phys-1004-lab-report-2-dc-circuits

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PHYS 1004 Lab report 2 "DC circuits" Introductory Electromagnetism and Wave Motion (Carleton University) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university PHYS 1004 Lab report 2 "DC circuits" Introductory Electromagnetism and Wave Motion (Carleton University) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Deelaka Appuhamy (deelakacarleton@gmail.com) lOMoARcPSD|31539259

PHYS 1004 Lab #2 “DC circuits” Name: Student Number: Date: Downloaded by Deelaka Appuhamy (deelakacarleton@gmail.com) lOMoARcPSD|31539259

PART A: A1: Deduce the color of the 3 remaining bands for R1, R2 and R3 using the nominal resistance in Table 1. Resistor R1 has two possible values, ≈ 270 Ω or ≈ 150 Ω , decide which one best matches your data in Part A of your Logger Pro file. A1 Band 1 Band 2 Band 3 Band 4 Nominal R± tolerance R 1 Red Purple Brown Gold 270±14 Ω R 2 Green Blue Black Gold 56±3 Ω R 3 Brown Black Brown Gold 100±5 Ω A2: Equivalent resistance calculation ( ? ? q ± ? ?? q ) (from logger pro file) R 1 = 271.3±0.2 Ω R 2 = 41.27±0.02 Ω R 3 = 68.47±0.02 Ω (1 / ? ? q ) = ((1/R 1 ) + (1/(R 2 + R 3 ))) (1 / ? ? q ) = ((1/271.3 Ω ) + (1/(41.27 Ω + 68.47 Ω ))) (1 / ? ? q ) = 0.012798 ? ? q = 78.134 Ω ? ?? q = √ (((dR eq / dR 1 ) 2 * ? R1 2 ) + ((dR eq / dR 2 ) 2 * ? R2 2 ) + ((dR eq / dR 3 ) 2 * ? R3 2 )) Apply Derivative ? ?? q = √ ((((R 3 + R 2 ) 2 / (R 1 + R 3 + R 2 ) 2 ) 2 * ? R1 2 ) + ((R 1 2 / (R 2 + R 3 + R 1 ) 2 ) 2 * ? R2 2 ) + ((R 1 2 / (R 3 + R 2 + R 1 ) 2 ) 2 * ? R3 2 )) Downloaded by Deelaka Appuhamy (deelakacarleton@gmail.com) lOMoARcPSD|31539259

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? ?? q = √ ((((68.47 Ω + 41.27 Ω ) 2 / (271.3 Ω + 68.47 Ω + 41.27 Ω ) 2 ) 2 * (0.2 Ω ) 2 ) + (((271.3 Ω ) 2 / (41.27 Ω + 68.47 Ω + 271.3 Ω ) 2 ) 2 * (0.02 Ω ) 2 ) + (((271.3 Ω ) 2 / (68.47 Ω + 41.27 Ω +271.3 Ω ) 2 ) 2 * (0.02 Ω ) 2 )) ? ?? q = √ (0.000275193 + 0.000102796 + 0.000102796) ? ?? q = 0.022 Ω Final Value for R eq = 78.134±0.022 Ω A3: From Logger Pro I vs V graph. Determine Slope ( ? ± ? ? ) Intercept ( ? ± ? ? ) m = 12.99±0.11mA/V b = -0.0205±0.3225mA A4: ( ? ? q m ± ? ?? q ? ) R eqm = (1 / m) * 1000 Ω R eqm = 77.0 Ω ? ?? qm = √ (((dR eqm / dm) 2 * ? m 2 )) * 1000 Ω Apply Derivative ? ?? qm = √ (((-1 / m 2 ) 2 * 0.11 2 )) * 1000 Ω ? ?? qm = √ (((-1 / (12.99) 2 ) 2 * 0.11 2 )) * 1000 Ω ? ?? qm = √ (4.16504E-7) *1000 Ω ? ?? qm = 0.6 Ω Downloaded by Deelaka Appuhamy (deelakacarleton@gmail.com) lOMoARcPSD|31539259

Final Value for R eqm = 77.0±0.6 Ω A5: t-test calculation ( ? ? qm ? s . ? ? q ). T = |R eq m - R eq | / √ ( ? Reqm 2 + ? Req 2 ) T = 1.134 / √ (0.3604) T = 1.889 Since T < 2, we can conclude that R eqm and R eq are measuring the same value Downloaded by Deelaka Appuhamy (deelakacarleton@gmail.com) lOMoARcPSD|31539259

PART B: B1: Determine the average diameter of the wire, ? ± ? ? . ? = (0.405 + 0.402 + 0.403 + 0.407 + 0.405) / 5 ? = 0.4044mm ? ? = (0.407 - 0.402) / √ 5 ? ? = 0.0022mm Final value for d = 0.4044±0.0022mm (0.04044±0.00022cm) B2: From Logger Pro R vs. L graph. Determine Slope ( ? ± ? ? ) Intercept ( ? ± ? ? ) m = 0.07902±0.00335 Ω /cm b = -0.5338±0.4494 Ω B3: Resistivity of the wire ( ? ± ? ? ) from the slope of R vs. I graph. ? = (m π d 2 / 4) / * (1m / 100cm) ? = ((0.07902 Ω /cm * π * (0.04044cm) 2 )/ 4) *(1m / 100cm) ? = 1.0149E-6 Ω m ? ? = ( √ (((d ρ / dm) 2 * ? m 2 ) + ((d ρ / dd) 2 * ? d 2 ))) * (1m / 100cm) Apply Derivative Downloaded by Deelaka Appuhamy (deelakacarleton@gmail.com) lOMoARcPSD|31539259

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? ? = ( √ ((( π d 2 / 4) 2 * ? m 2 ) + ((m π d / 2) 2 * ? d 2 ))) * (1m / 100cm) ? ? = ( √ ((( π * (0.04044cm) 2 / 4) 2 * (0.00335 Ω /cm) 2 ) + (((0.07902 Ω /cm) * π * (0.04044cm) / 2) 2 * (0.00022cm) 2 ))) * (1m/100cm) ? ? = ( √ (1.851E-11 + 1.22E-12)) * (1m / 100cm) ? ? = 4.4418E-8 B4: Final value for ? ± ? ? ? = 1.0149E-6±0.0442E-6 Ω m Downloaded by Deelaka Appuhamy (deelakacarleton@gmail.com) lOMoARcPSD|31539259

Discussion Questions: Q: In the first part of the experiment, you were working with individual resistance values collected using a DMM. The corresponding reading errors were provided in the data analysis file. According to their colour codes, each of the resistors was stated to have a 5% tolerance by the manufacturer. Which is more precise, the manufacturer’s tolerance or the uncertainty on the DMM? Which one is more relevant for the calculations? Explain your answer. A: The uncertainty in the DMM is much more precise than the 5% tolerance provided by the manufacturer, as the ± 0.2 reading error on R1 represents a 0.0737% variation in the resistance of the resistor, compared to a possible 5% variation stated by the manufacturer. The reading error is also much more relevant to calculations because the 5% tolerance provided by the manufacturer is not equivalent to saying that the resistor has an error equal to ± 5% of its value, rather, it is a guarantee by the manufacturer that the value of a given resistor will lie within 5% of the stated value. For example, if the manufacturer makes a 50 ohm resistor, the 5% tolerance states that the resistor's true value could be anywhere in between 47.5 ohms and 52.5 ohms. If we were to then measure the value of the resistor directly we would find that its actual value is in that range and the reading error from the instrument would serve as the true error on the resistor to be used in calculations. Q: Explain the benefit of determining the equivalent resistance graphically with a set of voltage/current data, rather than collecting a single ? and ? measurement and plugging them into ? = ?? to find ? . A: The greatest benefit of determining equivalent resistance via a current vs voltage graph as opposed to applying ohm's law directly to an element is because with a graph, the equivalent resistance (represented by the inverse of the slope of the graph) is determined from the average of multiple current and resistance values. This will result in a much more precise answer when compared to taking only one voltage and current value to apply ohm's law directly. Downloaded by Deelaka Appuhamy (deelakacarleton@gmail.com) lOMoARcPSD|31539259

Q: Based on your results, are equations (2.8) and (2.9) appropriate for calculating equivalent resistances for resistors in series and parallel configurations? How can you account for any discrepancies? A: Yes, the equations for parallel and series equivalent resistances are valid for calculating equivalent resistances, as the value we calculated using these formulas was reasonably similar to the value calculated using experimental data according to the T test. However there still exists a small discrepancy between the values. This can be accounted for by the fact that the equations for equivalent resistances assume that there is no resistance in the wires, which is not the case in real life. Q:Describe the difference between resistance and resistivity. Is the resistivity of a metal wire dependent on the temperature? A: Resistivity is an innate property of a material, while resistance represents the amount a specific component resists the flow of electric charge. The resistivity of a metal does vary with temperature (high temperature = high resistivity) Q:From your resistivity result (B4), can you conclude what material the wire is made of? You must include a reference with your answer. A: Based on the final value of ρ , the material the wire is made of is most likely a nichrome alloy. References: https://www.electronics-notes.com/articles/basic_concepts/resistance/electrical-r esistivity-table-materials.php https://www.thoughtco.com/table-of-electrical-resistivity-conductivity-608499 Downloaded by Deelaka Appuhamy (deelakacarleton@gmail.com) lOMoARcPSD|31539259

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Q:A piece of pure metal wire has a diameter (d) of 1 mm, a length (l) of 7 cm and a resistance (R) of 20 Ω . It is then melted down to form a new wire of length 1 cm. What is the diameter of this new wire if the resistance is now measured to be 6 Ω ? A: A = π r 2 (r = d / 2), d = 1mm A = 0.7853mm 2 = 7.853E-7m 2 ρ = (R * A) / L ρ = (20 Ω * 7.853E-7m 2 ) / 0.07m ρ = 2.244E-4 Ω m A = ( ρ * L) / R A = (2.244E-4 Ω m * 0.01m) / 6 Ω A = 3.740E-7m 2 = 0.374mm 2 = π r 2 , therefore r = √ (A / π ) r = √ (0.374 / π ) = 0.345mm, r*2 = d, d = 0.69mm Downloaded by Deelaka Appuhamy (deelakacarleton@gmail.com) lOMoARcPSD|31539259