Resistance phys 4b formal lab report

pdf

School

University of California, Berkeley *

*We aren’t endorsed by this school

Course

Subject

Electrical Engineering

Date

Apr 3, 2024

Type

pdf

Pages

Uploaded by KidSkunk3605

1 Resistance Lab Report By: Tong Chen, Hector Venegas, Favian Loza Professor Paul Haitkin PHYS 4B 40774 3 May, 2022

2 Statement of purpose: Part 1: In part 1, we were learning how to measure the current and voltage through certain material. We used the multimeters to measure current and voltage of each resistor. The resistance of the coil was found by Since we already know the 𝑅 = 𝑉 𝐼 . data of each resistance coil, we can get their theoretical resistance through the formula . 𝑅 = ρ𝑙 𝐴 ...... 1 ...... 2

3 Part 2: Measuring the resistance of a pair of wires using the multimeter. Designate one nichrome resistor to be R1, use the multimeter to measure the resistance of R1 and remember to subtract the resistance of the wires. Change the battery voltage to 0,0.75,1.5,2.25 and 3 and record the voltage of R1 and compare those two sets of datas. Repeat the previous steps for R2 which is another nichrome. Explore the relationship between the battery voltage and resistance voltage. Data Table for Part 1: *Coil #1 - #6 : Nickel-Silver *Coil #7: Copper V (voltage) I (mA) R (Ω) ρ(Rho) Coil #1 .21 +/- 0.05 54.5 +/- 0.5 3.86 +/- 0.0009 4.88 * 10 −5 Coil #2 .39 +/- 0.05 49.2 +/- 0.5 7.92 +/- 0.001 5.02 * 10 −5 Coil #3 .54 +/- 0.05 44.8 +/- 0.5 12.05 +/- 0.001 5.09 * 10 −5 Coil #4 .66 +/- 0.05 41.8 +/- 0.5 15.8 +/- 0.001 5.0 * 10 −5 Coil #5 .750 +/- 0.05 37.7 +/- 0.5 19.9 +/- 0.001 5.04 * 10 −5 Coil #6 .506 +/- 0.05 42.8 +/- 0.5 11.8 +/- 0.001 4.75 * 10 −5

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 Coil #7 .344 +/- 0.05 50.0 +/- 0.5 6.88 +/- 0.001 1.74 * 10 −6 Avg: 𝑐𝑜𝑖𝑙 #1−#6 6 = 2.987 * 10 −4 6 = 4. 96 * 10 −5 Analysis part 1: While testing various resistors, throughout the first part, we were able to see how different variables that result in resistance affect the current of the system and the amount of voltage passing through a resistor. In the first five trials, a material of Nickel-Silver was used as a resistor in which the diameter of the wire was consistent 0.0254 cm. However the length of the wire was increased by 40 cm after each trial. Throughout this we can see an increase of the voltage passing by the resistor, however we would see the current decrease as the length increased. We can see that since the length is directly proportional to the resistance which in turn also causes an increase in voltage passing through the resistor and decreasing the current. In Trial 1 as the length of the wire is 40 cm the multimeters read as I= 54.4 +/- 0.5 mA and V= .21 +/- 0.05 V, as wires length is increased for Trial 5 with a length of 200 cm the multimeter read as I= 37.7 +/- 0.5 and a V= 0.750 +/- 0.05 V. For Coil #6, due the increase of the diameter we see the opposite effect as the length of the coil, due to this the current through the system increases I= 42.8 +/-0.5 and voltage decreases V=0.506 +/- 0.05 V. Lastly, we are able to analyze the correlation between the resistivity of the materials used, throughout the first six trials Nickel Silver was used and for the final trial copper was used. Based on the information that was found in the textbook the resistivity of Nickel Silver is = 28 x 10^(-6) and the resistivity of Copper is =1.72 x 10^(-6). ρ ρ

5 Through this change it is noted that the smaller the resistivity the greater the current of the system but lower voltage running through the resistor. The uncertainty for our Voltage and Current were determined based on the multimeter and how precise it was able to be. Then for the propagation of uncertainty for Resistance was +/-0.001 due to this it can be concluded that this portion of the experiment was a success. Propagation of uncertainty for Part 1: 𝑅 = 𝑉 𝑅 δ𝑅 δ𝑉 = 1 𝑉 δ𝑅 δ𝐼 = −𝑉 𝐼 2 δ𝑅 = ( δ𝑅 δ𝑉 * δ𝑉) 2 + ( δ𝑅 δ𝐼 * δ𝐼) 2 * For every coil, the uncertainty is around ≈ 0. 001 Data Table for Part 2: * Voltages of approximately 0, 0.75, 1.50, 2.25, 3.00 volts. The objective for part 2 is to measure the voltage across through the resistor. (R1)

6 V(R1) Total V 0 0 0.75 0.8 1.50 1.6 2.25 2.5 3.00 3.3 * Nexty, the task was to find the voltage through a different resistor. (R2) V(R2) Total V 0 0

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

7 0.75 0.8 1.50 1.7 2.25 2.6 3.00 3.5 * After, we made R1 and R2 into a series. The task for this part is to find the voltage across both resistors. V(R2) Total V 0 0 0.75 0.8 1.50 1.5 2.25 2.4 3.00 3.2

8 ● 4 volts I (mA) V(R1) V(R2) V total 90.7 2.76 1.02 3.79 * Lastly, are now connecting R1 and R2 in parallel. Not only are we finding the voltages across both resistors, we are also finding the current through the circuit. V(R2) Total V I (current) 0 0 0 0.75 0.8 0.09 1.50 1.5 0.18 2.25 2.4 0.28 3.00 3.2 0.37

9 *4 volts I total (mA) I(R1) (mA) I(R2) (mA) V(R1&2) (Volts) 390 123.6 302.7 3.73 Anaylsis part 2: In the second part of this experiment, we analyzed how to arrange the resistors in different variations as far as how the circuit is built and how that would affect the voltage and the current. The resistors were arranged in series and in order to see if the voltage across the resistors would differ. This would also be done with the resistors in parallel. It is noted that when the resistors are in series it is observed that the current going through both of the resistors will be the same I=90.7 +/- 0.5 mA. However the voltage between the two will be different depending on their resistance. It is observed that the V(R1)= 2.76 +/- 0.05 V and V(R2)= 1.02 +/-0.05 V. As we change the circuit so the resistors are in parallel, it is observed that the voltages between the two resistors will be the same V(R1&R2)= 3.73 +/- 0.05 V, however the resistors being in parallel will cause for the two resistors to receive different amounts of current. As I(R1)= 123.6 +/- 0.5 mA and I(R2)= 302.7 +/- 0.5 mA. This is due to the current flowing splitting between the two wires as it travels through both resistors that are in parallel.Since the wires also have some resistance,this error is under permission. The uncertainty for this part was based off our devices since the data was all measured values, we were able to determine that our uncertainty for Current was +/- 0.5

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

10 mA, Voltage was +/- 0.05 V and the resistance that was determined was +/- 0.001 Ohms. Overall, the changes that were observed as we changed the circuits around were expected and led to a successful experiment. Propagation of uncertainty for Part 2: The uncertainty for this part was based off our devices since the data was all measured values, we were able to determine that our uncertainty was +/- 0.05 for our voltage. Conclusion: In the experiment conducted the relationship between voltage, current and resistance was analyzed, V=IR. Also analyzing the relationship of the resistors cross sectional area, length and objects resistivity, R= . From the help of the multimeters, it will help establish the current that ρ𝑙 𝐴 is flowing through the circuit and the change in electrical potential(voltage) through a resistor, depending on the different resistors for part 1 and the amount of voltage turned in thermal energy in the resistor in part 2. The main purpose of this lab is to calculate resistance of the provided sets of wires and comparing the experimental values with the theoretical values for part 1.