Lab_5

doc

School

Bergen Community College *

*We aren’t endorsed by this school

Course

PHY 222

Subject

Physics

Date

Apr 3, 2024

Type

doc

Pages

Uploaded by MagistrateFlag4534

Lab Assignment 5: Capacitance and Ohm’s Law Instructor’s Overview This set of lab experiments focuses on capacitance and Ohm’s Law. These concepts are used in the design of virtually every industrial and consumer electronic system or device. This lab is based on Labs 21 and 22 of your eScience Lab kit. We will be performing all of Lab 21 and Experiment 1 of Lab 22. Our lab consists of two main components. Here is a high-level view:  In Lab 21, you will be building your own circuits using the Snap Circuit kit and analyzing them with a digital multimeter. The focus of Lab 21 will be assessing the performance of RC circuits that include a light emitting diode (LED).  In Lab 22 Experiment 1, you will be performing an experiment to directly demonstrate Ohm’s Law. Take detailed notes as you perform the experiment and fill out the sections below. This document serves as your lab report. Please include detailed descriptions of your experimental methods and observations. What you’ll need for this lab activity Experiment Tips/Comments :  We will be using the digital multimeter for both lab activities. To set up the multimeter, plug in the black probe into COM port of the meter and plug the red probe into the V  mA port of the meter. For the capacitance lab, set the dial to the 20 setting in the DCV (direct current and voltage) section of the multimeter. JWH 1 Physics II Physics Physics II II From the eScience kit  Snap Circuit kit  Batteries  Stopwatch  Digital multimeter  Alligator clips  Masking tape  Permanent marker  Ruler

Date : 03/08/24 Student : Anthony Putrino Abstract Current is established in a circuit when electrons flow through it, and the potential difference is the energy required for current to move from one point to another in the circuit. The mounting of capacitors determines both the capacitance and the potential difference. In the following laboratory we will study the capacitors in series and in parallel, thanks to a pressure circuit kit, which contains 2 capacitors of different capacitance, and a resistance, showing that the capacitors in parallel have a higher equivalent capacitance (because is the sum of all the capacitors present) than the series capacitors. This was evidenced by the fact that the intensity and the LED turn-off time is greater for a circuit with capacitors in parallel than in a circuit with capacitors in series. Ohm's law and its linear relationship will also be studied, showing that as the voltage increases, adding more batteries to the circuit, their current also increases. Introduction Capacitors are made up of two conductors or armatures, generally in the form of plates or sheets separated by a dielectric material (materials that do not conduct electricity) that, subjected to a potential difference, acquire a certain electrical charge (due to this storage property of charge is called capacity or capacitance). The relationship between the accumulated electrical charges and the voltage on the capacitor is a constantly called capacity, the analysis of capacitance is highly important for the construction of a capacitor that is very useful when in the presence of charges. Ohm's law establishes a relationship between the potential or voltage difference (V), the current intensity (I) in ampere, and the resistance (R), which unites in its main postulate: “The flow of current that circulates through a closed electrical circuit is directly proportional to the voltage between the ends of a conductor and inversely proportional to the resistance of the same conductor Material and Methods  Snap Circuit kit  Batteries  Stopwatch  Digital multimeter  Alligator clips  Masking tape  Permanent marker  Ruler JWH 2 Physics II

Lab 21 Experiment 1: Capacitors in Series and Parallel Results/Observations It was observed that depending on the configuration of the capacitors, that is if they were in series or in parallel, the duration of the LED on-time varied. For example, in the column titled Capacitance 4, we can see that the average for which the LED was on was the longest time, followed by the column titled Capacitance 1, the second-best average time for which the LED was longer was observed. time on. Finally, the columns capacitance 3 and capacitance 2 were the lowest times, respectively, for which the LED was on for a very short time Table 1: Snap Circuits Capacitance Data Trial Capacitance 1 Time (s) Capacitance 2 Time (s) Capacitance 3 Time (s) Capacitance 4 Time (s) 1 1.98 0.54 0.63 2.27 2 2.00 0.64 0.78 2.18 3 1.99 0.70 0.73 2.24 4 1.96 0.59 0.70 2.40 5 2.10 0.56 0.67 2.17 6 2.06 0.69 0.62 2.35 7 2.18 0.56 0.66 2.19 8 2.30 0.55 0.70 2.25 9 1.98 0.67 0.75 2.33 10 1.97 0.65 0.64 2.37 AVERAGE 2.05 0.62 0.69 2.28 JWH 3 Physics II

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Create diagrams of each capacitor circuit The first circuit diagram is provided as an example. Please complete the other three. Circuit from Part 1 (single capacitor) Circuit from Part 1 (two capacitors) Circuit from Part 2 (single capacitor) Circuit from Part 2 (two capacitors) JWH 4 Physics II

Digital multimeter measurements of the capacitor circuits Use your digital multimeter to make the following voltage measurements: Circuit from Part 1 (single capacitor) Element Voltage drop (V) Battery pack (+ to -) 3V 1 k  resistor 1.32V LED 1.69V From + of LED to – of battery 1.69V 470  F capacitor 3V Add the voltage drops of the resistor and the LED. How does this sum compare with the source voltage? How does it compare with the voltage drop across the capacitor? Explain your answers. Answer The LED behaves like a Resistor, this means that when adding the voltage across the Resistor (1.32 V) and the Led (1.69 V) it is equal to the voltage supplied by the battery (3V). Also, the sum of the voltage drops of the resistor and the LED is equal to the voltage drop across the capacitor. The small differences are due to factors such as dissipated heat, a product of electrical energy flowing through the components present in the circuit. Circuit from Part 1 (two capacitors) Element Voltage drop (V) 1 k  resistor 1.32V JWH 5 Physics II

LED 1.69V 470  F capacitor 0V 100  F capacitor 3V Add the voltage drops across the two capacitors. Comment on this sum. Are the capacitors in series or parallel? How can you tell from your voltage measurements? Answer The capacitors are in series, therefore the sum of the voltage drops across both capacitors must be equal to the voltage across the battery (3V) Circuit from Part 2 (single capacitor) Element Voltage drop (V) 1 k  resistor 1.32V LED 1.69V 100  F capacitor 3V Element Voltage drop (V) Battery pack (+ to -) 3V 1 k  resistor 1.32V LED 1.69V From + of LED to – of battery 1.69V 470  F capacitor 3V How does this circuit compare with the initial circuit from part 1? Answer It behaves in the same way as the circuit in part 1, since it contains the same potential difference, thus verifying Ohm's Law, which tells us that voltage is directly proportional to the product of current and resistance. Circuit from Part 2 (two capacitors) Element Voltage drop (V) 1 k  resistor 1.32V LED 1.69V JWH 6 Physics II

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

470  F capacitor 3V 100  F capacitor 3V Are the two capacitors in series or parallel? How can you prove this with your voltage measurements? Answer The two capacitors are in parallel, and it can be tested because the voltage measured in each of the capacitors is equal (3V) JWH 7 Physics II

Lab 22 Experiment 1: Ohm’s Law Results/Observations Ohm's law could be verified since, thanks to the measurements, its linear dependence between voltage and current was observed. The following table shows us how as the voltage increases, due to the addition of more batteries to the circuit, the current also increases, thus verifying Ohm's Law Voltage Readings of Different AA Battery Combinations Battery Combination Voltage (V) Current (A) 1 1.5 0.00154 1 and 2 3.1 0.00308 1, 2, and 3 4.6 0.00473 1, 2, 3, and 4 6.1 0.00634 Resistor Data Resistance (  ) Slope (V/A) Percent difference 999 Ω 952.9 Ω 4.6 % JWH 8 Physics II

Lab 21 Experiment 1: Capacitors in Series and Parallel – Analysis and Discussion Based on your experimental results, please answer the following questions: 1. How did the behavior of the LED differ when you added the second capacitor in Part 1? Answer Before adding the second capacitor, a dimmer light was observed, and after adding the second capacitor the LED turned off faster. 2. After adding the 100 μF capacitor in Part 1, are the capacitors in series or in parallel? Use data to support your answer. Calculate the equivalent capacitance. Show all of your work. Answer The capacitors are connected in series, yielding an equivalent capacitance that has less capacity available to store charge, for this reason, it turned off faster. The equivalent capacitance is obtained as follows: 3. How did the behavior of the LED differ when you added the second capacitor in Part 2? Give an explanation of this behavior. Answer Due to the connection of the capacitors in parallel, the circuit has an equivalent capacitance that is equal to the sum of all the capacitances present, therefore it has a greater capacity to store charge, which causes the LED to take longer to turn off. , which in part 1. 4. After adding the 470 μF capacitor in Part 2, are the capacitors in series or in parallel? Use data to support your answer. Answer The capacitors are connected in parallel, and this is evidenced by the longer amount of time the LED was on. The equivalent capacitance is as follows: JWH 9 Physics II

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

It is clearly observed that the equivalent capacitance is larger, therefore it has a greater capacity to store charge. 5. The longer the light bulb is lit, the larger the capacitance. Calculate the capacitance for each circuit arrangement. Show all work. Rank your capacitances in order from greatest to least based off of the average time the light bulb was lit. Does your ranking agree with your calculations? Answer  For part 1 the equivalent capacitance is 470μF since it only has a 470μF capacitor  For part 2, there are the two capacitors connected in series, obtaining an equivalent capacitance of 82.5μF  For part 3, the circuit consists of a single capacitor, therefore the equivalent capacitance is 100μF  For part 4, the two capacitors are connected in parallel, and the equivalent capacitance is equal to the sum of the capacitors, obtaining an equivalent capacitance of 570μF The calculated capacitances coincide with the average times of the LED on, that is, its classification from highest to lowest is Part 4 > Part 1 > Part 3 > Part 2 JWH 10 Physics II

Lab 22 Experiment 1: Ohm’s Law – Analysis and Discussion Based on your experimental results, please answer the following questions: 1. Using an application such as Microsoft Excel, create a graph of the Voltage Readings of Different AA Battery Combinations table. Insert the plot below. [X-axis = current, Y-axis = voltage] Answer 2. What is the relationship between voltage and current? Use data from your graph to support your answer. Answer The relationship between voltage and current is linear, and it can be observed in the graph that as the voltage increases, the current also increases, being I am consistent with Ohm's law, which describes the proportional relationship between voltage and stream 3. Execute a linear regression fit in Excel to calculate the slope of the line and record it in the Resistor Data table. Answer The slope represents the Resistance that is 952.9Ω JWH 11 Physics II

4. How does the value of the slope compare to the resistance you measured? Calculate the percent difference. Answer The value of the measured resistance was 999Ω while the value of the experimental resistance is 952.9Ω To calculate the percentage difference we have: 5. Use the results of your experiment to verify Ohm’s Law. Answer Using the results from the table, we can verify Ohm's law, since: JWH 12 Physics II

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Conclusions Overall it was clear that we were able to verify the objectives set for this lab, which were to observe the effects of the capacitors, either in series or in parallel, and Ohm's law. In a series configuration, having a lower capacity to store charge, the LED was lit for less time and with a less bright light, while for the parallel configuration, the opposite occurred, since having a greater capacity to store charge, the LED was on for longer and brighter light. It was possible to verify Ohm's law that establishes the linear relationship between three physical quantities such as current, voltage, and resistance. The only errors observed were human errors since for example a timer operated by the student was used, and the reaction time was maybe not the most accurate, however the results were good. References Walker, James S. (2016). Physics. Fifth ed. San Francisco, CA: Pearson Addison Wesley, Pearson Education, Inc. JWH 13 Physics II

Lab_5

Related Documents