Question

Problem 4 In a previous lab, we designed a circuit to mimic stress relaxation using concepts of a 200 2 wwwm voltage divider and our general exponential decay of an RC circuit. To simplify it, we CoOmF stated that we could treat the time constant as simply R1C. However, that is NOT the case, and the time constant is more complicated than simply that. The circuit shown is a O-t modified version of that circuit. [Part A] Using whatever method, find the VALUE of the system time constant for this circuit [Part B] Given a step input from 0 to 10V, what is the voltage value at 1 tau?

Answer 1

Part (A): - The time constant tau is given by tau = R_th c where R_th is the Thevenin resistance seen across capacitor terminals circuit for finding R_th: - R_th = (200 ohm) Vertical-bar Vertical-bar 800 ohm R_th = 160 ohm tau = R_th c = 160 (60 times 10^-3) = 9.6 sec Part B: - For a step input of 10 v, The steady state value of capacitor voltage is given by the circuit below. V_c (infinity) = 10 (800)/800 + 200 V_c (infinity) = 8 v The capacitor voltage as a function of time is given by V_c (t) = V_c (infinity) + (V_c (0) - V_c (infinity)) e^-t/tau Assuming Capacitor is initially uncharged, V_c (0) = 0v V_c (t) = 8 + (0 - 8) e^-t/9.6 V_c (t) = 8 (1 - e^-t/9.6) at t = tau i.e at t = 9.6 sec, the voltage is given by V_c (t) = 8 (1 - e^-1) V_c (t) = 5.056 v