor with resistance are attached to the terminals of the battery, the voltage across the resistor is the same as that between the terminals of the battery. Part M How much work does the battery connected to the 21.0-ohm resistor perform in one minute? Hint M.1 How to approach the problem Hint not displayed Hint M.2 Find the charge Hint not displayed Express your answer in joules. Use three significant figures. ANSWER: = 360 Correct How a Real Voltmeter Works A nonideal voltmeter. Unlike the idealized voltmeter, a real voltmeter has a resistance that is not infinitely large. Part A A voltmeter with resistance is connected across the terminals of a battery of emf and internal resistance . Find the potential difference measured by the voltmeter. Hint A.1 How to approach the problem A real-world voltmeter can be thought of as an ideal ammeter (i.e., its resistance is 0 ohms), in series with a large resistance (see the figure). The large resistance ensures that the voltmeter draws very little current from the circuit. The ammeter is calibrated so that instead of displaying the current through the voltmeter, it displays , which is the potential difference across the voltmeter (since the ammeter is ideal, the potential drop across it is 0 volts). The goal of this problem is to illustrate the effect of placing a voltmeter in a circuit. Specifically, you should see that the value of is different from what it would be if the voltmeter were absent. If the voltmeter's resistance is large enough, then this change will be small. In the next part you will calculate how large a typical voltmeter resistance needs to be. Hint A.2 How to find the potential between points a and b Hint not displayed Hint A.3 An expression for Hint not displayed Hint A.4 Using Kirchhoff's loop rule Hint not displayed Express your answer in terms of , , and . ANSWER: = Correct With a little algebraic manipulation, the answer can also be written as . 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 is a good idea to check that the answer gives the correct result in the limit that . Part B If = volts and ohms, find the minimum value of the voltmeter resistance for which the voltmeter reading is within 1.0% of the emf of the battery. Hint B.1What is meant by "within 1%" Look at the above expression for the potential difference measured by the voltmeter: From this expression, you can see that the percentage difference between the potential difference measured by the voltmeter and the actual emf of the battery is . The specifications require that this expression be less than 1%. Express your answer numerically (in ohms) to at least three significant digits. ANSWER: = 44.6 = Correct Typical voltmeters have a range of possible resistances, some of which are much larger than the value you just obtained (on the order of megaohms). This allows reasonably accurate measurements of much larger resistances to be made. 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. 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. Kirchhoff's junction rule: The algebraic sum of the currents into (or out of) any junction in the circuit is zero. The figure shows a circuit that illustrates the concept of loops, which are colored red and labeled loop 1 and loop 2. Loop 1 is the loop around the entire circuit, whereas loop 2 is the smaller loop on the right. To apply the loop rule you would add the voltage changes of all circuit elements around the chosen loop. The figure contains two junctions (where three or more wires meet)--they are at the ends of the resistor labeled . The battery supplies a constant voltage , and the resistors are labeled with their resistances. The ammeters are ideal meters that read and respectively. The direction of each loop and the direction of each current arrow that you draw on your own circuits are arbitrary. Just assign voltage drops consistently and sum both voltage drops and currents algebraically and you will get correct equations. If the actual current is in the opposite direction from your current arrow, your answer for that current will be negative. 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.e., an inconsequential change in overall sign of the equation because it equals zero). 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. Hint A.1 At the junction Think of the analogy with water flow. If a certain current of water comes to a split in the pipe, what can you say (mathematically) about the sum of the three water currents at this junction? If this were not true, water would accumulate at the junction. 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.1Elements in series The current through resistance is not labeled. You should recognize that the current passing through the ammeter also passes through resistance because there is no junction in between the resistor and the ammeter that could allow it to go elsewhere. Similarly, the current passing through the battery must be also. Circuit elements connected in a string like this are said to be in series and the same current must pass through each element. This fact greatly reduces the number of independent current values in any practical circuit. Answer in terms of given quantities, together with the meter readings and and the current . ANSWER: Correct If you apply the juncion rule to the junction above , you should find that the ezpression you get is equivalent to what you just obtained for the junction labeled 1. Obviously the conservation of charge or current flow enforces the same relationship among the currents when they separate as when they recombine. Part C Apply the loop rule to loop 2 (the smaller loop on the right). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow. Remember that the current meter is ideal. Hint C.1 Elements in series have same current Hint not displayed Hint C.2 Sign of voltage across resistors Hint not displayed Hint C.3 Voltage drop across ammeter Hint not displayed Express the voltage drops in terms of , , , the given resistances, and any other given quantities. ANSWER: Correct Part D Now apply the loop rule to loop 1 (the larger loop spanning the entire circuit). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow. Express the voltage drops in terms of , , , the given resistances, and any other given quantities. ANSWER: Correct There is one more loop in this circuit, the inner loop through the battery, both ammeters, and resistors and . If you apply Kirchhoff's loop rule to this additional loop, you will generate an extra equation that is redundant with the other two. In general, you can get enough equations to solve a circuit by either 1. selecting all of the internal loops (loops with no circuit elements inside the loop) or 2. using a number of loops (not necessarily internal) equal to the number of internal loops, with the extra proviso that at least one loop pass through each circuit element. Networks of Resistors Consider the network of four resistors shown in the diagram, where = 4.00 , = 7.50 , = 1.00 , and = 7.00 . The resistors are connected to a constant voltage of magnitude . Part A Find the equivalent resistance of the resistor network. Hint A.1 How to reduce the network of resistors Hint not displayed Hint A.2 Find the resistance equivalent to and Hint not displayed Hint A.3 Three resistors in series Hint not displayed Express your answer in ohms. ANSWER: 10.6 Correct Part B Two resistors of resistance = 3.00 and = 3.50 are added to the network, and an additional resistor of resistance = 5.00 is connected by a switch, as shown in the diagram.. Find the equivalent resistance of the new resistor network when the switch is open. Hint B.1How 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 . Hint B.2Find the resistance equivalent to , , and Find the resistance equivalent to the resistor connection with , , and . Hint B.2.1Find the resistance equivalent to and Find the resistance equivalent to the connection between and . Hint B.2.1.1Two resistors in series Consider two 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. ANSWER: = 7.50 Correct If you replace the resistors and with their equivalent resistor (of resistance ), the resistor will result in parallel with . Hint B.2.2Two 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. ANSWER: = Answer not displayed Hint B.3Four resistors in series Hint not displayed Express your answer in ohms. ANSWER: 14.8 Correct Part C Find the equivalent resistance of the resistor network described in Part B when the switch is closed. Hint C.1 How to reduce the network of resistors when the switch is closed Hint not displayed Hint C.2 Find the resistance equivalent to and Hint not displayed Hint C.3 Four resistors in series Hint not displayed Express your answer in ohms. ANSWER: 10.7 Correct Which Bulb is Brightest? 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. Hint A.1 How to approach the problem Hint not displayed Hint A.2 Comparing bulb A to bulb B Hint not displayed Hint A.3 Comparing bulb D to bulb E Hint not displayed Hint A.4 Comparing bulb C to bulb D or E Hint not displayed Hint A.5 Comparing bulb C to bulb A or B Hint not displayed Rank from brightest to dimmest. To rank items as equivalent, overlap them. ANSWER: View Correct Now consider what happens when a switch in the circuit is opened. Part B What happens to the brightness of bulb A? Hint B.1How to approach this part Hint not displayed Hint B.2Consider changes in resistance Hint not displayed ANSWER: It gets dimmer. It gets brighter. There is no change. Correct Part C If the resistance of each light bulb is 25 , what was the resistance of the entire network of bulbs before the switch was opened? ANSWER: 29.2 Correct Part D If the battery voltage is 5 , what was the current through bulb C before the switch was opened? Recall that the resistance of each bulb is 25 . ANSWER: 0.114 Correct A Part E If the resistance of each light bulb is 25 , what was the resistance of the entire network of bulbs after the switch was opened? ANSWER: 37.5 Correct Part F If the battery voltage is 5 , what was the current through bulb C after the switch was opened? Recall that the resistance of each bulb is 25 . ANSWER: 0.133 Correct A Part G What happens to bulb C? Hint G.1 How to approach this part Hint not displayed ANSWER: It gets dimmer. It gets brighter. There is no change. Correct 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. Problem 26.62 Part A What must the emf in the figure be in order for the current through the resistor to be 1.78 ? Each emf source has negligible internal resistance. ANSWER: = 8.33 Correct Score Summary: Your score on this assignment is 99.5%. You received 49.77 out of a possible total of 50 points. clockwork MasteringPhysics: Assignment Print View

Back

Next

About this note

Author: Anonymous

Textbook: University Physics Vol 2 (Chapters 21-37) (12th Edition) (Chapters 21-37 v. 2)

Created: 2010-01-23

Updated: 2010-01-23

File Size: 9 page(s)

Views: 11345

Textbook: University Physics Vol 2 (Chapters 21-37) (12th Edition) (Chapters 21-37 v. 2)

Created: 2010-01-23

Updated: 2010-01-23

File Size: 9 page(s)

Views: 11345

Simply amazing. The flashcards are smooth, there are
many different types of
studying tools, and there is
a great search engine. I praise
you on the awesomeness.
- Dennis

I have been getting MUCH
better grades on all my tests
for school. Flash cards, notes,
and quizzes are great on here.
Thanks!
- Kathy

I was destroying whole rain forests with my flashcard production, but YOU, StudyBlue, have saved the ozone layer. The earth thanks you.
- Lindsey

This is the greatest app on my phone!! Thanks so much for making it easier to study. This has helped me a lot!
- Tyson

StudyBlue is not sponsored or endorsed by any college, university, or instructor.

© 2014 StudyBlue Inc. All rights reserved.

© 2014 StudyBlue Inc. All rights reserved.