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