Question

You charge an initially uncharged 89.9-mF capacitor through a 35.7-? resistor by means of a 9.00-V battery having negligible internal resistance. Find the time constant of the circuit. What is the charge of the capacitor 1.09 time constants after the circuit is closed?

Answer 1

t = RC

= 3.21 *10^-3 Seconds

Q= CV (1-e(-t/RC)) = 0.809*0.6637 = 0.537 mC

Answer 2

Time constanrt = R*C = 35.7*89.9*10^-3 = 3.2 s

Q = Qo*(1 - e^(-t/T))

= C*Vo*(1 - e^(-1.09))

= 89.9*10^-3*9*(1 - e^(-1.09))

= 5.37*10^-4 C

Answer 3