Question

Exercise 26.44 14 of 26 Part A Constants A 12.0 ?F capacitor is connected through a 0.885 ?? resistor to a constant potential difference of 60.0 v. Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0s Express your answers using two significant figures. Enter your answers numerically separated by commas. Submit ? Part B Compute the charging currents at the same instants Express your answers using two significant figures. Enter your answers numerically separated by commas. io, i5.0, iio, i20, i1oo Submit

Answer 1

Time constant for the circuit, tau = T times C = (0.885 times 10^6 Ohm) (12 times 10^6 F) = 10.62 s Charge at the capacitor is, q = CV_0 (1 - e^t/r) = (12 times 10^-6) (60) (1 - e^t/r) = 720 times 10^-6 (1 - e^- t/r) part A: q = 720 times 10^-6 (1 - e^- t/10.62) q_0 = 720 times 10^-6 (1 - e^- 0/10.62) = 0 q_5 = 720 times 10^-6 (1 - e^-5/10.62) = 2.699 times 10^-4 = 2.7 times 10^-4 C q_10 = 720 times 10^-6 (1 - e^-10/10.62) = 4.392 times 10^-4 C q_20 = 720 times 10^-6 (1 - e^-20/10.62) = -1.353 times 10^-3 = -1.4 TIMES 10^-3 c Q_100 = 720 times 1-^-6 (1 - e^-100/10.62) = -5.1415 times 10^-3 c = -5.1 times 10^-3 C q_0, q_5, q_10, q_20, q_100 = 0,2.7 times 10^-4 C, 4.4 times 10^-4 C, -1.4 times 10^-3 C, -5.1 times 10^-3 C part b: current at instant t: I = V_0/R (e^-t/r) = (60)/0.885