Question

29. (10 pts.) A capacitor and a25 ko resistor are connected in series to a 10vbatery. The voltage across the capacitor has risen to 7.0 V at a time 4.7 s after the capacitor begins charging a. What is the time constant t? (5 pts.)

Answer 1

Given, R = 25 k Ohm = 25 times 10^3 Ohm v_0 = 10 v v_c = 7.0 v at t = 4.7 s then, Time constant, T = ? and capacitance = ? step 1. voltage across the capacitor at any instance in time during charging is given by the equation: - v_c = v_0 (1 - e^-t/T) put all the known values in equation, Rightarrow 7 = 10 (1 - e^-t/T) Rightarrow 7/10 = 1 - e^-t/T Therefore Rightarrow e^-t/3 = 1 - 7/10 = 1 - 0.7 Rightarrow e^-t/T = 0.3 Taking log on both sides to cancel e. \r\n log [e^-t/T] = log ((10.3) - t/T = -0.5228 rightarrow Rounding about 3 figures. and t = 4.7 54.7/T = 0.523 Therefore T = 4.7/0.523 T = 8.9865 s. step 2. Time constant is also given by the equation, T = R times C where, R = Resistance in Ohm C = Capacitance in Fasad (F) Therefore 8.986 = 25 times 10^3 times C Therefore C = 8.986/25