PES2150_Salberg_KirchhoffPrelab

docx

School

University of Colorado, Colorado Springs *

*We aren’t endorsed by this school

Course

2150

Subject

Electrical Engineering

Date

Apr 3, 2024

Type

docx

Pages

Uploaded by SuperHumanJayPerson915

UNIVERSITY OF COLORADO – COLORADO SPRINGS Kirchhoff’s Laws Name: Abigail Salberg PES 2150 Prelab Questions 1.) Explain how to measure the following quantities with a DMM. a.) Voltage Touch the DMM’s leads to either side of the component, in parallel. Set the DMM to voltage mode and input the appropriate range. b.) Current Put the DMM in series with the component, the current must flow through the DMM as well as the component. Set the DMM to current mode and input the appropriate range. c.) Resistance Remove the resistor from the circuit and touch the DMM’s leads to either end. Set the DMM to resistance mode and input the appropriate range. Note: For Prelab questions 2, 3, and 4, go to the website: https://phet.colorado.edu/sims/html/circuit-construction-kit-dc-virtual-lab/latest/circuit- construction-kit-dc-virtual-lab_en.html Open the Circuit Construction Kit simulation.

P E S 2 1 5 0 - P H Y S I C S L A B O R A T O R Y I I 2.) Once the Circuit Construction Kit program has loaded, begin constructing the circuit shown below: Use the ‘Schematic’ visual setting to make sure your circuit matches the diagram above. When the circuit is set up and running, answer the following questions: a.) Why do the electrons in the program flow in the opposite direction as the ‘conventional’ current indicated in the figure above? Because electrons will flow towards the + side due their – charge. Conventional direction of flow if from + to -, so this is why they are opposite . b.) Right click on the battery and resistor to set them to exactly E = 10 V and R = 15 Ω respectively (Use the ‘Show Values’ check box to make sure of this). Then, use the voltmeter and ammeter to complete the following table. (Hint: How should you measure voltage and current with these meters?) The current here is slower, the voltage is at 10.00V, and the current is 0.67A. c.) Repeat Part (b) with the battery set at 10 V and the resistor at 5 Ω . What happens to the current here compared to Part (b)? Explain. The current here is much faster than in Part b. This is because there is less resistance to their movement. Battery Voltage Voltage Across Current I Resistance R Ohm’s Law Kirchhoff Prelab - 2 Schematic View ‘Lifelike’ View

P E S 2 1 5 0 - P H Y S I C S L A B O R A T O R Y I I E ( V ) Resistor V ( V ) ( A ) ( Ω ) R = V / I ( Ω ) 10 10.00 0.67 15 15 5 10.00 2.00 5 5 3.) Again, using the Circuit Construction Kit you used in the previous question, build the following circuit. The battery voltage should be exactly E = 10 V and the resistances should be exactly 15 Ω and 5 Ω respectively. a.) Once the circuit is set up, look at the flow of electrons in the program. Qualitatively, how does the current through the 15 Ω resistor compared to the current through the 5 Ω resistor? Why is this so? The current through the 5 Ω resistor is greater and faster than the 15 Ω resistor. This is because there is less resistance in the 5 Ω resistor, and therefore the electrons can flow more freely. b.) Using the Circuit Construction Kit program, measure the following quantities: Measured Current through the 15 Ω Resistor ( I 1 ) 0.67 A Measured Current through the 5 Ω Resistor ( I 2 ) 2.00 A Measured Voltage across the 15 Ω Resistor ( V ) 10.00 V Measured Voltage across the 5 Ω Resistor ( V ) 10.00 V Kirchhoff Prelab - 3 R = V/I R = 10.00/0.67 R = 15

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

P E S 2 1 5 0 - P H Y S I C S L A B O R A T O R Y I I c.) Compute the equivalent resistance of this circuit from: R = V I total where I total is the sum of the currents through each resistor. d.) Compute the equivalent resistance of this circuit from: 1 R = 1 R 1 + 1 R 2 and compare your answer here to your answer in Part (c). This has a percent different of 1.34%. 4.) Again, using the Circuit Construction Kit you used in the previous question, build the following circuit. The battery voltage should be exactly E = 10 V and the resistances should be exactly 15 Ω and 5 Ω respectively. a.) Once the circuit is set up, look at the flow of electrons in the program. Kirchhoff Prelab - 4 R = V/I total R = 10.00/2.67 R = 3.75 1/R = 1/15+1/5 1/R = 0.27 1 = 0.27R R = 3.70 R eq = 3.75 R eq = 3.70

P E S 2 1 5 0 - P H Y S I C S L A B O R A T O R Y I I Qualitatively, how does the current through the 15 Ω resistor compared to the current through the 5 Ω resistor? Why is this so? The current is the same. This is because they are in line with each other. b.) Using the Circuit Construction Kit program, measure the following quantities: Measured Current through the 15 Ω Resistor ( I ) 0.50 A Measured Current through the 5 Ω Resistor ( I ) 0.50 A Measured Voltage across the 15 Ω Resistor ( V 1 ) 7.50 V Measured Voltage across the 5 Ω Resistor ( V 2 ) 2.50 V c.) Compute the equivalent resistance of this circuit from: R = V total I where V total is the sum of the voltages across each resistor. d.) Compute the equivalent resistance of this circuit from: R = R 1 + R 2 and compare your answer here to your answer in Part (c). These have a percent different of 0%. Kirchhoff Prelab - 5 R = V total /I R = 10/0.50 R = 20 R = R1+R2 R = 15+5 R = 20 R eq = 20 R eq = 20

P E S 2 1 5 0 - P H Y S I C S L A B O R A T O R Y I I Kirchhoff Prelab - 6

Your preview ends here

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