Answer
a) the charge stays the same
b) less charge time
c) $ t=99ms $
Work Step by Step
(a) We have: $ R_{eq}=R_1+R_2$. Thus, a lower $ R_1$ gives a lower overall resistance for the circuit. Since $ V $ doesn't change, a higher current makes up for the reduced resistance and $ Q=CV $, however the charge stays the same.
(b) We know that
$ q(t)=C\epsilon(1-e^{-\frac{t}{\tau}})$
$\implies 0.8C\epsilon=C\epsilon(1-e^{-\frac{t}{\tau}})$
$\implies 0.8=1-e^{-\frac{t}{\tau}}$
$\implies e^{(-\frac{t}{\tau})}=0.2$
$\implies -\frac{t}{\tau}=-1.6094$
$\implies t=(1.6094)(RC)$ (where $\tau=RC $)
As time is directly proportional to resistance, a lower resistance will indicate less time needed to charge the capacitor. Hence, more current can flow towards the capacitor to put charge on it more quickly, leading to less charge time.
(c) We can find the required time as follows:
$ t=(1.6094)(63+275)(182\times 10^{-6})$
$ t=(1.6094)(338)(182\times 10^{-6})$
$ t=99ms $
