A 14.1 μF capacitor is charged to a potential of 45.0 V and then discharged through an 80.0 Ω resistor.

(a) How long after discharge begins does it take for the capacitor to lose 90.0% of the following?
(i) its initial charge in seconds

(ii) its initial energy in seconds

(b) What is the current through the resistor at both times in part (a)?
(i) at tcharge in Ampere

(ii) at tenergy in Ampere
PLEASE SHOW WORK AND SAY WHAT IS THE FINAL ANSWER IN PARTS A AND B WITH i AND ii IN BOTH A AND B

2 answers

T = R*C = 0.08k * 14.1uF = 1.13 Milliseconds.

a. 45/e^x = 4.5, e^x = 45/4.5 = 10,
X = 2.303 = t/T = t/1.13, t = 1.13 * 2.303 = 2.60 mS = 2.6*10^-3 s.

b. i = v/R = 4.5/80 = 0.056 Amps.
Energy = 0.5C*V^2 = 0.5*14.1*45^2
= 14,276 uJ. = 0.01428 J.

Energy = 0.5*14.1*V^2 = 0.001428,
V^2 = 2.03*10^-4, Vc = 1.425 Volts when 90% of it's energy is lost.

45/e^x = 1.425, e^x = 31.6, X = 3.45 = t/RC = t/1.13, 1.13*3.45 = 3.90 mS. = 3.90*10^-3 s. = Time
to loose 90% of it's initial energy.