Compute the charge on the capacitor at the following times a

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 100.0 s. Express your answers using two significant figures. Enter your answers numerically separated by commas. Compute the charging currents at the same instants. Express your answers using two significant figures. Enter your answers numerically separated by commas.

Solution

A)

Charge on the capacitor is given by:

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

where:

Qo = final charge = C*V = 12.8*10^-6*60 = 7.68*10^-4 C

So, at t = 0s,

charge , qo = Qo*(1-e^(-0)) = 0 <----answer

At t = 5s,

q(5) = 7.68*10^-4*(1-e^(-5/(12.8*10^-6*0.885*10^6)))

= 2.74*10^-4 C <-----answer

At t = 10s,

q(10) = 7.68*10^-4*(1-e^(-10/(12.8*10^-6*0.885*10^6)))

= 4.5*10^-4 C <------answre

At t = 20s,

q(20) = 7.68*10^-4*(1-e^(-20/(12.8*10^-6*0.885*10^6)))

= 6.47*10^-4 C

At t = 100s

q(100) = 7.68*10^-4*(1-e^(-100/(12.8*10^-6*0.885*10^6)))

= 7.68*10^-4 C

B)

Current in the circuit is given by:

I = Io*(e^(-t/RC))

where Io = V/R = 60/(0.885*10^6) = 6.78*10^-5 A

So, for t = 0s,

i(0)= 6.78*10^-5*(e^0) = 6.78*10^-5 A <-------answer

for t = 5s,

i(5) = 6.78*10^-5*e^(-5/(12.8*10^-6*0.885*10^6)) = 4.36*10^-5 A <----answer

for t = 10s,

i(10) = 6.78*10^-5*e^(-10/(12.8*10^-6*0.885*10^6)) = 2.8*10^-5 A <-----answer

for t = 20s,

i(20) =  6.78*10^-5*e^(-20/(12.8*10^-6*0.885*10^6)) = 1.16*10^-5 A <------answer

for i = 100s,

i(100) =  6.78*10^-5*e^(-100/(12.8*10^-6*0.885*10^6)) = 9.94*10^-9 A <-----answer