Kirchoff’s Rules
Inquiry Lab Kit for AP® Physics 2
Materials Included In Kit
Capacitors, axial electrolytic, 1000 μF, 25 V, 30 Cord connectors with alligator clips, red and black, 80 Parallel circuit pins, 64 Resistors,220 Ω, 1*
Resistors,620 Ω, 1* Resistors,1.1 kΩ, 1* Resistors,8.2 kΩ, 1
Additional Materials Required
(for each lab group) Multimeter with capacity meter function
Power supply or equivalent, 6V
Prelab Preparation
 Prepare a bag of electrical components for each student group by placing one of each resistor and two capacitors into a bag. Each bag should contain four resistors—220 Ω, 620 Ω, 1.1 kΩ and 8.2 kΩ—and two 1000 μF capacitors.
 Photocopy enough of the Supplementary Material information for each group.
Safety Precautions
If a resistor begins to darken or smoke, immediately disconnect the circuit and do not touch the resistor—it will be hot. Allow it ample time to cool. Handle pins with caution, as they are sharp. Please follow all laboratory safety guidelines.
Disposal
All materials may be saved and stored for future use.
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).
 Enough materials are provided in this kit for 24 students working in groups of three or 8 groups of students with extra capacitors and resitors. This investigation can be reasonably completed in two 50minute class periods.
 Students may require a review of how to use a multimeter. Consider drawing the following figures on the board to serve as models of how the probes of a multimeter are connected to measure potential difference and current, respectively.
 In the Opportunities for Inquiry portion of the lab, students may make a circuit involving resistors and capacitors in various combinations of series and parallel. Check the circuit designs and assembled circuits before students connect them to a battery. Students should measure the change in current over time for their circuit to assess how the time constant is affected.
 When gathering data it is important for the circuit to be in a steady state, meaning that the capacitor is either fully charged or fully discharged. The amount of time for steadystate to be reached is generally five times the RC value of the circuit.
Teacher Tips
 Encourage students to determine how the effective capacitance can be related to the formula for capacitance. Capacitance is dependent on the dielectric material used, the surface area of the plates and the distance separating the plates. Assuming the dielectric material is unchanged, then capacitance has a direction relation to the surface area of the plates and indirect relation to the separation distance. When placing capacitors in series, the effective capacitance decreases because the separation of the positive and negative plates increases. When placing capacitors in parallel, the effective capacitance increases because the surface area of the positive and negative plates increases.
 Knowledge of how to derive formulas to track the change in current and charge in an RC circuit are beyond the scope of the course. However, knowledge of how a capacitor affects the current in a circuit is integral to understanding their importance in RC circuits.
Further Extensions
Opportunities for Inquiry Design an experiment to investigate initial and steadystate current in circuits of varying combinations of resistors and capacitors.
Alignment to the Curriculum Framework for AP^{®} Physics 2
Enduring Understandings and Essential Knowledge The electric and magnetic properties of a system can change in response to the presence of, or changes in, other objects or systems. (4E) 4E5: The values of currents and electric potential differences in an electric circuit are determined by the properties and arrangements of the individual circuit elements such as sources of emf, resistors, and capacitors.
The energy of a system is conserved. (5B) 5B9: Kirchhoff’s loop rule describes conservation of energy in electrical circuits. (The application of Kirchhoff’s laws to circuits is introduced in Physics 1 and further developed in Physics 2 in the context of more complex circuits, including those with capacitors.)
 Energy changes in simple electrical circuits are conveniently represented in terms of energy change per charge moving through a battery and resistor.
 Since electric potential difference times charge is energy, and energy is conserved, the sum of the potential differences about any closed loop must add to zero.
 The electric potential difference across a resistor is given by the product of the current and the resistance.
 The rate at which energy is transferred from a resistor is equal to the product of the electric potential difference across the resistor and the current through the resistor.
 Energy conservation can be applied o combinations of resistors and capacitors in series and parallel 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. (Physics 1: covers circuits with resistors in series, with at most one parallel branch, one battery only. Physics 2: includes capacitors in steadystate situations. For circuits with capacitors, situations should be limited to open circuit, just after circuit is closed, and a long time after the circuit is closed.) Learning Objectives 4E5.1: The student is able to make and justify a quantitative prediction of the effect of a change in values or arrangements of one or two circuit elements on the currents and potential differences in a circuit containing a small number of sources of emf, resistors, capacitors, and switches in series and/or parallel. 4E5.2: The student is able to make and justify a qualitative prediction of the effect of a change in values or arrangements of one or two circuit elements on currents and potential differences in a circuit containing a small number of sources of emf, resistors, capacitors, and switches in series and/or parallel. 4E5.3: The student is able to plan data collection strategies and perform data analysis to examine the values of currents and potential differences in an electric circuit that is modified by changing or rearranging circuit elements, including sources of emf, resistors, and capacitors. 5B9.5: The student is able to use conservation of energy principles (Kirchhoff ’s loop rule) to describe and make predictions regarding electrical potential difference, charge, and current in steadystate circuits composed of various combinations of resistors and capacitors. 5B9.6: The student is able to mathematically express the changes in electric potential energy of a loop in a multiloop electrical circuit and justify this expression using the principle of the conservation of energy. 5C3.6: The student is able to determine missing values and direction of electric current in branches of a circuit with both resistors and capacitors from values and directions of current in other branches of the circuit through appropriate selection of nodes and application of the junction rule. 5C3.7: The student is able to determine missing values, direction of electric current, charge of capacitors at steady state, and potential differences within a circuit with resistors and capacitors from values and directions of current in other branches of the circuit. Science Practices 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. 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. 6.1 The student can 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
Asking questions and defining problems Developing and using models Planning and carrying out investigations Engaging in argument from evidence Obtaining, evaluation, and communicating information
Disciplinary Core Ideas
MSPS2.A: Forces and Motion HSPS2.A: Forces and Motion
Crosscutting Concepts
Energy and matter Systems and system models Patterns
Performance Expectations
MSPS23. Ask questions about data to determine the factors that affect the strength of electric and magnetic forces
Answers to Prelab Questions
 Examine Figure 1 of a battery connected to two resistors in series.
 Explain how Kirchoff’s loop rule can be applied to solve for the potential difference (voltage drop) across resistor R_{1}.
Kirchoff ’s loop rule states that the overall change in electric potential in a circuit must equate to zero. The voltage of the battery summed with the voltage drop across the effective resistance of the circuit is equal to zero. Since the resistors are in series, the effective resistance is the sum of the resistances. Therefore, the voltage of the battery summed with the voltage drops across resistor R_{1} and R_{2} is equal to zero. If we know the voltage drop across resistor R_{2}, then the voltage drop across R_{1} is equal to the voltage of the battery minus the drop across R_{2} or “the drop across R_{2} subtracted from the voltage of the battery.”
 Explain how Kirchoff’s junction rule can be applied to solve for the current through the resistors, R_{1} and R_{2}.
Kirchoff ’s junction rule states that the amount of current flowing into a node must equal the amount of current leaving that same node. Since there are no nodes in the circuit, then the current is constant throughout and the current flowing through resistors R_{1} and R_{2} is equal to the voltage of the battery divided by the effective resistance of the circuit.
 Examine Figure 1 of a battery connected to two resistors in parallel.
 Explain how Kirchoff’s loop rule can be applied to solve for the potential difference (voltage drop) across resistor R_{1}.
Kirchoff ’s loop rule states that the overall change in electric potential in a circuit must equate to zero. The voltage of the battery summed with the voltage drop across the effective resistance of the circuit is equal to zero. Since the resistors are in parallel, the drop in voltage across them is equal. Since the parallel branch flows directly into the negative terminal of the battery we know that the voltage drop across both resistors must equal the voltage of the battery.
 Explain how Kirchoff’s junction rule can be applied to solve for the current through the resistors, R_{1} and R_{2}.
Kirchoff ’s junction rule states that the amount of current flowing into a node must equal the amount of current leaving that same node. Since the voltage drops are the same across both resistors and are equal to the voltage of the battery, one can simply divide the voltage of the battery by the respective resistance of each resistor to solve for the respective current through each resistor.
Sample Data
Introductory Activity
Resistors in Series
{14014_Data_Table_1}
Resistors in Parallel
{14014_Data_Table_2}
Analyze the Results In both the series and parallel arrangements, the measured resistance values for the resistors fall within the 5% tolerance set by the manufacturer (R_{1} = 220 ± 11, R_{2} = 620 ± 31). The effective resistance values for the series and parallel arrangements agree with Equations 2 and 3, respectively. The effective resistance value for resistors in series is the sum of the component resistance values and has greater resistance than either resistor in the arrangement. The effective resistance value for resistors in parallel is equal to the sum of the inverse of the resistance values and has less resistance than either resistor in the arrangement. GuidedInquiry Activity Part B. Capacitors in Series Analyze the Results
 Present your data in a table that shows the capacitance of each individual capacitor, the effective capacitance of the circuit, and the voltage readings for the resistors and capacitors used.
For a simple circuit containing a single resistor and capacitor:
Resistance of resistor used, R = 8.2 kΩ Capacitance of capacitor used, C_{1} = 990 μF
{14014_Data_Table_3}
For a circuit containing a single resistor and two capacitors in series:
Resistance used = 8.2 kΩ Capacitance of first capacitor, C_{1} = 990 μF Capacitance of second capacitor, C_{2} = 999μF Measured effective capacitance C_{eff} = 494 μF
{14014_Data_Table_4}
Charge on capacitors used:
{14014_Data_Table_5}
 Summarize your findings in several sentences while discussing charge, voltage, resistance and effective capacitance versus individual capacitance of the resistors.
The sum of the potential differences across C_{1} and C_{2} is nearly equal to the potential difference across both in series. The sum of the potential difference for the resistor and two capacitors used never surpassed the potential difference provided by the power supply, which aligns with Kirchoff ’s loop rule. The charge on the separate capacitors, C_{1} and C_{2} and across the effective capacitor is nearly identical, which aligns with Kirchoff ’s junction rule. Because there are no junctions, the current does not split in the circuit and all capacitors accumulate the same quantity of charge. The measured effective capacitance of the circuit compared to the individual capacitances of each capacitor measured aligns almost perfectly with the relationship derived in step 9 for capacitors in series. The calculated effective capacitance is 497.23 μF. The resistor appears to have no effect on the capacitance in the circuit.
Part C. Capacitors in ParallelAnalyze the Results
 Present your data in a table that shows the capacitance of each individual capacitor, the effective capacitance of the circuit, and the voltage readings for the resistors and capacitors used.
For a circuit containing a single resistor and two capacitors in parallel:
Resistance used = 8.2 kΩ Capacitance of first capacitor, C_{1} = 980 μF Capacitance of second capacitor, C_{2} = 960 μF Measured effective capacitance C_{eff} = 1960 Μf
{14014_Data_Table_6}
Charge on capacitors used:
{14014_Data_Table_7}
 Summarize your findings in several sentences while discussing charge, voltage, resistance and effective capacitance versus individual capacitance of the resistors.
After an extended time and once the circuit is closed, the potential differences across the individual capacitors 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 the battery. The sum of the charge on C_{1} and C_{2} is equal to the charge on the effective capacitor which agrees completely with the relationship derived in step 7. There is a junction where the current must split. The charge is divided according to the values of capacitance, the larger the capacitance, the larger the charge stored. The charges stored on the individual capacitors, although different, must combine to be equal to the charge stored due to the effective capacitance. The observations show that this is the case and align with Kirchoff ’s junction rule. The calculated effective capacitance is 1940 μF, nearly identical to the measured effective capacitance. The resistor does not appear to affect the capacitance of the circuit.
Answers to Questions
Part A. Capacitor Overview
 Recall that current is measured in amperes and is a derived unit. What SI units is the ampere derived from? Explain what a measurement of 2.0 A means with respect to the SI units.
Amperes are derived from coulombs (the unit of electrical charge) and seconds (a unit of time). An ampere is a measure of coulombs per second. A measurement of 2.0 A means that 2.0 coulombs of charge are moving through a given point in a circuit every second.
 Figure 9 shows a simple circuit made with a battery, a resistor and a capacitor. As the capacitor C_{1} charges, it creates an electric field between the two plates that opposes the voltage applied across it.
{14014_Answers_Figure_9}
Figure 10 shows the relationship between the charging current, I, and the charge on the capacitor, Q. In general, describe what happens to the current and charge, respectively, on the capacitor after t = 0?
{14014_Answers_Figure_10}
The charge on the capacitor starts at zero, increases steadily and then begins to slow. Eventually, the charge on the capacitor reaches a maximum (asymptote) at Q_{max}. The current starts at a maximum, I_{max}, decreases steadily, and eventually goes to zero (asymptote).
 Figure 11 shows a typical circuit used for investigating the properties of capacitors. Consider the following conditions and list the electronic components encountered by a charge moving in the circuit and identify its motion as clockwise or counterclockwise.
{14014_Answers_Figure_11}
 Assume the charge starts at point A and follows convention in the following instances:
 The switch is not in contact with wire W nor wire Z.
The charge does not move from point A. There is no complete circuit for the charge to follow.
 The switch is in contact with wire W. The charge moves in a clockwise manner.
The charge first encounters the resistor R, then capacitor C, and finally the battery, V_{batt}.
 The switch is in contact with wire Z. The charge does not move from point A.
There is no complete circuit for the charge to follow.
 The charge now starts at the top plate of capacitor C, the capacitor is fully charged and convention is followed:
 The switch is not in contact with wire W nor wire Z.
The charge does not move from the top place of C. There is no complete circuit for the charge to follow.
 The switch is in contact with wire W.
The charge does not move from the top plate of C. The fully charged capacitor completely opposes the applied voltage from the battery and no charge moves.
 The switch is in contact with wire Z.
The charge moves in a counterclockwise manner. It first encounters the resistor R and then the bottom plate of capacitor C.
Part B. Capacitors in Series
 Consider a capacitor connected directly to a battery. If the top plate accumulates a charge of +Q, what charge is accumulated on the bottom plate? Explain.
If the top plate accumulates +Q charge, the bottom plate must accumulate –Q charge. The amount of charge accumulated on the plates must be equal, however the signs of the charge will be opposite.
Figure 7 is a schematic of a circuit with capacitors C _{2} and C _{3} in series. Examine Figure 7 and answer the following questions.
 Which plates in the capacitors are directly connected to the battery terminals? Predict the expected charge (+Q or –Q) on these plates after an extended time once the switch is closed.
The top plate of C_{2} is connected directly to the positive terminal of the battery and will likely carry a +Q charge. The bottom plate of C_{3} is directly connected to the negative terminal of the battery and will likely carry a –Q charge.
 Note which plates are not directly connected to the battery terminals (i.e., there are no wires from the plates to the battery). Predict the expected charge on these plates after an extended time once the switch is closed.
The bottom plate of C_{2} and the top plate of C_{3} do not have wires connecting them to the battery terminals. Despite not being connected to the battery terminals, these plates will respond to the accumulated charges on their partner plates. The bottom plate of C_{2} will accumulate a –Q charge to match the +Q charge on the top plate of C_{2} . The top plate of C_{3} will accumulate a +Q charge to match the –Q charge on the bottom plate of C_{3}.
 Treating the plates that are not connected to the battery as a unit, what would you notice about the net charge on the plates? Does this align with the law of conservation of charge?
The twoplate unit is charge neutral. The positive and negative charges are equal in amount and cancel out. This agrees with the law of conservation of charge because there is no source of charge or a method of discharge for the plates, so the amount of charge on the plates cannot be increased or decreased.
 What is the relationship between the charge accumulated on the effective capacitor and the charge on the individual capacitors, C_{2} and C_{3}? Explain.
The charge on the effective capacitor is equal to the charge on the individual capacitors. Whatever amount of charge accumulates on the top plate of the first capacitor, the remaining plates will carry the same amount of charge (not necessarily the same sign) because the law of conservation of charge does not allow for charge to be created or lost in the process. Therefore, Q = Q_{2} = Q_{3}.
 What is the voltage drop (potential difference) between points D and F? Explain how you determined this in terms of Kirchoff’s loop rule.
The potential difference between points D and F is 6.00 V. The battery provides a potential of 6.00 V. The potential must decrease by an equal amount so that the sum of the voltages is zero. Therefore, the decrease across points D and F, and across the capacitors C_{2} and C_{3}, must be equal to –6.00 V.
 Write a mathematical relationship between V_{battery} and the potential differences caused by the components of the circuit in Figure 7.
V_{battery} = V_{2} + V_{3}
 What is the effective capacitance, C_{eff}, of the circuit according to Equation 5?
C_{eff} = Q/V_{battery}
 Considering your answers to the previous questions, write a formula that relates the effective capacitance, C_{eff}, to the individual capacitors.
V_{battery} = V_{2} + V_{3} = Q /C_{eff} Q = Q_{2} = Q_{3} Q /C_{eff} = Q /C_{2} + Q /C_{3} 1/C_{eff} = 1/C_{2} + 1/C_{3}
Part C. Capacitors in ParallelFigure 12 is a schematic of a circuit with capacitors C _{1} and C _{2} in parallel.
{14014_Answers_Figure_12}
 How many paths (loops) can a charge follow from the positive terminal of the battery to the negative terminal?
Charges can follow two paths in the circuit. The charges can flow from point D to I through points E and G or through points F and H. Either path will let the charges flow from the positive terminal battery to the negative terminal.
 What is the voltage drop (potential difference) between points D and I? 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 potential difference between points D and I is equal to 6.00 V, the potential of the battery. Following a path through any closedloop circuit, the sum of the changes in voltage must equal zero. Between points D and I, 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 E and G and points F and H.
 After point D, there is a junction where current can flow in C_{1} or C_{2} and splits into I_{1} and I_{2}. Will the current split equally between C_{1} and C_{2}? Explain why or why not.
The current will only split evenly if the two capacitors have the same capacitance. An equal split of current will supply the same amount of charge per second to the top plates of the capacitors, charging them at an equal rate. If the capacitance values are unequal, then the current will split unequally. However, the current through each of the capacitors will be such that charge accumulates at the same rate.
 Write a mathematical relationship between V_{battery} and the potential differences caused by the components of the circuit in Figure 12.
V_{battery} = V_{4} = V_{5}
 What is the overall effective capacitance, C_{eff}, of the circuit according to Equation 5?
C_{eff} = Q / V_{battery}
 Consider the distribution of charge between C_{1} and C_{2} and its relation to the charge on C_{eff}. How do these values compare?
The charge on C_{eff} must be equal to the charge distributed on C_{1} and C_{2}. Therefore, Q_{total} = Q_{1} + Q_{2} . There cannot be more charge accumulated on the capacitors than accumulated on the effective capacitor because charge must be conserved at the junction after point D. The total charge entering point D must be accounted for on C_{1} and C_{2}.
 Considering your answers to the previous questions, write a formula that relates the effective capacitor, C_{eff}, to the individual capacitors, C_{1} and C_{2}.
C_{eff} = Q_{total}/ V_{battery} = (Q_{1} + Q_{2}) / V_{battery} C_{eff} = Q_{1} /V_{battery} + Q_{2} /V_{battery} C_{eff} = C_{1} + C_{2}
Review Questions for AP^{®} Physics 2
 The capacitance through one branch of a circuit is measured to be 200 μF. A capacitor is added to the circuit. The effective capacitance is now measured to be 120 μF.
 Was the new capacitor added in series or in parallel to the circuit? Explain how you made your determination.
The new capacitor was added in series to the 200μF capacitor. When capacitors are added in series, the effective capacitance is decreased because the inverse capacitances are summed. If the capacitor was added in parallel, the effective capacitance would increase because the capacitances are summative.
 What was the value of the added capacitor?
1/C_{eff} = 1/C_{1} + 1/C_{2} 1/120 μF = 1/200 μF + 1/C_{2} 1/C_{2} = 1/120 μF – 1/200 μF 1/C_{2} = 2/600 μF C_{2} = 300 μF
 The time constant for any given RC circuit is the value of the effective resistance of the circuit multiplied by the effective capacitance of the circuit. What is the value of the time constant for the circuit?
{14014_Answers_Figure_13}
{14014_Answers_Figure_14}
The circuit can be simplified down by using the known relationships for resistors and capacitors in series and in parallel:
 The 20 kΩ and 5 kΩ resistors in parallel equal a 4 kΩ resistor by 1/20kΩ + 1/5kΩ = 1/4kΩ.
 The 500 μF and 2000 μF capacitors in series become 400 μF by 1/500μF + 1/2000μF = 1/400μF.
 The pair of 800 μF capacitors in series in the parallel branch following the first 1 kΩ resistor become 400 μF by 1/800 μF + 1/800 μF = 1/400 μF.
 The calculated 400 μF capacitor is in parallel with the 2000 μF capacitor, which add to become 2400 μF.
 The two remaining 1 kΩ resistors are in series with the 4 kΩ resistor, which add to become 6 kΩ.
 The calculated 400 μF capacitor is in series with the calculated 2400 μF capacitor and become 342.857 by 1/400 μF + 1/2400 μF = 1/342.857 μF.
Therefore, R x C = 6,000 Ω x (342.857 x 10^{–6} F) = 2.057 seconds.
 For the following circuit, the switch is closed and steadystate conditions have been reached, meaning the capacitor is fully charged.
{14014_Answers_Figure_15}
 What is the voltage across the 4 Ω resistor?
The voltage is zero because there is no current running through the branch since the capacitor is fully charged.
 What is the voltage across the 6 μF capacitor?
First, one must solve for current. We use Ohm’s law and Kirchoff ’s loop rule for each component of loop ABA to make the sum of the voltage increase due to the battery and voltage drops due to the resistors equate to zero. 12 V – 10 Ω(I) – 3 Ω(I) – 2 Ω(I) = 0 12 – 15(I) =0 15(I) = 12 I = 12/15 =4/5 or 0.8 A
To find the capacitor voltage we can simply use Kirchoff ’s loop rule for loop BCB and use the known current calculated above for the 3 Ω resistor. 4 Ω(0) – V_{C} + 3 Ω(I) = 0 V_{C} = 3 Ω(4/5 A) V_{C} = 2.4 V
 How much charge is stored on the capacitor?
Q = CV C = 6 μF V = 2.4 V Q = (6 × 10^{–6} F)(2.4 V) = 1.44 × 10^{–5} C
References
AP^{®} Physics 1: AlgebraBased and Physics 2: AlgebraBased Curriculum Framework; The College Board: New York, NY, 2014.
