Lab 6 - Resistors (1)

pdf

School

Florida International University *

*We aren’t endorsed by this school

Course

MISC

Subject

Electrical Engineering

Date

Apr 3, 2024

Type

pdf

Pages

Uploaded by MinisterMandrillMaster120

R1 R2 R1 Name Date RESISTOR SIMULATION Introduction Resistors are passive electronic devices which have fixed values. The resistance follows Ohm's Law I = 1 V R The SI unit of resistance is the ohm, 1 Ω = 1 V/A. The schematic symbol for a resistor is Resistors can be connected in an electronic circuit in a series or parallel combination. Series: In a series connection the components are connected at a single point, end to end as shown below: For a series combination the current through each resistor is the same while the voltage drops will add. Using this we can find the equivalent capacitance, R eq R EQ = R 1 + R 2 Parallel: In the parallel connection, the components are connected together at both ends as shown below: R2 For a parallel combination the current through each resistor will add while the voltage drops are equal. Using this we can find the equivalent capacitance, R eq 1 R EQ = 1 R 1 + 1 R 2

1 – Ohms Law To gain some familiarity with Ohm’s law open the simulation (https://phet.colorado.edu/sims/html/ohms-law/latest/ohms-law_en.html). The simulation allows you change the voltage or resistance and see how this effects the current using the sliders on the page. 1. How does the current change as you increase the voltage? The current also increases. 2. How does the current change as you decrease the resistance? The current increases as the resistance decreases. 3. With V = 3.5 V and R = 703 Ω what is the current? (mA) (m-milli 10 -3 ) The current is 5.0 mA 2 – Current vs Voltage Ohm’s Law written as I = V/R states that current goes linearly with the voltage. 4. What does the slope m of the line correspond to? The slope of the line corresponds to 1/R Suppose a resistor is connected to a power supply. You vary the voltage and measure the current passing through the resistor. The table below displays the data: the voltage in V and the current in mA. V 1 2 3 4 5 6 7 8 9 10 I 0.952 2.155 2.838 4.382 5.142 6.629 7.478 8.125 9.051 11.0 Make a plot of I vs V (Remember to set the y intercept to zero). 5. Fitting the data with a line what is the value of slope? (1/Ω) The equation of the line is y = 0.1819x. The value of the slope is m = 0.1819 Ω -1 6. What is the resistance of the resistor? (Ω) The resistance of the resistor is 5.5 Ω

3 – Joule Heating When current passes through a resistive element (for example a light bulb) power is dissipated (the light bulbs emits light). The power dissipated is P=IV . Using Ohm’s law this can be rewritten in two other ways P = I 2 R = V 2 /R. Open the link (https://phet.colorado.edu/en/simulation/legacy/battery-resistor-circuit) and download/run the applet or use https://phet.colorado.edu/sims/cheerpj/battery-resistor-circuit/latest/battery-resistor-circuit.html. The simulation shows a resistor and battery. The process of dissipate power causes the restive element to heat up (the light bulbs get hot); that’s the name Joule heating. There is ammeter which shows the current and a bar displays how hot the resistor gets. The controls on the right allow you to change the voltage of the battery and the value of the resistance. 7. How does the temperature of the resistor change as you increase the voltage? The increase in power leads to an increase in temperature, causing the resistor to heat up. Therefore, as you increase the voltage, the temperature of the resistor is expected to rise. 8. How does the temperature of the resistor change as you increase the resistance? The temperature of the resistor increases. When current flows through a resistor, it generates heat due to the resistive properties of the material. With increased resistance and reduced current, the same amount of power is dissipated across the resistor (P = VI). The higher voltage drop across the resistor leads to increased heating. Set the resistance to R = 0.53Ω and the voltage of the battery to V = 12V. 9. What is the power being dissipated through the resistor? (J/s) 271.7 J/s 10. What is the current through the resistor? (A) 22.64 A

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

4 – Series and Parallel Combinations Open the link ( https://phet.colorado.edu/en/simulation/legacy/circuit-construction-kit-ac ) and download/run the simulation. Or use http://phet.colorado.edu/sims/html/circuit-construction-kit-ac/latest/circuit-construction-kit-ac_en.html You will see a blank canvas Click the schematic button on the right hand side. This simulation will let you construct circuits of different kinds (become familiar with this simulation since you will use it again several times). The circuit elements are displayed in bar on the right hand side and can be dragged into the blue canvas. Once you have a circuit element in the canvas it can be modified by right clicking on the object. This will let you change the value of voltage of the batteries or resistance of the resistors. Click the voltmeter button. This will give you a meter to measure the voltage across elements in your circuit. To do this move the two probes (1 red 1 black) to touch each side of the wire connecting to a circuit element, the voltmeter will display the voltage across that object. An example of this is shown in the circuit diagram below with a measurement being conducted of the voltage across R 2 . Construct the following circuit in the simulation. Choose the following: R 1 = 25Ω, R 2 = 50Ω, V = 5V 11. What is the voltage measured across R 1 ? (V) V1 = IR1 = (1/15)(25) = (5/3)V = 1.67 V 12. What is the voltage measured across R 2 ? (V) V2 = IR2 = (1/15)(50) = (10/3)V = 3.33 V 13. Using Ohm’s law what is the current through R 1 ? (A) R1 = I = 1/15 A 14. What is the current through R 2 ? (A) R2 = I = 1/15A 15. What is the equivalent resistance R eq in this case? (Ω) Req = 25 Ω + 50 Ω = 75 Ω 16. What is the current flowing through the equivalent resistor? (A) I = 1/15 A

Construct the following circuit in the simulation by adding a third resistor in parallel with resistor 2. Choose the following:R 1 = 25Ω, R 2 = 50Ω, R 3 = 100Ω, V = 5V 17. What is the voltage measured across R 1 ? (V) R23 = (50 x 100)/(50+100) = 33.333 ; I = 5/(25+33.333) = 0.0857 A; V1 = I1R1 = 0.0857 x 25 = 2.143 V 18. What is the voltage measured across R 2 ? (V) V2 = 5 – 2.143 = 2.857 V 19. What is the voltage measured across R 3 ? (V) Since R2 and R3 are parallel voltage V2 = V3 = 2.857 V 20. Using Ohm’s law what is the current flowing through R 1 ? (A) I = 0.0857 A. Hence, current flow through R1 = 0.0857 A 21. What is the current through R 2 ? (A) I2 = V2/R2 = 2.857/50 = 0.0571 A 22. What is the current through R 3 ? (A) I3 = V3/R3 = 2.857/100 = 0.02857 A 23. What is the equivalent resistance R 23 of the parallel combination of R 2 and R 3 ? (Ω) R23 = (50 x 100)/(50+100) = 33.333 Ω 24. What is the equivalent resistance R eq of the whole system? i.e. the series combination of R 1 with R 23 ? (Ω) Req = R1 + R23 = 25 + 33.333 = 58.333 Ω 25. What is the current flowing through the equivalent resistor R 23 ? (A) I23 = 5/(25+33.333) = 0.0857 A 26. What is the current flowing through the equivalent resistance R eq in this case? i.e. the total current? (A) I23 = 0.0857 A

Lab 6 - Resistors (1)

Related Documents