ELTR 105 (DC 2), section 1

ELTR 105 (DC 2), section 1 Recommended schedule Day 1 Day 2 Day 3 Day 4 Day 5 Topics: Series-parallel circuit analysis Questions: 1 through 15 Lab Exercise: Kirchhoff s Voltage Law (question 61) Topics: Series-parallel circuits and Wheatstone bridges Questions: 16 through 30 Lab Exercise: Wheatstone bridge (question 62) Topics: Series-parallel circuits, safety grounding, and troubleshooting Questions: 31 through 45 Lab Exercise: Series-parallel DC resistor circuit (question 63) Topics: Loaded voltage dividers Questions: 46 through 60 Lab Exercise: Loaded voltage divider (question 64) Exam 1: includes Series-parallel DC resistor circuit performance assessment Lab Exercise: Troubleshooting practice (loaded voltage divider circuit question 64) Practice and challenge problems Questions: 67 through the end of the worksheet Impending deadlines Troubleshooting assessment (voltage divider) due at end of ELTR105, Section 3 Question 65: Troubleshooting log Question 66: Sample troubleshooting assessment grading criteria 1

ELTR 105 (DC 2), section 1 Skill standards addressed by this course section EIA Raising the Standard; Electronics Technician Skills for Today and Tomorrow, June 1994 B Technical Skills DC circuits B.03 Demonstrate an understanding of the meaning of and relationships among and between voltage, current, resistance and power in DC circuits. B.05 Demonstrate an understanding of application of Ohm s Law to series, parallel, and series-parallel circuits. B.14 Understand the principles and operations of DC series-parallel and bridge circuits. B.15 Fabricate and demonstrate DC series-parallel and bridge circuits. B.16 Troubleshoot and repair series-parallel and bridge circuits. B.17 Understand the principles and operations of the Wheatstone bridge. B.18 Understand principles and operations of DC voltage divider circuits (loaded and unloaded). B.19 Fabricate and demonstrate DC voltage divider circuits (loaded and unloaded). B.20 Troubleshoot and repair DC voltage divider circuits (loaded and unloaded). C Technical Skills AC circuits C.07 Demonstrate an understanding of the principle and operation of safety grounding systems: (lightning arrestors, ground fault interrupters, etc.). B Basic and Practical Skills Communicating on the Job B.01 Use effective written and other communication skills. Met by group discussion and completion of labwork. B.03 Employ appropriate skills for gathering and retaining information. Met by research and preparation prior to group discussion. B.04 Interpret written, graphic, and oral instructions. Met by completion of labwork. B.06 Use language appropriate to the situation. Met by group discussion and in explaining completed labwork. B.07 Participate in meetings in a positive and constructive manner. Met by group discussion. B.08 Use job-related terminology. Met by group discussion and in explaining completed labwork. B.10 Document work projects, procedures, tests, and equipment failures. Met by project construction and/or troubleshooting assessments. C Basic and Practical Skills Solving Problems and Critical Thinking C.01 Identify the problem. Met by research and preparation prior to group discussion. C.03 Identify available solutions and their impact including evaluating credibility of information, and locating information. Met by research and preparation prior to group discussion. C.07 Organize personal workloads. Met by daily labwork, preparatory research, and project management. C.08 Participate in brainstorming sessions to generate new ideas and solve problems. Met by group discussion. D Basic and Practical Skills Reading D.01 Read and apply various sources of technical information (e.g. manufacturer literature, codes, and regulations). Met by research and preparation prior to group discussion. E Basic and Practical Skills Proficiency in Mathematics E.01 Determine if a solution is reasonable. E.02 Demonstrate ability to use a simple electronic calculator. E.05 Solve problems and [sic] make applications involving integers, fractions, decimals, percentages, and ratios using order of operations. E.06 Translate written and/or verbal statements into mathematical expressions. E.12 Interpret and use tables, charts, maps, and/or graphs. E.13 Identify patterns, note trends, and/or draw conclusions from tables, charts, maps, and/or graphs. E.15 Simplify and solve algebraic expressions and formulas. E.16 Select and use formulas appropriately. E.17 Understand and use scientific notation. 2

ELTR 105 (DC 2), section 1 Common areas of confusion for students Difficult concept: Respective rules for series versus parallel circuits. By themselves, series circuits should be fairly easy for you to analyze at this point. The same may be said for parallel circuits by themselves. The respective rules for each (voltages same across parallel components; currents same through series components; parallel currents add; series voltages add; etc.) are not that difficult to remember, especially if you understand why each one is true. Things become much trickier, though, when we begin to mix series networks of resistors with parallel networks of resistors, as we jump from one rule set to the other in our analysis. My best advice here is to avoid trying to memorize the rules by rote, and rather work on comprehending why each rule is as it is for each type of circuit. And of course, practice, practice, practice! Difficult concept: Using Ohm s Law in context. When applying Ohm s Law (E = IR ; I = E R ; R = E I ) to circuits containing multiple resistances, students often mix contexts of voltage, current, and resistance. Whenever you use any equation describing a physical phenomenon, be sure that each variable of that equation relates to the proper real-life value in the problem you re working on solving. For example, when calculating the current through resistor R 2, you must be sure that the values for voltage and resistance are appropriate for that resistor and not some other resistor in the circuit. If you are calculating I R2 using the Ohm s Law equation I = E R, then you must use the value of that resistor s voltage (E R2 ) and that resistor s resistance (R 2 ), not some other voltage and/or resistance value(s). Some students have an unfortunate tendency to overlook context when seeking values to substitute in place of variables in Ohm s Law problems, and this leads to incorrect results. Common mistake: Carelessness when analyzing series-parallel networks. When analyzing a series-parallel circuit to determine component voltages and currents, one must be very careful to reduce the circuit step-by-step, section-by-section, into equivalent resistances. Students new to this process typically see all the work that is involved and try to save effort by taking shortcuts, which paradoxically causes more work and more confusion for them later on. Use lots of paper to document your work when you reduce series-parallel resistor networks to equivalent resistances, re-drawing the circuit for each reduction. This helps reduce the number of mistakes, and also makes it easier to transfer calculated values of voltage and current to the correct components (see the difficult concept shown above: using Ohm s Law in context). Difficult concept: Qualitative analysis of circuits. Most students find that qualitative analysis of electric circuits is much more difficult than quantitative analysis. In other words, it is easier to use a calculator to compute how much current goes through a particular resistor in a circuit than it is to figure out if that current will increase, decrease, or remain the same given a certain change in the circuit. What this requires is a feel for how variables in Ohm s Law and Kirchhoff s Laws relate to each other, which is a very different skill than numerical calculation. A good way to build this skill is to practice qualitative analysis on every quantitative circuit question you encounter. Even though the question only asks for a numerical answer, you can challenge yourself by asking whether or not a particular variable will change (and which direction that change will be in) given any particular change in the circuit (more or less supply voltage, a wire breaking open, a resistor failing open, or a resistor failing shorted). Then, you may re-calculate the numerical answers to see if your qualitative predictions are correct. Computer-based circuit simulation programs are excellent for this, as they allow one to skip the steps of numerical calculation to rapidly see the effects of certain circuit faults. 3

Question 1 Identify which of these components are connected directly in series with each other, and which are connected directly in parallel with each other: Figure 1 Figure 2 Figure 3 R1 R1 R1 R2 R3 R2 R3 R2 R3 Figure 4 Figure 5 Figure 6 R1 R1 R3 R1 R3 R2 R4 R2 R4 R2 R3 Assume that the open wire ends are connection points to a power source. file 01752 Answer 1 Figure 1: R2 in parallel with R3. Figure 2: R1 in series with R2. Figure 3: R2 in series with R3. Figure 4: R1 in series with R2; R3 in series with R4. Figure 5: R1 in parallel with R3; R2 in parallel with R4. Figure 6: R1 in series with R2. 4

Notes 1 Work with your students to clearly identify rules by which series and parallel connections may be identified. This is extremely important for students to grasp if they are to be successful analyzing seriesparallel networks of any kind. The most common problems I encounter as an electronics instructor with reference to series-parallel are invariably related to students lack of ability to consistently distinguish series sub-networks and parallel sub-networks in series-parallel combination circuits. 5

Question 2 Identify which of these components are connected directly in series with each other, and which are connected directly in parallel with each other: Figure 1 Figure 2 Figure 3 SW1 R2 R4 C1 C1 R1 R1 R3 R2 L1 R1 Figure 4 Figure 5 Figure 6 C1 X1 L1 R2 R4 R2 L1 L2 C1 R1 R3 R1 Assume that the open wire ends are connection points to a power source. In circuits where ground symbols appear, consider ground as the other side of the power source. file 01753 6

Answer 2 Figure 1: R1 in series with SW1. Figure 2: R1 in series with R2; R3 in parallel with R4. Figure 3: R1 parallel with R2. Figure 4: R1 parallel with R2. Figure 5: L1 in series with C1. Figure 6: R3 in parallel with R4. Challenge question: if you compare figures 2 and 6, you see how merely changing the location(s) where the power supply connects to the network can alter the series/parallel relationships of the components. But, exactly what is it that is altered? If two components are in series with each other in one power source configuration, can that series relationship change by moving the power supply connection points? How about parallel connections? If two components are in parallel with each other, can that parallel relationship become altered merely by moving the points where the power source connects to the network? Explain. Notes 2 Work with your students to clearly identify rules by which series and parallel connections may be identified. This is extremely important for students to grasp if they are to be successful analyzing seriesparallel networks of any kind. The most common problems I encounter as an electronics instructor with reference to series-parallel are invariably related to students lack of ability to consistently distinguish series sub-networks and parallel sub-networks in series-parallel combination circuits. 7

Question 3 Identify which of these components are connected directly in series with each other, and which are connected directly in parallel with each other: Printed circuit board R1 Lamp - - C1 D1 - + Battery file 00031 Answer 3 Connected directly in series: Battery and R1. R1 and Battery in series R1 Lamp - - C1 D1 - + Battery Connected directly in parallel: Lamp, C1, and D1 Lamp, C1, and D1 in parallel R1 Lamp - - C1 D1 - + Battery 8

Notes 3 Students must have a firm understanding of what constitutes series versus parallel in real circuits. Here is a place where some students will feel uncomfortable because the textbook definitions they memorized are easier said than applied. It is imperative that students have a strong working knowledge of terms, and do not simply memorize definitions. 9

Question 4 In a series circuit, certain general rules may be stated with regard to quantities of voltage, current, resistance, and power. Express these rules, using your own words: In a series circuit, voltage... In a series circuit, current... In a series circuit, resistance... In a series circuit, power... For each of these rules, explain why it is true. file 00291 Answer 4 In a series circuit, voltage drops add to equal the total. In a series circuit, current is equal through all components. In a series circuit, resistances add to equal the total. In a series circuit, power dissipations add to equal the total. Notes 4 Rules of series and parallel circuits are very important for students to comprehend. However, a trend I have noticed in many students is the habit of memorizing rather than understanding these rules. Students will work hard to memorize the rules without really comprehending why the rules are true, and therefore often fail to recall or apply the rules properly. An illustrative technique I have found very useful is to have students create their own example circuits in which to test these rules. Simple series and parallel circuits pose little challenge to construct, and therefore serve as excellent learning tools. What could be better, or more authoritative, than learning principles of circuits from real experiments? This is known as primary research, and it constitutes the foundation of scientific inquiry. The greatest problem you will have as an instructor is encouraging your students to take the initiative to build these demonstration circuits on their own, because they are so used to having teachers simply tell them how things work. This is a shame, and it reflects poorly on the state of modern education. 10

Question 5 In a parallel circuit, certain general rules may be stated with regard to quantities of voltage, current, resistance, and power. Express these rules, using your own words: In a parallel circuit, voltage... In a parallel circuit, current... In a parallel circuit, resistance... In a parallel circuit, power... For each of these rules, explain why it is true. file 00292 Answer 5 In a parallel circuit, voltage is equal across all components. In a parallel circuit, currents add to equal the total. In a parallel circuit, resistances diminish to equal the total. In a parallel circuit, power dissipations add to equal the total. Notes 5 Rules of series and parallel circuits are very important for students to comprehend. However, a trend I have noticed in many students is the habit of memorizing rather than understanding these rules. Students will work hard to memorize the rules without really comprehending why the rules are true, and therefore often fail to recall or apply the rules properly. An illustrative technique I have found very useful is to have students create their own example circuits in which to test these rules. Simple series and parallel circuits pose little challenge to construct, and therefore serve as excellent learning tools. What could be better, or more authoritative, than learning principles of circuits from real experiments? This is known as primary research, and it constitutes the foundation of scientific inquiry. The greatest problem you will have as an instructor is encouraging your students to take the initiative to build these demonstration circuits on their own, because they are so used to having teachers simply tell them how things work. This is a shame, and it reflects poorly on the state of modern education. 11

Question 6 Rank these three light bulb assemblies according to their total electrical resistance (in order of least to greatest), assuming that each of the bulbs is the same type and rating: A B C Explain how you determined the relative resistances of these light bulb networks. file 00030 Answer 6 C (least total resistance) A B (greatest total resistance) Notes 6 I prefer to enter discussion on series and parallel circuits prior to introducing Ohm s Law. Conceptual analysis tends to be more difficult than numerical analysis in electric circuits, but is a skill worthwhile to build, especially for the sake of effective troubleshooting. It is effective after conceptual (qualitative) analysis, though, to go through a numerical (quantitative) analysis of a circuit like this to prove that the concepts are correct, if the students are advanced enough at this point to do series-parallel resistance calculations. 12

Question 7 Which components are guaranteed to share the exact same voltage by virtue of their connections with each other? Which components are guaranteed to share the exact same current by virtue of their connections with each other? Printed circuit board R1 Lamp - - C1 D1 - + Battery file 00033 Answer 7 The lamp, C1, and D1 are all guaranteed to share the exact same voltage. The battery and R1 are both guaranteed to share the exact same current. Notes 7 Here, the important relations between voltage, current, and component connection patterns are explored. This serves to further define, in practical ways, what the terms series and parallel really mean. 13

Question 8 Which components in this partial automobile schematic diagram are guaranteed to share the exact same voltage by virtue of their connections with each other? Which components are guaranteed to share the exact same current by virtue of their connections with each other? Brake pedal switch Headlight #1 Light switch Ammeter A Fusible link Brake lights Battery Gen Generator Headlight #2 file 00034 Answer 8 The two headlights are guaranteed to share the same voltage. So are the two brake lights. However, the voltage across the brake lights may not be the same as the voltage across the headlights at any given time! So long as the fusible link is not blown, the generator and battery will share approximately the same voltage. The ammeter, fusible link, and generator are all guaranteed to share the same current. Notes 8 Here, the important relations between voltage, current, and component connection patterns are explored. This serves to further define, in practical ways, what the terms series and parallel really mean. This question also affords the opportunity of discussing what a fusible link is, and how it compares to fuses and circuit breakers as an overcurrent protection device. 14

Question 9 In this series-parallel circuit, resistors R1 and R2 are in series with each other, but resistor R3 is neither in series nor in parallel with either R1 or R2: R1 R2 R3 Normally, the first step in mathematically analyzing a circuit such as this is to determine the total circuit resistance. In other words, we need to calculate how much resistance the voltage source sees in the network formed by R1, R2, and R3. If the circuit were a simple series configuration, our task would be easy: R1 R2 R3 Likewise, if the circuit were a simple parallel configuration, we would have no difficulty at all calculating total resistance: R1 R2 R3 Due to the fact that our given circuit is neither purely series nor purely parallel, though, calculation of total resistance is not a simple one-step operation. However, there is a way we could simplify the circuit to something that is either simple series or simple parallel. Describe how that might be done, and demonstrate using numerical values for resistors R1, R2, and R3. file 01755 15

Answer 9 Suppose we had these resistor values: R1 = 3000 Ω R2 = 2000 Ω R3 = 5000 Ω Notes 9 The total resistance in this case would be 2500 Ω. I ll let you figure out how to do this! Hint: 2.5k is exactly one-half of 5k Figuring out how to calculate total resistance in a series-parallel network is an exercise in problemsolving. Students must determine how to convert a complex problem into multiple, simpler problems which they can then solve with the tools they have. This sort of exercise is also helpful in getting students to think in terms of incremental problem-solving. Being able to take sections of a circuit and reduce them to equivalent component values so that the circuit becomes simpler and simpler to analyze is a very important skill in electronics. 16

Question 10 Rank these five light bulb assemblies according to their total electrical resistance (in order of least to greatest), assuming that each of the bulbs is the same type and rating: A D B C E Explain how you determined the relative resistances of these light bulb networks. file 00039 Answer 10 C (least total resistance) D A E B (greatest total resistance) Notes 10 I prefer to enter discussion on series and parallel circuits prior to introducing Ohm s Law. Conceptual analysis tends to be more difficult than numerical analysis in electric circuits, but is a skill worthwhile to build, especially for the sake of effective troubleshooting. It is effective after conceptual (qualitative) analysis, though, to go through a numerical (quantitative) analysis of a circuit like this to prove that the concepts are correct, if the students are advanced enough at this point to do series-parallel resistance calculations. 17

Question 11 Determine the amount of electrical resistance indicated by an ohmmeter connected between the following points in this circuit: Ω A 3k3 B A V Ω COM 4k7 2k7 C 1k5 D Between points A and B = Between points A and C = Between points C and D = Between points D and B = Between points B and C = Explain whether or not it makes any sense to speak of a total resistance for this network. file 01601 Answer 11 Between points A and B = 2.41 kω Between points A and C = 2.89 kω Between points C and D = 1.32 kω Between points D and B = 2.10 kω Between points B and C = 2.75 kω Notes 11 The purpose of this question is to get students to realize that the resistance looking into different areas of a resistive network depends on what those areas are. 18

Question 12 Calculate the resistance between points A and B (R AB ) for the following resistor networks: Figure 1 Figure 2 Figure 3 All resistors 500 Ω All resistors 1 kω B A A 2 kω 5 kω A B 100 Ω 470 Ω B Figure 4 Figure 5 Figure 6 250 Ω All resistors 2.2 kω B A 470 Ω A 220 Ω 470 Ω 100 Ω A 940 Ω B 330 Ω B file 01757 Answer 12 Figure 1: R AB = 500 Ω Figure 2: R AB = 750 Ω Figure 3: R AB = 1.511 kω Figure 4: R AB = 940 Ω Figure 5: R AB = 880 Ω Figure 6: R AB = 80.54 Ω 19

Notes 12 Note that the circuit in figure 4 is a trick: two of the resistors contribute absolutely nothing to R AB! Be sure to discuss why this is with your students. Discuss with your students how they approached each of these problems, and let the entire class participate in the reasoning process. The point of this question, like most of the questions in the Socratic Electronics project, is not merely to obtain the correct answers, but to stimulate understanding of how to solve problems such as these. 20

Question 13 Calculate the amount of voltage dropped across resistor R 2 : 1k5 R 1 R2 R3 24 V 2k2 5k Also, note the direction of current through it and the polarity of the voltage drop across it. file 01759 Answer 13 V R2 = 12.11 volts, positive on top and negative on bottom. If you follow conventional flow notation, this means current goes down through resistor R 2. The actual flow of electrons through R 2, however, is up. Notes 13 Discuss with your students how they obtained their answers for this question. The reasoning and procedures are far more important than the actual answer itself. Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don t have to actually do the math, but rather outline the steps they would take, in the order they would take them. By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these open thinking activities is how to check your work to see if any mistakes were made. 21

Question 14 Complete the table of values for this circuit: 220 Ω 470 Ω R 1 R 3 R 2 130 Ω 12 volts R 1 R 2 R 3 Total V I R P 220 Ω 130 Ω 470 Ω file 01758 Answer 14 R 1 R 2 R 3 Total V I R P 1.778 V 1.778 V 10.22 V 8.079 ma 13.67 ma 21.75 ma 220 Ω 130 Ω 470 Ω 12 V 21.75 ma 551.7 Ω 14.36 mw 24.30 mw 222.3 mw 261.0 mw Notes 14 Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work. Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don t have to actually do the math, but rather outline the steps they would take, in the order they would take them. By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these open thinking activities is how to check your work to see if any mistakes were made. 22

Question 15 Antique American automobiles often used 6 volt electrical systems instead of the 12 volt components found in more modern cars and trucks. People who restore these old vehicles may have difficulty finding old 6-volt generators and batteries to replace the defective, original units. An easy solution is to update the vehicle s generator and battery with modern (12 volt) components, but then another problem arises. A 12 volt generator and 12 volt battery will overpower the old 6 volt headlights, brake lights, and other electrical loads in the vehicle. A solution used by antique automobile restorers is to connect resistors between the 12-volt generator system and the 6-volt loads, like this: Brake pedal switch Light switch 6 volts 6 volts Brake lights Headlight #1 6 volts A Ammeter Fusible link Headlight #2 6 volts Battery 12 volts Gen 12 volts Generator Explain why this solution works, and also discuss some of the disadvantages of using resistors to adapt the new (12 volt) to the old (6 volt) components. file 00062 Answer 15 The purpose of the resistors is to drop half the voltage supplied by the generator and battery, so that the loads (light bulbs, in this case) receive only 6 volts instead of 12 volts. A disadvantage of using resistors to do this is that the resistors waste a lot of electrical power in the form of heat. Notes 15 Make sure your students understand the concept of a load: any electrical or electronic component that uses power from an electrical source. Usually, loads are the end-use components of a circuit: light bulbs, motors, solenoids, speakers, etc. In this case, the resistors could be considered loads as well as the light bulbs, but since the light bulbs are the only components performing useful work from the power source, it is customary to think of them when the word load is used, rather than the resistors. 23

Question 16 Draw a schematic diagram of this breadboarded circuit: R1 R2 R4 R3 + - file 01760 Answer 16 R1 R2 R3 R4 Notes 16 If your students are not yet aware of how solderless breadboard holes are connected together, this is a good time to introduce them! 24

Question 17 From observation of this circuit (with components attached to a terminal strip ), draw an appropriate schematic diagram: Light bulb Resistor Motor Pushbutton switch Battery + - file 00115 Answer 17 Mtr Notes 17 This type of question is one that lends itself well to students drawing their answers on the board in front of class. The skill of transferring a real circuit into a cleanly-drawn schematic is one that some students struggle mightily with, but it is important. Those students will want to know what technique(s) may be used to make the transfer. Students who are more spatially adept will probably have a couple of different ways to approach a problem such as this. Allow them to explain to the rest of the class their technique(s) for tracing the real circuit s wiring into a schematic diagram. Giving students the opportunity to teach their peers is a powerful instructional method, and should be encouraged at all times! 25

Question 18 Think of a way to re-wire the electrical system of this old automobile (with 6-volt light bulbs) so as to not require resistors between the loads and the generator/battery portion of the circuit (operating at 12 volts each). Brake pedal switch Light switch 6 volts 6 volts Brake lights Headlight #1 6 volts A Ammeter Fusible link Headlight #2 6 volts Battery 12 volts Gen 12 volts Generator file 00063 Answer 18 Connect the light bulb pairs in series instead of parallel. This way, each light bulb will receive 6 volts, with a total system voltage of 12 volts. Follow-up question: there is a disadvantage of this strategy, though, and it concerns the safety of operating the automobile. Explain what this disadvantage is. Notes 18 This solution works only because the load sets are in pairs, and because 6 + 6 = 12. One benefit of this solution is greater efficiency, as there are no resistors in the circuit to waste power by dissipating it in the form of heat. However, there is a disadvantage to doing things this way, as indicated by the follow-up question. Discuss this disadvantage with your students, reinforcing the idea that the most efficient engineering solutions may not be the best when assessed from other perspectives, such as safety! 26

Question 19 Calculate the voltage drops V AB, V BC, and V CD in the following circuit: + - 1200 Ω A B C 150 Ω 470 Ω D 1.5 VDC file 01764 Answer 19 V AB = 461 mv V BC = 0 V V CD = 1.039 V Follow-up question: explain why the voltage between points A and B (V AB ) would increase if the 1200 Ω resistor were to fail shorted. Hint: imagine a jumper wire connected across that resistor to simulate a shorted failure. Challenge question: explain how you can calculate these same answers without ever having to calculate total circuit current. Notes 19 Ask your students how they could tell V BC must be zero, just by examining the circuit (without doing any math). If some students experience difficulty answering this question on their own, have them translate the drawing into a proper schematic diagram. Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don t have to actually do the math, but rather outline the steps they would take, in the order they would take them. By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these open thinking activities is how to check your work to see if any mistakes were made. 27

Question 20 Calculate the voltage magnitude and polarity between points A and D in this circuit, assuming a power supply output voltage of 10.5 volts: Power supply + - A 3k3 B 4k7 2k7 C 1k5 D Also, calculate the total current output by the power supply as it energizes this resistor network. file 01765 Answer 20 V AD = 7.31 volts, A positive and D negative. The total power supply current is 4.36 ma. Follow-up question: explain why the voltage across the 4.7 kω resistor would go to zero if the 1.5 kω resistor were to fail open. Notes 20 Though some students might not realize it at first, there is no series-parallel analysis necessary to obtain the voltage drop V AD. Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don t have to actually do the math, but rather outline the steps they would take, in the order they would take them. By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these open thinking activities is how to check your work to see if any mistakes were made. As a follow-up to the follow-up question, ask your students what other resistor in this circuit would completely lose voltage given an open failure of the 1.5 kω resistor. 28

Question 21 Calculate the power supply s output (total) current: Power supply + - A 3k3 B 4k7 2k7 C 1k5 D V A V OFF A A COM file 01766 Answer 21 I total = 4.69 ma Follow-up question: explain why the voltage across the 1500 Ω resistor would remain unchanged if the 4700 Ω resistor were to fail open. Challenge question: what crucial assumptions underlie the calculated figure for current shown here? In other words, what unknown quantities can affect the accuracy of our predicted current value? Notes 21 This is an interesting series-parallel circuit problem to solve, and it shows once again how a good understanding of circuit theory enables unmeasured variables to be inferred. Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don t have to actually do the math, but rather outline the steps they would take, in the order they would take them. By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these open thinking activities is how to check your work to see if any mistakes were made. 29

Question 22 How much voltage needs to be dropped across resistor R 1 in order to make voltage V AB equal to zero? R 1 10 kω 36 VDC A V AB B 60 kω 30 kω How much resistance must R 1 possess in order to drop that amount of voltage? file 00543 Answer 22 V R1 = 9 V R 1 = 20 kω Follow-up question: what do you notice about the four resistors values in this condition where V AB = 0? Pair up these four resistors into two sets of two pairs, and calculate the ratios of those pairs. What do you notice about these ratios? Notes 22 The follow-up question regarding ratios is a good introduction to the fundamental principle of balanced bridge circuits. Having students work through the calculations together is a good way for them to see the principle for themselves. It is also important to note in this circuit which ratios are not in agreement with each other. You can t just divide these four resistors into any set of two pairs and expect the ratios to equal each other! It is very important for students to see this, as well. 30

Question 23 In general terms, describe what must be done to balance this bridge circuit. What, exactly, does the term balance mean in this context? R 1 R 3 + V - R 2 R 4 Also, write an equation containing only the four resistor values (R 1, R 2, R 3, and R 4 ) showing their relationship to one another in a balanced condition. file 00544 Answer 23 For a bridge circuit to be balanced means that there is zero voltage between the two opposite corners of the circuit (where the battery does not) connect. Achieving a condition of balance in a bridge circuit requires that the resistance ratios of the four arms of the circuit be in proportion: R 1 R 3 = R 2 R 4 Follow-up question: the bridge-balance equation shown above may also be written in a slightly different form: R 1 R 2 = R 3 R 4 Show algebraically how the first equation may be manipulated to take the form of the second equation, thus demonstrating these two equations equivalence. Notes 23 Challenge your students to write a balance equation describing how the ratios must relate to each other in order to achieve balance. 31

Question 24 Identify the most important qualification for the null meter used to balance a bridge circuit. In other words, describe what type of meter we would be looking for if we were to select one for use as a null meter. Describe why this particular quality is important. file 00545 Answer 24 Notes 24 Above all, a null meter must be sensitive. Discuss with your students the definition of sensitivity with regard to meter movements, and why null meters have to be sensitive in order for the bridge circuit to be accurately balanced. If your students have studied meter movement design, you may wish to challenge them with a question on exactly how a null meter movement might be constructed (i.e. what would have to be done to maximize its sensitivity?). 32

Question 25 What will happen to the voltage between points A and B if the power supply voltage increases? 50 kω 10 kω A B 50 kω 10 kω file 00546 Answer 25 Notes 25 V AB will remain the same as V supply increases. This question highlights another important concept of bridge circuits, namely that balance is irrespective of supply voltage. 33

Question 26 Explain how this bridge circuit is capable of being balanced for any values of R 1 and R 2 : R 1 G R pot R 2 file 03474 Answer 26 The potentiometer acts as a complementary pair of resistors: moving the wiper one direction increases the value of one while decrease the value of the other. Thus, it forms a voltage divider with an infinitely adjustable division ratio of 0% to 100%, inclusive. Notes 26 This question showcases one more use of the potentiometer: as a voltage divider used specifically to balance a bridge circuit for any arbitrary values of fixed resistances. If students have difficulty seeing how this is possible, you might want to try representing the pot as a pair of fixed resistors (R 3 and R 4 ), the wiper position determining the balance of those two resistance values (R pot = R 1 + R 2 ). 34

Question 27 A strain gauge is a device used to measure the strain (compression or expansion) of a solid object by producing a resistance change proportional to the amount of strain. As the gauge is strained, its electrical resistance alters slightly due to changes in wire cross-section and length. The following strain gauge is shown connected in a quarter-bridge circuit (meaning only one-quarter of the bridge actively senses strain, while the other three-quarters of the bridge are fixed in resistance): Metal specimen A V B Strain gauge (glued to specimen) Explain what would happen to the voltage measured across this bridge circuit (V AB ) if the strain gauge were to be compressed, assuming that the bridge begins in a balanced condition with no strain on the gauge. file 00547 Answer 27 The bridge circuit will become more unbalanced, with more strain experienced by the strain gauge. I will not tell you what the voltmeter s polarity will be, however! Notes 27 Be sure to have your students explain how they arrived at their answers for polarity across the voltmeter terminals. This is the most important part of the question! 35

Question 28 A strain gauge is a device used to measure the strain (compression or expansion) of a solid object by producing a resistance change proportional to the amount of strain: Metal specimen A V B Strain gauge (glued to specimen) Increased temperature The bridge circuit is supposed to respond to changes in specimen strain, but explain what will happen to the voltage measured across this bridge circuit (V AB ) if the specimen s temperature increases (with no stress applied), assuming that the bridge begins in a balanced condition with no strain on the gauge, at room temperature. Assume a positive α value for the strain gauge conductors. What does this indicate about the effectiveness of this device as a strain-measuring instrument? file 00548 Answer 28 If the specimen heats up, a voltage will develop between points A and B, with A being positive and B being negative. Notes 28 Be sure to have your students explain how they arrived at their answers for polarity across the voltmeter terminals. Ask your students whether or not the fact of the circuit s sensitivity to temperature invalidates its use as a strain-measuring system. Is it impossible to obtain a reliable measurement of strain, if we know temperature also affects the circuit output voltage? How could we compensate for the effects of temperature on the system? 36

Question 29 Explain how this strain gauge circuit exploits a property of bridge circuits to provide automatic temperature compensation (so that changes in specimen temperature do not compromise strain measurement accuracy): A V B Strain gauge "Dummy" gauge file 00550 Answer 29 The dummy gauge is attached to the specimen in such a way that it is not subjected to strain like the working gauge is. It it merely exposed to the same specimen temperature. The action of this circuit is easiest to comprehend in a scenario where there is no stress applied to the specimen, but its temperature changes. Follow-up question: suppose the dummy strain gauge develops an open failure, so no current may pass through it. Identify the polarity of the voltage drop that will develop across the voltmeter as a result of this fault. Notes 29 Because bridge circuits are inherently differential circuits, it is possible to perform neat tricks such as this where the effects of the undesired influence (temperature) become canceled. Incidentally, the principle of cancellation by differential measurement is one that is very common in electronic systems, especially instrumentation systems. 37

Question 30 This bridge circuit is suppose to generate an output voltage proportional to the difference between light exposure on the two photocells: R 1 R 2 + V - R 3 R 4 However something has failed in this circuit, because the voltmeter is pegged fully negative and will not change with varying light exposures on the two cells. Identify at least two possible failures that could cause the voltmeter to over-range in the negative direction. file 00504 Answer 30 Here are two failures, although they are not the only possibilities: R 1 could have failed shorted. Photocell R 3 could have failed open. Notes 30 Be sure to ask your students to describe failures other than the two mentioned in the answer. And, for all answers given, be sure to ask students how they determined those faults would cause the observed negative pegging of the voltmeter. As usual, the method of solution is much more important than the actual answer in this question. 38

Question 31 Complete the table of values for this circuit: 2.2 kω 6.8 kω 1 kω R 1 R 3 470 Ω R 2 R 4 6 volts R 1 R 2 R 3 R 4 Total V I R P 1 kω 2.2 kω 470 Ω 6.8 kω file 01756 Answer 31 R 1 R 2 R 3 R 4 Total V I R P 1.649 V 3.627 V 725 mv 725 mv 1.649 ma 1.649 ma 1.542 ma 107 µa 1 kω 2.2 kω 470 Ω 6.8 kω 6 V 1.649 ma 3.64 kω 2.718 mw 5.979 mw 1.117 mw 77.24 µw 9.891 mw Notes 31 Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work. Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don t have to actually do the math, but rather outline the steps they would take, in the order they would take them. By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these open thinking activities is how to check your work to see if any mistakes were made. 39

Question 32 Complete the table of values for this circuit: 220 Ω 100 Ω R 1 R 4 470 Ω R 3 18 V 130 Ω 270 Ω R 2 R 5 R 1 R 2 R 3 R 4 R 5 Total V I R P 220 Ω 130 Ω 470 Ω 100 Ω 270 Ω file 01767 Answer 32 R 1 R 2 R 3 R 4 R 5 Total V 3.978 V 2.351 V 6.328 V 3.155 V 8.517 V 18 V I 18.08 ma 18.08 ma 13.46 ma 31.55 ma 31.55 ma 31.55 ma R 220 Ω 130 Ω 470 Ω 100 Ω 270 Ω 570.6 Ω P 71.92 mw 42.50 mw 85.21 mw 99.51 mw 268.7 mw 567.8 mw Notes 32 Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work. 40

Question 33 Complete the table of values for this circuit: 790 Ω 2.2 kω R 1 8.6 kω 1 kω R 3 32 V R 4 630 Ω R 2 R 5 R 1 R 2 R 3 R 4 R 5 Total V I R 790 Ω 1 kω 8.6 kω 2.2 kω 630 Ω P file 01768 Answer 33 R 1 R 2 R 3 R 4 R 5 Total V I R P 13.43 V 18.57 V 13.43 V 17.00 ma 18.57 ma 1.562 ma 32 V 14.55 ma 0 V 0 ma 790 Ω 1 kω 8.6 kω 2.2 kω 630 Ω 228.4 mw 344.7 mw 20.98 mw 465.5 mw 0 mw 32 V 33.11 ma 966.4 Ω 1.06 W Challenge question: what circuit parameters will change if the diagonal wire in the right-hand side of the circuit is cut? 790 Ω 2.2 kω R 1 R 4 8.6 kω 1 kω R 3 32 V break 630 Ω R 2 R 5 41

Notes 33 Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work. 42

Question 34 Complete the table of values for this circuit: V A V OFF A A COM + - R1 R2 R3 R4 R5 R 1 R 2 R 3 R 4 R 5 Total V I R P 2 kω 1 kω 3.3 kω 4.7 kω 4.7 kω file 01769 Answer 34 R 1 R 2 R 3 R 4 R 5 Total V I R P 4.500 V 676.6 mv 2.233 V 1.590 V 1.590 V 4.500 V 2.250 ma 676.7 µa 2 kω 1 kω 3.3 kω 4.7 kω 4.7 kω 10.12 mw 676.7 µa 457.9 µw 1.511 mw 338.3 µa 338.3 µa 538.0 µw 538.0 µw 2.927 ma 1.538 kω 13.17 mw 43

Notes 34 Ask your students to identify components in this series-parallel circuit that are guaranteed to share the same voltage, and components that are guaranteed to share the same current, without reference to any calculations. This is a good exercise in identifying parallel and series interconnections, respectively. Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don t have to actually do the math, but rather outline the steps they would take, in the order they would take them. By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these open thinking activities is how to check your work to see if any mistakes were made. 44

Question 35 In simple AC power systems, one of the two conductors is typically called the hot, while the other conductor is typically called the neutral. What distinguishes the hot conductor from the neutral conductor in such a system? In other words, what exactly determines whether a conductor will be called either hot or neutral? 120 VAC Which one of these conductors is "hot" and which one is "neutral"? file 00312 Answer 35 The neutral conductor in an AC power system is always made electrically common with the earth, while the non-grounded conductor in an AC power system is called the hot conductor: Hot 120 VAC Earth ground Neutral Follow-up question #1: in a power circuit such as this, there is typically a single fuse or circuit breaker for overcurrent protection. Identify the best location to place this fuse or circuit breaker in the circuit. Follow-up question #2: explain how to use a multimeter to identify the hot and neutral conductors of a power system such as this. Notes 35 Ask your students what safety issues surround the hot and neutral conductors, respectively. What do the names imply about relative hazard, and why is this? Ask them to explain how these safety considerations impact the placement of the overcurrent protection device. 45