Math265 Course Project Part A

Math265 Project Part A Name: In this part of the project, you will be modelling the behavior of a resistor-capacitor (RC) circuit that is being charged by a DC voltage source. In this circuit, initially, the capacitor is uncharged. When switch ‘A’ closes, the voltage source (ε) begins charging the capacitor (C) through the resistor (R). Applying Kirchhoff’s Voltage Law results in the equation ε − V R − V C = 0 Using the definition of capacitance, C = q V C , where ‘q’ is the charge on the capacitor and ‘V C ’ is the voltage across the capacitor, we can solve for the voltage as V C = qC . Next, using Ohm’s law, the voltage drop across the resistor is V R = IR . The current through any portion of the circuit is defined as the rate of change of the charge: I = dq dt . Inserting these values into our equation above, we have ε − R dq dt − qC = 0 This differential equation can be solved to determine an expression for the charge on the capacitor as a function of time: Q ( t ) = Q 0 [ 1 − e − t τ ] where Q 0 = Cε is the final charge on the capacitor and τ = RC is known as the time constant I. Graphing Charge vs Time (25 points) Consider an RC circuit that is being charged with a voltage source ε = 5 V . The charge on the capacitor as a function of time is shown in the table below, with time in seconds (s) and charge in Coulombs (C). 1 | P a g e

Time (s) Charge (C) 0 0 0.0005 1.11×10 -6 0.001 1.97×10 -6 0.0015 2.64×10 -6 0.002 3.16×10 -6 0.0025 3.57×10 -6 0.003 3.88×10 -6 0.0035 4.13×10 -6 0.004 4.32×10 -6 0.0045 4.47×10 -6 0.005 4.59×10 -6 Adjust the fitting parameters ‘a’ and ‘b’ using the sliders so that the curve matches the points and record your fitting parameters below: a=.000005 b=500 Noting that a = Q 0 = εC , where ε = 5 V , determine the capacitance of the capacitor. C=.000005/5=.000001 Noting that b = 1 τ = 1 RC , determine the resistance of the resistor. 500=1/r.000001=R=2000 Take a screenshot of Desmos showing charge versus time and your data and paste below. 2 | P a g e

II. Summary of Section I. (10 points) Write a two paragraph summary of your findings from Section I. Explain the setup and the results. The setup is a circuit with no values at first. We get values for time and charge and have and have to plot the equations to get a graphical explanation. The graph shows the amount of charge as the time passes in seconds. With those values we get our A and B which is determining the min and max. from there we 4 | P a g e

find our capacitance of the capacitor of Which is C our A/B values. From there we have to find our resistance with an equation of B=1/RC. The results show the relations between the switch, the capacitor, the DC voltage, and resistor. The voltage is not sent to any of this with the switch open as soon as the switch is closed the voltage is sent through the circuit. The voltage goes through a resistor in to the capacitor which charges/stores the voltage in the capacitor. When the voltage no longer needs to be stored in the capacitor the voltage gets discharged. The capacitor discharges the voltage through the resistor and is drained of any voltage stored by the capacitor. III. Summary of Section II. (10 points) Write a two paragraph summary of your findings from Section II. Explain the setup and the results. The setup was plugging in the values to calculate current at each time interval. We divided .0005 which each equation because of the time lapse between each recording. This gave us a graph representation of the current flow through the capacitor. It also showed the relationship between each component in the circuit if you look closely at it. The graphs show the discharge of the capacitor sending out the current up to the end when you see the peak at the top at .009 Amps. This is showing the capacitor is stored this current and is not discharging any at the time for whatever the reason not being demanded or turned off. This also give the other side of the relationship between the capacitor, the voltage, the resistor, and the switch. 5 | P a g e