Materials Included In Kit

Additional Materials Required

Prelab Preparation

1. Use the wire stripper to create two 2 inch long lengths of bare wire for each lab group.
2. Use masking tape to cover the color band of all 8.2 kΩ resistors.
3. (Optional) Practice making a capacitor as instructed in the Introductory Activity and measuring the capacitance with the multimeter.
4. After cutting enough wire for the Introductory Activity, cut multiple 2 to 4 cm long pieces of wire (only stripping the ends of each segment, leaving the middle insulated) for use in the Guided-Inquiry Activity.
5. Assign half the lab groups to Group A and the other half to Group B for the Introductory Activity.

Safety Precautions

Although 9-V batteries and power supplies are not powerful enough to deliver an electric shock under normal conditions, please follow all proper safety precautions when working with electronics. Never allow battery terminals to touch or short circuit, as this may create a fire hazard. Never handle batteries or electrical components with wet hands. 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 50-minute class periods. It is important to allow time between the Introductory Activity and the Guided-Inquiry Activity for students to discuss and design the guided-inquiry procedures. Also, all student-designed 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.
• In order to obtain the necessary data in the Introductory Activity, a multimeter capable of measuring capacitance is required. Multimeters are available for purchase from Flinn Scientific, Catalog No. AP9000.
• Diodes are circuit components that only allow current to pass in one direction. A light emitting diode (LED) is a special type of diode that will light up when a critical voltage is reached. Because of the low current requirement, LEDs are ideal for use in this application, requiring only their critical voltage to remain illuminated. The critical voltage for these diodes is within the range 1.7–2.2 V.
• It is advisable to check students’ circuits to ensure there will be no short circuit that could possibly damage some of the electronic components, such as the LED.
• If students cannot obtain a reading for capacitance in the Introductory Activity, soldering the wires to the aluminum foil will provide a better connection. Care should be taken to tightly roll the capacitors.

Teacher Tips

• Some introduction may be required regarding the functionality of the solderless breadboard prior to the Guided-Inquiry Activity.
• Before performing the Guided-Inquiry Activity, make sure that students are aware of the current direction dependence of the capacitor used. Have students inspect the capacitor to determine which lead is the negative lead. This will be represented by a black band with a “–” symbol, along with arrows pointing in the direction of the negative lead.
• If students are unable to fully deduct a method to solve for the unknown resistance in the Guided-Inquiry Activity, the Analyze the Results section is designed to guide students to this answer. Students are likely to collect data for the time the LED stays lit; this is enough for them to be able to move on to the Analyze the Results section.

Further Extensions

Opportunities for Inquiry 
An extension would be to incorporate different dielectric materials in the construction of a capacitor and attempt to determine the capacitance when incorporated with a known large resistor and LED in an RC circuit.

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 change in, other objects or systems. (4E)
4E4: The resistance of a resistor and the capacitance of a capacitor can be understood from the basic properties of electric fields and forces as well as the properties of materials and their geometry.

1. The resistance of a resistor is proportional to its length and inversely proportional to its cross-sectional area. The constant proportionality is the resistivity of the material.
2. The capacitance of a parallel plate capacitor is proportional to the area of one of its plates and inversely proportional to the separation between its plates. The constant of proportionality is the product of the dielectric constant, k, of the material between the plates and the electric permittivity, ε₀.
3. The current through a resistor is equal to the potential difference across the resistor divided by its resistance.
4. The magnitude of charge of one of the plates of a parallel plate capacitor is directly proportional to the product of the potential difference across the capacitor and the capacitance. The plates have equal amounts of charge of opposite sign.

4E5: The values of currents and electric potential differences in an electric circuit are determined by the properties and arrangement of individual circuit elements such as sources of emf, resistors or capacitors.

Learning Objectives
4E4.1: The student is able to make predictions about the properties of resistors and/or capacitors when placed in a simple circuit, based on the geometry of the circuit element and supported by scientific theories and mathematical relationships.
4E4.2: The student is able to design a plan for the collection of data to determine the effect of changing the geometry and/or materials on the resistance or capacitance of a circuit element and relate results to the basic properties of resistors and capacitors.
4E4.3: The student is able to analyze data to determine the effect of changing the geometry and/or materials on the resistance or capacitance of a circuit element and relate results to the basic properties of resistors and capacitors.
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 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.
4E5.3: 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.

Science Practices
2.2 The student can apply mathematical routines to quantities that describe natural phenomena.
2.3 The student can estimate numerically quantities that describe natural phenomena.
3.1 The student can pose scientific questions.
4.1 The student can justify the selection of the kind of data needed to answer a 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 and relationships.
5.2 The student can refine observations and measurements based on data analysis.
6.1 The student can justify claims with evidence.
6.2 The student can construct explanations of phenomena based on evidence produced through scientific practices.
6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models.
7.2 The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understanding and/or big ideas.

Answers to Prelab Questions

1. The units for the time constant, RC, are seconds. Derive this from the units of capacitance and resistance.

   The units for capacitance are farads, or C/V (coulombs/volt). The units for resistance are ohms, or V/A (volt/amperes). Amps are C/s (coulombs/second). So C/V x V/A = C/C/s = s.
   The ohm, the unit for resistance, is volts/amperes as described in Ohm’s law learned in AP^® Physics 1: V = IR (voltage = current x resistance)

2. A 250 μF capacitor is connected in series with a 10 k Ω resistor and a 9.00-V battery. What is the time constant?

   R x C = (10 k Ω) × (250 μF) = 10,000 Ω x (25 x 10-5 F) = 2.5 seconds

3. Two parallel metal plates are separated by a distance of 5 cm. There is a uniform electric field of 3 N/C between the plates.
   1. What is the voltage across the plates? Hint: Convert cm to m.

      V = E x d
      V = 3 N/C x 0.05 m
      V = 0.15 V

   2. How much charge is stored on each plate if the capacitance of the plates is 20 x 10^–6 F?

      q = C x V
      q = 20 x 10^–6 F x 0.15 V
      q = 3 x 10^–6 C

4. In a circuit containing a capacitor, raising or lowering the voltage does not change the value of the capacitance. Why? Justify your answer using Equation 2.

   The value of the capacitance of a capacitor depends only on the geometric and material properties of the capacitor. The capacitance in Equation 2 is a constant and changing the voltage would change the amount of charge stored on the plates of the capacitor accordingly. The ratio of C/V would remain the same, and therefore the value of capacitance would not change.

Sample Data

Introductory Activity

Analyze the Results

• Compare your observations to those of the opposite group (group A or B). How are the results different?

  The capacitance of the capacitors made by group B was higher than the capacitance of the capacitors made by group A.

• Did the reading change when the capacitor was squeezed? If yes, explain why using Equation 1.

  Yes, the capacitance increased when the capacitor was squeezed. This relates directly to the distance between the plates being reduced. Per Equation 1, if the distance between the plates is reduced, then the capacitance increases in value, which is confirmed by this observation.

• Qualitatively compare the charge stored on each capacitor.

  Student answers will vary. Almost all the capacitors in group B gave a higher capacitance than those in group A. The capacitors in group B that did not hold higher values appeared to not be as tightly rolled and that could have affected results. Likewise, comparing within groups, those capacitors that appeared more tightly rolled had higher capacitance values.

• If a 30" x 10" piece of paper was used with large aluminum foil pieces, how would you expect the capacitance to compare to those of groups A and B?

  The capacitance would be larger than both groups A and B because increasing the area of the plates will increase the capacitance as shown by Equation 1.

Guided-Inquiry Activity

Sample Procedure
Replace the 47 kΩ resistor with the unknown value resistor in the breadboard. Using the same technique as previously done with the first three resistors, run multiple trials to collect multiple data points for the time the LED remains lit. Analyze the Results

• In your own words, fully describe the motion of charge in the circuit when the capacitor is discharging.

  The electrons are the only free charged particles able to move in conductors. When the capacitor is fully charged, the bottom plate has excess electrons. When the capacitor is discharging, due to the electric field between the plates, electrons flow counterclockwise through the right hand circuit loop in Figure 7. They continue to flow counterclockwise to the top plate of the capacitor until the charges balance out and there is no longer a potential difference across the plates.

• What is the correlation between the calculated time constant and the amount of time the LED remained lit?

  The larger the time constant, the longer the LED remained lit.

• How does changing the resistor affect the motion of charge in the circuit? How does this affect the charging rate of the capacitor?

  Changing a resistor will change the current in the circuit. Current is defined as coulombs per second and is a measure of the rate of flow of charge in a circuit. Changing the resistor will change the rate at which the charges move. This will directly affect the charging rate of the capacitor; the higher the current, the faster the charging rate, and vice versa.

• Ranking the resistors in order of smallest to largest, where does the unknown resistor rank?

  The unknown resistor would rank as the third highest value resistor.

• How must the time constant in the circuit with the unknown resistor compare with the time constant in the circuits using the known resistors?

  The time constant for the circuit using the unknown resistor is the third highest among the four circuits tested.

Answers to Questions

Guided-Inquiry Activity

1. Breadboards are convenient devices used to build circuits without long and inconvenient cords, or the need to solder. The provided breadboard is separated by two sides: columns a, b, c, d and e on one side and columns f, g. h, i, j on the other side. Each row is electrically independent. However, each column per row is electrically connected. This means that pinholes 1a, 1b, 1c, 1d and 1e are electrically connected. The f–j columns are independent of the a–e columns. There is also a positive and negative (ground) column designated by red plus and blue minus signs, respectively. The entire positive column is connected and the entire negative column is connected. This means that if a positive input of 3 V is plugged into the top of the column, the bottom pinhole of the same column would also provide the same 3 V.
   1. Using the breadboard and materials available, set up the circuit in Figure 3 of the Background. Note: Do not connect the circuit to power. Connect the circuit to the power input and output columns without connecting an external power supply to the respective breadboard columns.
   2. Have your instructor confirm proper set up of the circuit.
2. Consider Figure 7.
   {14013_Data_Figure_7}
   1. Label the components in the circuit diagram above.
   2. Use the 620 ohm resistor to construct the circuit above without connecting to power using the 620 ohm resistor. Confirm proper set up with your instructor.
3. Using the multimeter, determine the exact resistance of the resistor. Note: Resistors have a tolerance range indicated by a gold or silver band. The gold band indicates that the resistor has a 5% tolerance, meaning the value would be within 5% of 620 Ω.

   Measured resistance of resistor is 611 Ω.

4. Calculate and record the RC value for this resistor-capacitor combination.

   R x C = 611 Ω x 1000 μF = 0.611 seconds

5. Close the circuit to the left (in regard to Figure 7) in order to complete the battery loop circuit. Leave this circuit closed for at least 20 seconds to ensure the capacitor is fully charged.
   1. In what direction is the current flowing?

      Clockwise.

   2. In what direction are the electrons flowing?

      Counterclockwise.

   3. Illustrate the accumulation of charge on the capacitor plates in Figure 7.

      The top plate is positively charged, the bottom plate is negatively charged.

Review Questions for AP^® Physics 2

1. Consider the following circuit.
   {14013_Answers_Figure_1}
   1. The time constant is 2 seconds. The resistor has a value of 1.0 k Ω. What is the value of the capacitor?

      2 s = R x C 2 s = 1000 Ω x C C = 0.002 F

2. Identify the direction of the current at the ammeter (up or down) in the circuits below.
   {14013_Answers_Figure_2}
   In a, the direction of the current at the ammeter is down. In b, the direction of the current at the ammeter is down. In c, the direction of the current at the ammeter us up. The direction of current is always follows the direction of positive charge, meaning from the positive terminal of a battery to the negative terminal. Or from the positive plate of a capacitor to the negative plate. At a the direction of the current at the ammeter is down. In b the direction of the current at the ammeter is down. In c the direction of the current at the ammeter us up.
3. Rank the time constant from smallest to largest in the following circuits.
   {14013_Answers_Figure_3}
   From smallest to largest: circuit c, circuit a, circuit d, circuit b. In circuit a, the time constant is R x C. In circuit b the resistors are in series and the overall resistance is R + R = 2R; the time constant is (R + R) x C or 2RC. In circuit c the resistors are in parallel so the overall resistance is 1/(1/R + 1/R) = R/2; the time constant is (R/2) x C or RC/2. In circuit d there is a resistor in series with resistors in parallel and the overall resistance is R + R/2 = 3R/2; the time constant is 3R/2 x C or 1.5RC.
4. Examine the following situation. Two parallel plates are connected to a battery, each plate with an area of 0.35 m², separated by a distance d = 0.05 m. Each plate has a hole at its center through which electrons can pass. After the plates are fully charged, energetic electrons emitted by an electron source enter the top plate with a speed v₀ = 4 x 10⁶ m/s, take 3 ns to travel between the plates, and leave the bottom plate with speed v_f = 7.6 x 10⁶m/s.
   {14013_Answers_Figure_4}
   1. Label the battery terminals, at the top and bottom of the battery, with positive and negative symbols. Support your answer with a reference to the direction of the electric field between the plates.

      The top terminal of the battery is the negative terminal and the bottom terminal is positive. Since the electron accelerates downwards towards the bottom plate (which is deduced from the increase in velocity), then the electric field between the plates points vertically upwards (electrons travel in a direction opposite the field due to their negative charge) from the positive bottom plate to the negative top plate.

   2. Recall that Ohm’s law is expressed with the equation E = F/q, where E = electric field, F = force and q = charge. Calculate the magnitude of the electric field between the plates.

      To find the magnitude of the electric field, the equation E = F/q is used.
      F = ma where a is (7.6 x 10⁶ – 4 x 10⁶m /s) / 3 x 10^–9 s = 1.2 x 10¹⁵ m/s².
      m is the mass of an electron, 9.1 x 10^–31 kg.
      So, F = 9.1 × 10^–31 kg × 1.2 × 10¹⁵m/s² = 1.092 x 10^–15 N.
      Here, q is the elementary charge, 1.602 x 10^–19 C.
      Therefore, E = 1.092 x 10^–15 N / 1.602 x 10^–19 = 6816 N/C.

   3. Calculate the voltage of the battery.

      The voltage across the fully charged plates is the same as the voltage across the battery.
      V = E x d where d = 0.05 m.
      V = 6,816 N/C x 0.05 m
      V = 340.8 V

   4. Calculate the magnitude of the charge on each plate.

      For a parallel plate capacitor the charge Q, on each plate = C x V. The capacitance is given by Equation 1.
      C = keA/d
      d = 0.05 m
      k = 1
      A = 0.35 m²
      e = 8.854 x 10^–12 F/m
      1(8.854 x 10^–12 F/m)(0.35 m²)/(0.05 m) = 6.195 x 10^–11F
      C = 6.195 x 10^–11 F

   5. If a dielectric was added between the plates, how would it affect the magnitude of charge on each plate?

      Adding a dielectric between the plates would increase the magnitude of charge on each plate. This can be shown by the direct correlation between capacitance and dielectric material value. The higher the dielectric constant, the higher the capacitance. In turn, a higher capacitance increases the amount of charge that can be stored on each plate. A dielectric would reduce the magnitude of the original electric field between the plates and therefore more charges would need to be stored on each plate in order for the potential difference across the plates to equal the potential difference of the connected battery.

   6. The plates are discharged and a resistor is added to the circuit. The plates are allowed to fully charge again. How does this affect the speed of the electrons leaving the bottom plate?

      This does not affect the speed of the electrons leaving the bottom plate. The resistor simply increases the amount of time needed for the plates to fully charge. Once fully charged, the electric field across the plates would have the same magnitude as in a scenario without a resistor.

References

AP^® Physics 1: Algebra-Based and Physics 2: Algebra-Based Curriculum Framework; The College Board: New York, NY, 2014.

Introduction

Capacitors are found nearly everywhere in electronics from camera-flash units to touchscreens and heart defibrillators. Investigate the electric field and geometric properties that allow capacitors to store electric potential energy.

Concepts

• Capacitance
• Circuit diagrams
• RC circuits
• Solderless breadboards
• Time constant

Background

Capacitors are circuit components designed to store electric energy. They function similarly to rechargeable batteries; once a voltage has been supplied, they will hold that voltage until a complete circuit loop is made, at which point the store charge is released. Capacitance is the measure of the stored charge in a capacitor.

A capacitor is able to store charge because of two parallel metal plates separated by an insulator called a dielectric. The dielectric can be any sort of insulating material such as glass, wool, rubber, ceramic or even air. When the two parallel conducting plates are connected to a battery (see Figure 1), electrons flow from the negative terminal of the battery towards the positive terminal; this is the electric current.

When the electrons encounter the first plate, they accumulate and cause a negative charge. The accumulation of negative charge on the first plate causes electrons to move away from the second plate, which gives the second plate a positive charge. This creates an electric field between the two plates (see Figure 2). Once the voltage of the electric field equals that of the battery, the capacitor is fully charged. At this point, the electrons are equally attracted to the positively-charged plate as they are to the positive terminal of the battery. When this happens, the current drops to zero.

The amount of charge a capacitor can store is dependent on the voltage of the battery and the set capacitance of the capacitor. The only way to alter the capacitance of a capacitor is to change its physical properties by altering the dielectric material in the capacitor, the surface area of the plates or the distance between the plates. The capacitance of a parallel plate capacitor can be expressed with the following equation: where

C = capacitance
k dielectric constant
ε₀ = permittivity of free space
A = area of one of the plates
d = separation distance between plates

C is measured in farads, F (1 farad = 1 coulomb per volt), A is measured in m², and d in meters. The permittivity of free space is a proportionality constant used to specify the strength of an electric field in a vacuum or empty space and has a value of 8.854 x 10^–12 F/m. The units for the permittivity of free space constant is Farads per meter.

Capacitance is determined by the physical geometry of the capacitor itself. The amount of charge in a capacitor at a specific voltage can be determined by where

q = charge stored on one plate
C = capacitance
V = applied voltage from the battery

Three variables can affect the charge capacity of a capacitor. The first is surface area. If a capacitor has plates with larger surface areas, then there is more space available for charge to be stored, so the capacitance is higher. The second variable is the distance between the plates. Consider a situation with two capacitors with the same surface area and the same charge on each plate. In this case, the capacitor with the plates closest to each other will have the greatest capacitance value. Recall that: where

V = voltage
E = the magnitude of the electric field between the plates
d distance between the plates

If ΔV is held constant, then a larger distance between the plates leads to a smaller magnitude of the electric field between the plates. The magnitude of E between the plates is directly proportional to the amount of charge on the plates. Therefore, the larger the separation between the plates, the smaller the magnitude of electric charge on the plates.

The third variable is the type of dielectric material. Dielectric material creates an electric field that opposes the field created by the plates, which enhances a capacitor by allowing it to accumulate more charge. The electric field in the dielectric material is created when polar molecules in the dielectric material align themselves with the external electric field provided by the plates: The positive side of the molecules face the negative plate, and the negative side of the molecules face the positive plate. This alignment of dipoles creates the electric field in the dielectric material to oppose the electric field created by the plates. This means that more charge needs to be accumulated on the plates in order to completely oppose the voltage from the battery. If the capacitor in Figure 3 is allowed to fully charge, and then the battery is removed from the circuit, the capacitor will discharge. In fact, the capacitor acts like a battery and the flow of current will be in the opposite direction as to what was provided by the battery. The time it takes for the capacitor to fully discharge depends on the resistance of the circuit and the capacitance of the circuit (R x C). Within a specific circuit, RC is constant throughout that circuit. This is referred to as the time constant (τ). where

R resistance in circuit
C = capacitance in circuit
τ = time constant

Experiment Overview

The objective of this advanced inquiry lab is to form an understanding of how capacitors work and what properties affect their function. First a capacitor will be constructed from simple materials in an introductory activity. A capacitance meter will be used to evaluate how different variables in the composition of a capacitor affect capacitance. The guided-inquiry activity uses the capacitor from the introduction activity along with known capacitors in a circuit. The circuit contains an LED to allow for a qualitative measurement of the time constant of an RC circuit.

Materials

Prelab Questions

1. The units for the time constant, RC, are seconds. Derive this from the units of capacitance and resistance.
2. A 250 μF capacitor is connected in series with a 10 k Ω resistor and a 9-V battery. What is the time constant?
3. Two parallel metal plates are separated by a distance of 5 cm. There is a uniform electric field of 3 N/C between the plates.
   1. What is the voltage across the plates? Hint: Convert cm to m.
   2. How much charge is stored on each plate if the capacitance of the plates is 20 x 10^–6 F?
4. In a circuit containing a capacitor, raising or lowering the voltage does not change the value of the capacitance. Why? Justify your answer using Equation 2.

Safety Precautions

Although 9-V batteries and power supplies are not powerful enough to deliver an electric shock under normal conditions, please follow all proper safety precautions when working with electronics. Check completed circuits with your instructor to ensure that you have not created a short circuit. Never allow battery terminals to touch or short circuit, as this may create a fire hazard. Never handle batteries or electrical components with wet hands. Please follow all laboratory safety guidelines.

Procedure

Introductory Activity

1. If in group A, cut a piece of paper to be a 6" x 3". If in group B, cut a piece of paper to be 9" x 4".
2. Fold the paper into thirds along its length (see Figure 4).
   {14013_Procedure_Figure_4}
3. Cut two pieces of aluminum foil so they fit within one of the end sections, leaving 5 mm between the edge of the foil and the paper (see Figure 4).
4. Use three pieces of tape to attach one of the foil pieces to the section on the left. Do not tape the top side of the foil (see Figure 4).
5. Loop a 1" piece of tape and place toward the top of the center section (see Figure 4).
6. Place the wire on the tape so that about half of the wire’s length is off the paper (see Figure 4).
7. Fold the first section over the center section so the aluminum foil piece makes contact with the wire.
8. Flip over and tape the second piece of foil on the left side of the folded piece of paper. See Figure 5.
9. Loop a 1" piece of tape and place on the remaining section of paper (see Figure 5).
   {14013_Procedure_Figure_5}
10. Place the wire on the tape so that about half of the wire’s length is off the paper (see Figure 5). (Each wire should protrude from the paper on opposite sides).
11. Fold the paper so that the wire and aluminum foil make a connection.
12. Roll the paper—the tighter the roll, the better the connection between wire and aluminum foil (see Figure 6).
    {14013_Procedure_Figure_6}
13. Tape the capacitor shut so that it does not unravel.
14. Set the multimeter to measure capacitance. Connect the multimeter leads to the capacitor and record the reading.
15. While connected to the multimeter, hold and squeeze the capacitor. Check the reading and record your observations.

Analyze the Results

• Compare your observations to those of the opposite group (group A or B). How are the results different?
• Did the reading change when the capacitor was squeezed? If yes, explain why using Equation 1.
• Qualitatively compare the charge stored on each capacitor.
• If a 30" x 10" piece of paper was used with large aluminum foil pieces, how would you expect the capacitance to compare to those of groups A and B?

Guided-Inquiry Design and Procedure

1. Breadboards are convenient devices used to build circuits without long and inconvenient cords, or the need to solder. The provided breadboard is separated by two sides: columns a, b, c, d and e on one side and columns f, g. h, i, j on the other side. Each row is electrically independent. However, each column per row is electrically connected. This means that pinholes 1a, 1b, 1c, 1d and 1e are electrically connected. The f–j columns are independent of the a–e columns. There is also a positive and negative (ground) column designated by red plus and blue minus signs, respectively. The entire positive column is connected and the entire negative column is connected. This means that if a positive input of 3 V is plugged into the top of the column, the bottom pinhole of the same column would also provide the same 3 V.
   1. Using the breadboard and materials available, set up the circuit in Figure 3 of the Background. Note: Do not connect the circuit to power. Connect the circuit to the power input and output columns without connecting an external power supply to the respective breadboard columns.
   2. Have your instructor confirm proper set up of the circuit.
2. Consider Figure 7.
   {14013_Procedure_Figure_7}
   1. Label the components in the circuit diagram.
   2. Use the 620 ohm resistor to construct the circuit above without connecting to power. Confirm proper set up with your instructor.
3. Using the multimeter, determine the exact resistance of the resistor. Note: Resistors have a tolerance range indicated by a gold or silver band. The gold band indicates that the resistor has a 5% tolerance, meaning the value would be within 5% of 620 Ω.
4. Calculate and record the RC value for this resistor-capacitor combination.
5. Close the circuit to the left (in regard to Figure 7) in order to complete the battery loop circuit. Leave this circuit closed for at least 20 seconds to ensure the capacitor is fully charged.
   1. In what direction is the current flowing?
   2. In what direction are the electrons flowing?
   3. Illustrate the accumulation of charge on the capacitor plates in Figure 5.
6. Obtain a timer. One person should be prepared to flip the switch to the LED loop and another to measure the amount of time from when the switch is closed to when the LED flickers out.
7. Close the switch toward the right, in order to complete the LED loop. Time how long the LED remains lit. Record all observations.
8. Once the LED goes out, leave the circuit closed for at least 30 seconds to ensure the capacitor is sufficiently discharged.
9. Repeat steps 5–8 at least twice more to average the results.
10. Repeat steps 3–9 for the 1.1 kΩ resistor. Note: Turn off the power supply when changing resistors.
11. Repeat steps 3–9 for the 47 kΩ resistor. Note: Turn off the power supply when changing resistors.
12. Using available materials and techniques, and without the aid of a multimeter, determine and carry out a procedure to find the relative resistance of the remaining unknown resistor.

Analyze the Results

• In your own words, fully describe the motion of charge in the circuit when the capacitor is discharging.
• What is the correlation between the calculated time constant and the amount of time the LED remained lit?
• How does changing the resistor affect the motion of charge in the circuit? How does this affect the charging rate of the capacitor?
• Ranking the resistors in order of smallest to largest, where does the unknown resistor rank?
• How must the time constant in the circuit with the unknown resistor compare with the time constant in the circuits using the known resistors?

Student Worksheet PDF

14013_Student1.pdf