Electrical Circuits
Inquiry Lab Kit for AP® Physics 1
Materials Included In Kit
Cord connectors, black, 40 Cord connectors, red, 40 Lamp receptacles, plastic, 10 Lightbulbs, miniature, 3.7V, 20 Parallel circuit pins, 32 Resistors, 39 Ω, 15
Resistors, 220 Ω, 15 Resistors, 620 Ω, 15 Resistors, 1.1 kΩ, 15 Resistors, 8.2 kΩ, 15 Resistors, 47 kΩ, 15
Additional Materials Required
(for each lab group) Battery, 6V or equivalent
Multimeter
Prelab Preparation
Prepare a bag of resistors for each student group by placing one of each resistor into a bag. Each bag should have six resistors—39 Ω, 220 Ω, 620 Ω, 1.1 kΩ, 8.2 kΩ and 47 kΩ.
Safety Precautions
Remind students to have caution when handling pins. In the case of a burned resistor, disconnect the circuit and ventilate the room if an unpleasant odor persists. Remind students to wash their hands thoroughly with soap and water before leaving the laboratory, and follow all laboratory safety guidelines.
Disposal
All materials may be saved and stored for future use. For resistors that have burned out, wait until they are cool and dispose of them in the regular trash.
Lab Hints
 This laboratory activity can be completed in two 50minute class periods. It is important to allow time between the Introductory Activity and the GuidedInquiry Activity for students to discuss and design the guidedinquiry procedures. Also, all studentdesigned procedures must be approved for safety before students are allowed to implement them in the lab. Prelab Questions may be completed before lab begins the first day, and analysis of the results may be completed the day after the lab or as homework. An additional lab period would be needed for students to complete an optional inquiry investigation (see Opportunities for Inquiry).
 In the Introductory Activity, it may be necessary for the students to adjust the multimeter to read the current at a smaller scale. As the value of the resistors increase, the current will decrease. In trials conducted in our labs, the current is best measured in milliamps (mA) up to 1.1 kΩ; beyond that, the current is best measured in microamps (μA).
 Other resistors may be used in this experiment. However, the resistors should be tested to ensure the multimeter can read precise data. In our trials, resistance values greater than 47 kΩ produced currents below the microamp (μA) range.
 The lamp is required in the Introductory Activity in order to protect the 39 Ω resistor from overheating. The 39 Ω resistor should not be used in a parallel circuit.
 The GuidedInquiry Design and Procedure portion of the lab is divided into two parts so students can focus on the unique characteristics of parallel and series circuits.
 The Supplementary Material Teacher PDFshould be copied for each student or each group in order to complete the Introductory Activity data analysis.
 It may be helpful to show the students how to use the parallel circuit pins to construct the junctions. The figure at right is one way the resistors can be connected in parallel.
 In the Opportunities for Inquiry portion of the lab, students should make a circuit involving a parallel branch in series with resistors. Students could also explore parallel braches that contain resistors in series on one branch.
 Extra resistors are included in case of any resistors failing.
{13795_Hints_Figure_6}
Teacher Tips
 The terms “voltage drop” and “change in potential” have similar meanings when discussing electrical circuits. Students should understand that by connecting the terminals of a battery together with a wire, a potential difference is applied to the wire. Charges will undergo changes in potential as they flow from one terminal to the other. A resistor impedes the current and has a potential difference across it. However, a resistor does not cause a change in potential.
 The resistance values for resistors in series is summative, in that R_{eff} = R_{1} + R_{2} + … + R_{n}.
 The resistance values for resistors in parallel is the sum of the inverse values, such that:
{13795_Tips_Equation_2}
Further Extensions
Alignment to the Curriculum Framework for AP^{®} Physics 1
Enduring Understandings and Essential Knowledge The energy of a system is conserved. (5B) 5B9: Kirchhoff’s loop rule describes conservation of energy in electrical circuits.
The electric charge of a system is conserved. (5C) 5C3: Kirchhoff’s junction rule describes the conservation of electric charge in electrical circuits. Since charge is conserved, current must be conserved at each junction in the circuit. Examples should include circuits that combine resistors in series and parallel.
Learning Objectives 5B9.1: The student is able to construct or interpret a graph of the energy changes within an electrical circuit with only a single battery and resistors in series and/or in, at most, one parallel branch as an application of the conservation of energy (Kirchhoff’s loop rule). 5B9.2: The student is able to apply conservation of energy concepts to the design of an experiment that will demonstrate the validity of Kirchhoff’s loop rule (ΣΔ V = 0) in a circuit with only a battery and resistors either in series or in, at most, one pair of parallel branches. 5B9.3: The student is able to apply conservation of energy (Kirchhoff’s loop rule) in calculations involving the total electric potential difference for complete circuit loops with only a single battery and resistors in series and/or in, at most, one parallel branch. 5C3.1: The student is able to apply conservation of electric charge (Kirchhoff’s junction rule) to the comparison of electric current in various segments of an electrical circuit with a single battery and resistors in series and in, at most, one parallel branch and predict how those values would change if configurations of the circuit are changed. 5C3.2: The student is able to design an investigation of an electrical circuit with one or more resistors in which evidence of conservation of electric charge can be collected and analyzed. 5C3.3: The student is able to use a description or schematic diagram of an electrical circuit to calculate unknown values of current in various segments or branches of the circuit.
Science Practices 1.1 The student can create representations and models of natural or manmade phenomena and systems in the domain. 1.4 The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively. 2.1 The student can justify the selection of a mathematical routine to solve problems. 2.2 The student can apply mathematical routines to quantities that describe natural phenomena. 3.1 The student can pose scientific questions. 3.2 The student can refine scientific questions. 3.3 The student can evaluate scientific questions. 4.1 The student can justify the selection of the kind of data needed to answer a particular scientific question. 4.2 The student can design a plan for collecting data to answer a particular scientific question. 4.3 The student can collect data to answer a particular scientific question. 5.1 The student can analyze data to identify patterns or relationships. 5.3 The student can evaluate the evidence provided by data sets in a particular scientific question. 6.1 The student justify claims with evidence. 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models.
Correlation to Next Generation Science Standards (NGSS)^{†}
Science & Engineering Practices
Analyzing and interpreting data Developing and using models Planning and carrying out investigations Asking questions and defining problems Using mathematics and computational thinking Constructing explanations and designing solutions Obtaining, evaluation, and communicating information
Disciplinary Core Ideas
HSPS3.A: Definitions of Energy HSPS3.B: Conservation of Energy and Energy Transfer HSPS4.A: Wave Properties HSPS4.B: Electromagnetic Radiation HSETS1.C: Optimizing the Design Solution
Crosscutting Concepts
Patterns Cause and effect Systems and system models Energy and matter Structure and function Stability and change
Performance Expectations
HSPS35. Develop and use a model of two objects interacting through electric or magnetic fields to illustrate the forces between objects and the changes in energy of the objects due to the interaction. HSPS41. Use mathematical representations to support a claim regarding relationships among the frequency, wavelength, and speed of waves traveling in various media.
Answers to Prelab Questions
 What is the voltage drop (potential difference) between points A and B? Explain how you determined this, referencing Kirchoff’s loop rule.
The voltage drop (change in potential) between points A and B is equal to 6.00 V, the potential of the battery. The sum of the potential changes in a closed circuit must equal to zero. Because the battery has a potential of 6.00 V, the decrease in potential across the resistor, R_{1}, must be equal to 6.00 V as well.
 The current at A is measured with an ammeter to be 125 mA. What would be the expected current at B? Explain how you made your determination referencing Kirchoff’s junction rule.
The current at point B would be 125 mA. The current at point A enters R_{1} and must leave with the same magnitude at point B. The current at point B enters the battery with the same magnitude as that which leaves the battery, entering point A. Because there are no nodes (junctions) in the circuit, the current cannot split at either point A or point B. Therefore, the currents at both points must be equivalent.
 What is the resistance of resistor R_{1}?
V = I x R
{13795_PreLabAnswers_Equation_3}
 Based on the information provided in the Background, draw an arrow indicating the direction the current is flowing in the circuit in Figure 2.
The current flows from the positive terminal of the battery to the negative terminal. The diagram should show an arrow pointing to the right.
 Explain how the graph supports Kirchoff’s loop rule.
In the graph, the electric potential increases across the battery from the negative terminal to the positive terminal. When the resistor R_{1} is encountered, there is a decrease in potential (voltage drop) equal to that of the battery. The graph supports Kirchoff’s loop rule as the sum of the changes in potential equals zero.
Sample Data
Introductory Activity
{13795_Data_Table_1}
Calculation and Analysis for Resistor 1
V = I x R
{13795_Data_Equation_4}
{13795_Data_Equation_5}
{13795_Data_Equation_6}
Percent Error = 2.05% Tolerance = Gold band = 5% The resistor falls within the tolerance of resistance set by the manufacturer. Series Circuit
{13795_Data_Table_2}
Analyze the Results
The sum of the potential differences across the individual resistors is nearly equal to the potential difference across both in series (5.32 V and 5.35 V, respectively). This aligns with Kirchoff’s loop rule in that the change in potential across both resistors is equivalent to that of the battery. Similarly, the current through both resistors is the same, confirming Kirchoff’s junction rule. Because there are no junctions (nodes), the current remains constant throughout the circuit. The measured effective resistance was 833 Ω and the individual resistances were 217 Ω and 612 Ω. The sum of the individual resistances is nearly equal to the effective resistance of the series, (829 Ω and 833 Ω, respectively) (i.e., R _{eff} = R _{1} + R _{2} + … + R _{n}). Parallel Circuit
{13795_Data_Table_3}
Analyze the Results
The potential differences across the individual resistors is equal to the potential difference across both in parallel. This aligns with Kirchoff’s loop rule in that the change in potential across both loops is equivalent to that of the battery. The current through the individual resistors is not the same. There is a junction (node) where the current must split to provide equal changes in potential across both resistors. According to Kirchoff’s junction rule, the currents must sum to be equal to the overall current of the circuit. The currents through the individual resistors, do indeed, sum to the overall current. The measured effective resistance was 163 Ω and the individual resistances were 221 Ω and 615 Ω. The effective resistance is less than the resistance of either path. The sum of the individual, inverse resistance values is equal to the inverse value of the effective resistance.
{13795_Data_Equation_7}
Answers to Questions
GuidedInquiry Discussion Questions
Part A. Series Circuits
 What is the voltage drop (potential difference) between points A and C? Explain how you determined this, referencing Kirchoff’s loop rule.
The drop in voltage between points A and C would be 6.00 V. The battery provides a potential of 6.00 V. The potential must be decreased by an equal amount so that the sum of the voltages is zero. Therefore, the decrease in potential across points A and C, and across the resistors, R_{2} and R_{3}, must be equal to 6.00 V.
 The current at point A is measured with an ammeter to be 15.5 mA. What would be the expected current at point B? What would be the expected current at point C? Explain how you made your determination in terms of Kirchoff’s junction rule.
The current at points B and C would be 15.5 mA. The current at point A enters R_{2} and must leave with the same magnitude at point B. The current at point B enters and leaves R_{3} with the same magnitude as at point A. The current at point C enters the battery with the same magnitude as that which leaves the battery, entering point A.
 What is the overall, effective resistance, R_{eff}, of the circuit according to Ohm’s law?
V = I x R_{eff}
{13795_Answers_Equation_8}
 Considering your answers to the previous questions, write a formula that relates the effective resistance, R_{eff}, to the individual resistors.
V = V_{2} + V_{3}I x R_{eff} = I_{2} x R_{2} + I_{3} x R_{3} I = I_{2} = I_{3} ∴R_{eff} = R_{2} + R_{3}
 Construct a graph of potential versus position for the circuit in Figure 4. Hint: See Figure 2 from Prelab Questions.
{13795_Answers_Figure_7}
 What other data would need to be collected in order to determine the resistance of the individual resistors, R_{2} and R_{3}? Explain.
In order to apply Ohm’s law to the individual resistors, the voltage drop across R_{2} would need to be measured. The sum of the voltage drops across both resistors is the potential of the battery. By measuring the voltage drop of one resistor, the other can be found by subtraction. The current through the resistors is the same, so only a single measurement is needed.
 Using the circuit diagram in Figure 4 as a template, draw schematics for how you would measure:
 Voltage drop across resistor R_{2}, resistor R_{3}, and the series resistor R_{2,3}
 Current through resistor R_{2} and resistor R_{3}
{13795_Answers_Figure_8}
Part B. Parallel Circuits
 How many paths (loops) can charges follow from the positive terminal of the battery to the negative terminal?
There are two paths charges can follow in the circuit. The charges can flow from point A to F through points B and D or through points C and E. Either path will let the charges flow from the positive terminal of the battery to the negative terminal.
 What is the voltage drop (potential difference) between points A and F? Would this potential difference be observed between any other points in the circuit? Explain how you determined this in terms of Kirchoff’s loop rule.
The voltage drop across points A and F is equal to 6.00 V, the potential of the battery. Following a path through any closed loop circuit, the sum of the changes in voltage must equal zero. Between points A and F, there must be a decrease in electric potential equal to that of the battery. This difference in electric potential would also be measured across points B and D and points C and E.
 The current at point A is measured with an ammeter to be 20.5 mA. What would be the expected current at point F? Explain how you made your determination referencing Kirchoff’s junction rule.
The current at point F is 20.5 mA. The current at point A is the amount leaving the battery and must enter the battery with the same magnitude. Although there are junctions between points A and F, charges cannot be lost or gained because that would violate the conservation of charge. Therefore, the current at F is equal to the current at A.
 After point A, there is a junction where current can flow into R_{4} or R_{5} and splits into I_{4} and I_{5}. Will the current split equally between R_{4} and R_{5}? Explain why or why not. Hint: Keep in mind the potential difference identified in Question 2.
In order to maintain the same difference in potential across R_{4} and R_{5}, the current must split in such a way that Ohm’s law for the individual resistors are equivalent (i.e., I_{4}R_{4} = V = I_{5}R_{5}). Because the resistors have the same difference in potential, the current split is based on the ratio of the resistances. If the resistance of the two resistors is the same, the split will be equal; otherwise, the split will be unequal.
 Prior to point F, there is a junction joining R_{4} and R_{5} together. What happens to the currents, I_{4} and I_{5}, at the junction? Explain your reasoning in terms of Kirchoff’s junction rule.
The currents I_{4} and I_{5} join together at the junction, becoming a single current again. This final current will be equal to 20.5 mA (see Question 3). Kirchoff’s current law states that the sum of the currents entering a junction is equal to the sum of the currents leaving. Therefore, I_{4} + I_{5} = I = 20.5 mA.
 What is the overall effective resistance, R_{eff}, of the circuit according to Ohm’s law?
V = I x R_{eff}
{13795_Answers_Equation_9}
 Considering your answers to the previous questions, write a formula that relates the effective resistance, R_{eff}, to the individual resistors.
I = I_{4} + I_{5}
{13795_Answers_Equation_10}
V = V_{4} = V_{5}
{13795_Answers_Equation_11}
 Construct a graph of potential versus position for the circuit in Figure 5. Hint: See Figure 2 From Prelab Questions.
{13795_Answers_Figure_9}
 What other data would need to be collected in order to determine the resistance of the individual resistors, R_{4} and R_{5}?
In order to apply Ohm’s law to the individual resistors, R_{4} and R_{5}, the current through each path of the circuit is needed. This would give the current through the separate resistors. The voltage drop is the same for each resistor.
 Using the circuit diagram in Figure 5 as a template, draw schematics for how you would measure:
 Voltage drop across resistor R_{4}, resistor R_{5} and the parallel resistor R_{4,5}
 Current entering and leaving parallel resistors
 Current through the individual resistors, R_{4} and R_{5}
{13795_Answers_Figure_10}
Review Questions for AP^{®} Physics 1
 The resistance through one branch of a circuit is measured to be 15.0 Ω. A resistor is added to the branch. The resistance is now measured to be 4.50 Ω.
 Was the new resistor added in series or parallel to the branch? Explain how you made your determination.
The new resistor was added in parallel to the 15.0 Ω resistor. When resistors are added in parallel, the effective resistance is decreased because the inverse resistances are summed. If the resistor were added in series, then the effective resistance would increase because the resistances are summative.
 What is the resistance value of the added resistor?
{13795_Answers_Equation_12}
 A circuit is constructed with a 12.0 volt battery and three resistors, R_{1}, R_{2}, and R_{3}. The resistors are connected in series. The resistance values of R_{1} and R_{2} are the same. The current entering R_{3} is 158 mA, and R_{3} has a resistance of 50.0 Ω.
 Calculate the resistance of R_{1} and R_{2}.
R_{1} = R_{2} V = V_{1} + V_{2} + V_{3} V_{1} = V_{2} because R_{1} = R_{2} V = 2V_{1} + V_{3} = 2 x IR_{1} + IR_{3} 2 x IR_{1} = V – IR_{3}
{13795_Answers_Equation_13}
R_{1} = R_{2} = 13.0 Ω
 Calculate the change in potential across the three resistors individually.
V_{1} = V_{2} = I x R_{1} V_{1} = 0.158 A x 13.0 Ω = 2.05 V V_{1} = V_{2} = 2.05 V V_{3} = I x R_{3} V_{3} = 0.158 A x 50 Ω = 7.9 V
 The circuit in Figure 6 is constructed using a voltage source of 110 V and five resistors of equal resistance, R. The current entering R_{1} is 2.01 A.
 Determine the value of R.
R = R_{1} = R_{2} = R_{3} = R_{4} = R_{5}V = I x R_{T}R_{T} = R_{1} + R_{eff}
{13795_Answers_Equation_14}
{13795_Answers_Equation_15}
 Calculate the change in potential across R_{1}.
V_{1} = I x R_{1} V_{1} = 2.01 A x 39.1 Ω = 78.6 V
References
AP Physics 1: AlgebraBased and Physics 2: AlgebraBased Curriculum Framework; The College Board: New York, NY, 2014.
