Question

A 15.9 μF capacitor is charged to a potential of 60.0 V and then discharged through a 75.0 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
s

(ii) its initial energy
s

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

(i) at t_charge
A

(ii) at t_energy
A

Answer 1

a )

given

C = 15.9 μF = 15.9 X 10^-6 F

v = 60 volts

R = 75.0 Ω

Q = C X V

Q = 15.9 X 10^-6 X 75

Q = 1.192 X 10^-3 C

given 90 % , so the new charge is 10 % is 1.192 X 10^-4 C

new voltage is = 6 v

V = v exp ( - t / RC )

6 = 60 X exp ( - t / 75 X 15.9 X 10^-6 )

- 2.3 = ( - t / 75 X 15.9 X 10^-6 )

t = 0.00274 s

energy stored is U = 1/2 C V²

= 0.5 X 15.9 X 10^-6 X 60²

= 0.028 J

new energy is 10 % so it is 0.0028 J

0.0028 = 0.5 X 15.9 X 10^-6 X v ²

v = 18.76 volts

18.76 = 60 X exp ( - t / 75 X 15.9 X 10^-6 )

- 1.16 = ( - t / 75 X 15.9 X 10^-6 )

t = 0.0013833 sec

b )

v = 18.76 volts

18.76 = 60 X exp ( - t / 75 X 15.9 X 10^-6 )

- 1.16 = ( - t / 75 X 15.9 X 10^-6 )

t = 0.0013833 sec

current through resistor part a 6 / 75 = 0.08 A

part b is   = 18.76 / 75 = 0.25 A