E5 Report

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School

Boston College *

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Course

S2051

Subject

Electrical Engineering

Date

Apr 3, 2024

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pdf

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Uploaded by LieutenantRain11995

E5 REPORT - SPRING 2023 DATA (2.5 points) (General instructions: Data should be in tables, properly labeled and with units; where appropriate, include data for repeated measurements, and also the calculated average and error values) Data for Circuit 1: Power Supply (V) Current (µA) Voltage (V) Resista nce of Voltmeter (Ω) 3 3 2.98 9.93 x 10 5 +/- 9.93 x 10 4 4.5 4.5 4.49 9.97 x 10 5 +/- 9.97 x 10 4 6 5.9 5.98 1.01 x10 6 +/- 1.01 x 10 5 Data for Circuit 2: Power Supply (V) Current (mA) Voltage (V) Resista nce of Ammeter (Ω) 3 29.2 0.11 3.77 +/- 0.377 4.5 44.1 0.18 4.08 +/- 0.408 6 58.6 0.20 3.41 +/- 0.341 Data for Circuit 3: I 0 = 87.4 mA Power Supply (V) Element Current (mA) Predicted Current (mA) Voltage (V) Predicted Voltage (V) 4.5 33 Ω Resistor 62.8 +/- 3.14 (agrees) 65.8 2.23 +/- 0.112 (agrees) 2.34 4.5 100 Ω Resistor 20.9 +/- 1.05 (agrees) 21.7 2.25 +/- 0.113 (agrees) 2.17 4.5 Light Bulb 86.4 +/- 4.32 (agrees) 87.4 2.24 +/- 0.112 (agrees) 2.17 Data for Circuit 4: I 0 = 153.4 mA

Power Supply (V) Element Current (mA) Predicted Current (mA) Voltage (V) Predicted Voltage (V) 4.5 33 Ω Resistor 34.2 +/- 1.71 (agrees) 33.8 1.11 +/- 0.056 (agrees) 1.12 4.5 100 Ω Resistor 34.2 +/- 1.71 (agrees) 33.8 3.38 +/- 0.169 (agrees) 3.38 4.5 Light Bulb 123.4 +/- 6.17 (agrees) 119.6 4.49 +/- 0.2245 (agrees) 4.5 Data for Circuit 5: I 0 = 90.7 mA Power Supply (V) Element Current (mA) Predicted Current (mA) Voltage (V) Predicted Voltage (V) 4.5 33 Ω Resistor 90.6 +/- 4.53 (agrees) 90.7 2.98 +/- 0.149 (agrees) 2.993 4.5 100 Ω Resistor 15.4+/- 0.77 (agrees) 15.06 1.50 +/- 0.075 (agrees) 1.506 4.5 Light Bulb 74.9 +/- 3.75 (agrees) 75.6 1.51+/- 0.076 (agrees) 1.506 PLOTS and FIGURES (1.5 points) (General instructions: include plot titles, axes labels and units; sketches and figures should also be properly labeled and have units, where appropriate) Draw neat sketches (circuit diagrams) of circuits 3, 4, and 5 showing where you connected the voltmeters and ammeters to measure voltage across, and current through, each circuit element. (circuit 5 is the circuit you designed).

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CALCULATIONS AND ERROR ANALYSIS (3 points) (General instructions:show all steps for each type of calculation or error analysis only once; must be typed) Circuits 1 and 2 Calculate the resistance of the ammeter and voltmeter from your data for circuits 1 and 2 (0.25 points). In order to calculate the resistance of the ammeter and voltmeter, it is necessary to use Ohm’s Law (V= IR): 𝑅 = 𝑉 𝐼 For example, for the first circuit at 3V power source: = 9.93 x 10 5 Ω 𝑅 𝑣 = 2.98 3𝑥10 −6 Calculate the absolute error in these resistances, assuming all measured voltage and current values have 5% error (0.5 points). In order to calculate absolute error in the resistance, the following formula is used: + δ𝑅 𝑅 = δ𝑉 𝑉 δ𝐼 𝐼 R(0.05 + 0.05) δ𝑅 = = R(0.1) δ𝑅 For example, for the first circuit at 3V power source: = (9.93 x 10 5 )(0.1) δ𝑅 = +/- 9.93 x 10 4 Ω δ𝑅 Circuits 3, 4, and 5 Calculate the absolute error in all measured values of current through and voltage across , and 𝑅 1 𝑅 2 the bulb, assuming there is a 5% relative error in the voltmeter and ammeter measurements. (0.5 points) To calculate the absolute error in current (same for error in voltage), each value is multiplied by the relative error (5% = 0.05) For example, for the measured current through the 33 Ohm Resistor: I = 62.8 mA I = 62.8 * 0.05 = +/- 3.15 mA δ

Using Kirchhoff's Rules and the measured values of the power supply current, power supply voltage, and the resistances of and , predict the current through and voltage across , and 𝑅 1 𝑅 2 𝑅 1 𝑅 2 the bulb (since these circuits are different, you will need to show the calculations for each circuit). (1.5 points) Circuit 3: For this, it is first necessary to calculate R eq of the resistors in series (R 1 and R 2 ): Resistance of resistors in series is calculated as shown below: + = + = 24.9 Ω 𝑅 𝑒𝑞 = 1 𝑅 1 1 𝑅 2 1 33.1 1 100.3 Then, we can find the resistance of the bulb using the loop rule: 4.5V - I(R eq ) - I(R bulb ) = 0 4.5 - I(R eq + R bulb ) = 0 4.5 - 0.0874(24.9 + R bulb ) = 0 26.77 Ω = predicted R bulb Because components in series have the same current and 87.4 A= I 0 (predicted current through bulb) : V= IR bulb = 0.0874 (26.77) = 2.34 V = predicted voltage across bulb V= IR eq = 0.0874 (24.8) = 2.17 V = predicted voltage across resistors R1 and R2 Because components in parallel have the same voltage: I R1 = V/R 1 = 2.17/33 = 65.8 mA = predicted current through R 1 I R2 = V/R 2 = 2.17/100 = 21.7 mA = predicted current through R 2 Circuit 4: We must first establish the first equation which uses the junction rule: I 0 = I 1 + I 2

0.1534 = I 1 + I 2 Because resistors in parallel have the same voltage and resistors 1 and 2 are in series: R 1 + R 2 = 4.5 V Predicted voltage across Bulb = 4.5 V Using the junction rule and the fact that resistors 1 and 2 are in series R 1 + R 2 = R eq = 100 + 33 = 133 Ω = R eq I 0 = I 1 + I 2 0.1534 = I 1 + ((V R1 + R2 )/R eq ) 0.1534 = I 1 + (4.5/133) 0.1534 = I 1 + 0.0338 (current through R 1 and R 2 (in A)) 119.6 mA = predicted current through bulb To find voltage across each resistor: V R1 = I 2 R 1 = 0.0338(33) = 1.12V = voltage across R 1 V R2 = I 2 R 2 = 0.0338(100) = 3.38 V = voltage across R 2 Circuit 5: V R1 = (I 0 )(R 1 ) = (.0907)(33) = 2.99V = V R1 Loop 1 (through R 1 and R 2 ): 4.5V - I 0 R 1 - I R2 R 2 = 0 4.5 - (90.7)(33) - I R2 (100) = 0 I R2 = 15.06 mA V R2 = I R2 R 2 = (0.01506)(100) = 1.506V = V R2 Because the light bulb and R 2 are in parallel, they have the same voltage across them: V bulb = 1.506V

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Because the bulb is after a junction, I R1 = I R2 + I bulb -> 90.7 - 15.06 = 75.6 mA = I bulb For each circuit, create a table showing measured (with error) and predicted values of voltage and current for each circuit element. Do the predicted voltage and currents from Kirchhoff's Rules agree with your measurements, within error? (0.25 points) **SHOWN IN TABLE SECTION The predicted voltage and currents from Kirchhoff's Rules all agree with our measurements within error. QUESTIONS (2 points) 1. Did the resistance of the ammeter change when you changed the power supply voltage? How about the resistance of the voltmeter? (0.25 points) The resistance of the ammeter changed slightly without any pattern when the power supply voltage was changed with the resistance first increasing and then decreasing. However, this is likely due to the ammeter not being an ideal ammeter which means that some voltage drops across it. Likewise, the resistance of the voltmeter changes slightly by increasing in small increments as the power supply voltage increases. 2. Is it reasonable to assume the meters are ideal? Explain. (0.5 points) The meters were not ideal due to the low resistance of the ammeter connected in series and the high resistance of the voltmeters connected in parallel. 3. Calculate the resistance of the light bulb in circuits 3, 4, and 5 using either the measured or predicted values of current and voltage. Should the resistance of the bulb be the same or different in each circuit? Explain. (1 point) The resistance of the lightbulb in the three circuits should be different because the resistance depends on the configuration of the circuit and the I 0 of each of the different circuits. 4. What are your conclusions regarding the usefulness of Kirchhoff's Rules? ( 0.25 points) A conclusion about the usefulness of Kirchhoff’s Rules is that it can be very helpful for analyzing circuits. The knowledge of the rules makes it possible to determine the change in voltage in a series as the sum of all the electric potential difference around the circuit is equal to 0. TIME STAMPED NOTES ( -2 points, if missing) (General Instructions: Include here, time-stamped pictures of all the notes you took during the lab)