Math265 Project Part B Name: In this part of the project, you will model the behavior of a resistor-inductor (RL) circuit in a transient state. Initially, the DC voltage source is off and there is no current in the circuit. After the source ‘ε’ is turned on, the current through the inductor quickly rises and produces an EMF that opposes the change in current: 𝑉𝐿 = 𝐿 𝑑𝑖 𝑑𝑡 where ‘L’ is the inductance Using Kirchhoff’s Voltage Law and Ohm’s Law results in the equation 𝜀 − 𝐼𝑅 − 𝐿 𝑑𝑖 =0 𝑑𝑡 This differential equation can be solved to determine an expression for the current through the circuit as a function of time: 𝑡 𝐼(𝑡) = 𝐼𝐹 [1 − 𝑒 − ⁄𝜏𝐿 ] 𝜀 where 𝐼𝐹 = 𝑅 is the final current and 𝜏𝐿 = 𝐿⁄𝑅 is known as the inductive time constant. The theoretical EMF produced by the inductor can be found as follows: 𝑉𝐿 = 𝐿 I. 𝑑𝑖 𝑑 𝑡 𝑡 = 𝐿 (𝐼𝐹 [1 − 𝑒 − ⁄𝜏𝐿 ]) = 𝜀𝑒 − ⁄𝜏𝐿 𝑑𝑡 𝑑𝑡 Graphing Current vs Time (25 points) Consider an RL circuit that is being charged with a voltage source ε. The voltage across the inductor is shown as a function of time in the table below, with time in seconds (s) and voltage in volts (V). The magnitude of the current at time tf can be determined from the voltage (v) as 1|Page 𝑡𝑓 1 𝐼(𝑡) = ∫ 𝑣 ∙ 𝑑𝑡 + 𝐼(𝑡𝑖 ) 𝐿 𝑡𝑖 where I(ti) is the current at time ti. We will use a technique to calculate this integral algebraically, to determine the current from the voltage. By doing so, we are finding the approximate area under the voltage versus time curve. To determine the current at a time ‘2’, where time ‘1’ is the previous interval, use 𝐼2 = (𝑉2 + 𝑉1 )/2 ∗ ∆𝑡 + 𝐼1 𝐿 The numerator is the average voltage multiplied by the time interval and is an algebraic approximation for the area underneath the voltage versus time curve between t1 and t2. In this example, for each interval, ∆𝑡 = 3 × 10−7 𝑠 𝑎𝑛𝑑 𝐿 = 1 × 10−3 𝐻. Referencing the voltage in the table below, to determine the current for the first cell and knowing that at time t=0s, 𝐼1 = 𝐼(𝑡 = 0𝑠) = 0𝐴 (2.74 + 5)/2 ∗ (3 × 10−7 ) 𝐼2 = + 0 = 0.0016𝐴 1 × 10−3 and for the next cell 𝐼3 = (1.51 + 2.74)/2 ∗ (3 × 10−7 ) + 0.0016𝐴 = 0.0018𝐴 1 × 10−3 Repeat the process to fill in all of the cells in the table maintaining 3 significant figures for the current. Time (s) 0.0E+00 3.0E-07 6.0E-07 9.0E-07 1.2E-06 1.5E-06 1.8E-06 2.1E-06 2.4E-06 2.7E-06 3.0E-06 Voltage (V) 5.00 2.74 1.51 0.83 0.45 0.25 0.14 0.07 0.04 0.02 0.01 I(A) 0 0.00116 0.00180 Open Desmos by copying and pasting the following link into your web browser or use control+click (https://www.desmos.com/calculator/lhdbhowyyh). Enter the data from the table for current in the column ‘y1’. 2|Page You will fit the data using the theoretical equation for current as a function of time. Adjust the fitting parameters ‘a’ and ‘b’ using the sliders so that the curve matches the points. Since we are using an approximating method for calculating the points, the curve will not be an exact match. To fit the curve as well as possible, note the R2 value. R2 is known as the coefficient of determination. It is a unit-less number between 0 and 1 and indicates how well the curve fits the points. The closer R2 is to 1, the better the fit. Adjust ‘a’ and ‘b’ so that R2 is as close to 1 as possible Record your fitting parameters below as well as your R2 value: a= b= R2= Comparing to the equation for current above, noting that 𝑎 = 𝜀⁄𝑅 and given that 𝜀 = 5𝑉, determine the resistor value. Show your work R= Noting that 𝑏 = 1⁄𝜏 = 𝑅⁄𝐿, determine the value of the inductor. Show your work. L= Take a screenshot of Desmos showing current versus time and your data and paste below. Screenshot of Desmos 3|Page II. Graphing Voltage vs Time (25 points) The magnitude of the voltage across the inductor was shown to be 𝑡 𝑉𝐿 = 𝜀𝑒 − ⁄𝜏𝐿 where 𝜏𝐿 = 𝐿⁄𝑅 Open Desmos by copying and pasting the following link into your web browser or use control+click (https://www.desmos.com/calculator/g3dkvlljti) Using Desmos and the quantities R, L and ε from Section I., plot the voltage as a function of time, using the equation above. In Desmos, type the values for ε, R and L to replace the # symbols in the equation provided. Copy and paste your graph with the theoretical voltage versus time below. Screenshot of Desmos: III. Summary of Section I (10 points) Write a two paragraph summary of your findings from Section I. Explain the setup and the results. Section I Summary 4|Page IV. Summary of Section II (10 points) Write a two paragraph summary of your findings from Section II. Explain the setup and the results. Section II Summary 5|Page
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