What we have?
[tex]R=12M\Omega=12*10^6\Omega\\c=1.75\mu F = 1.75*10^{-6}F\\V=13.50V\\[/tex]
a) We obtain the time constant
[tex]\tau = Rc\\\tau= (12.10^6)(1.75*10^{-6})\\\tau=21s[/tex]
b) We have the time and this time is t=48.5s
With the charge on capacitor Q(t) whe can calculate the final charge, so
[tex]Q(t)=Q_f(1-e^{-t/Rc})\\Q(t)=Q_f(1-^{48.5/21})\\Q(t)= Q_F(0.900691)[/tex]
Through the relation
[tex]Q(t)=Q_f(0.9)=0.9Q_f\\\frac{Q(t)}{Q_f}=0.9[/tex]
We conclude that the fraction of final charge Qf is on the capacitor at t=48.5s is 0.9
c) To calculate the current we have,
[tex]I(t) = I_0*e^{-t/Rc}\\I(48.5)=I_0*e^{-48.5/2}\\I(t)=I_0(0.099308)\\I(t)=0.099I_0\\\frac{I(t)}{I_0}=0.099[/tex]
The fraction of the initial current I:0 is still flowing at t=58.5 is 0.099
