\f\fVoltage Across Capacitor Trial 2 0 y = -106.21x - 0.045 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 -0.5 -1 -1.5 ............................................................................"es...... -2 -2.5 -3 Capacitance Value:10 Micro -3.5 Farads Resistance Value: 100k Ohms -4 Time\f\fStep 16 a. The time constant for this circuit is T = RC. What form should the graphs of the time constant vs. resistance and the time constant vs. capacitance have? Is this what you see in your graphs? b. What should be the slope of the graph of the time constant vs. resistance? What should be the slope of the graph of the time constant vs. capacitance? Determine these slopes from your graphs (use linear trendlines with the equations in Excel), and compare them to the expected values. c. Show that the time constant 1' = RC has units of seconds ifR has units of Ohms and C has units of Farads. Tip: units of resistance and capacitance are related to other units like Coulombs, Volts, and Amps by the equations V = [R and Q = CV. d. As stated in the lab manual, the voltage of the capacitor as a function of time is Vc(t) = VoleUT. The voltage is related to the charge on the capacitor as a function of time: q(t) = qoe'UT. (qo is the initial charge on the capacitor when it is at its starting voltage of V0: q0 = CV0.) This equation for the capacitor charge is found by solving the differential equation: dq q R n = _ _ dt C Show that q(t) = qoe't/T is a solution to this differential equation ifr = RC. Tips: plug q(t) into the left and right sides and show that the two sides are equal; when taking the derivative of q(t) with respect to t, go and 1' are constant