Answer
(a) $9.75ms$
(b) $668\mu C$
(c) $68.6mA$
Work Step by Step
(a) We can find the time constant as
$t=RC$
We plug in the known values to obtain:
$t=(175\Omega)(55.7\times 10^{-6}F)$
$t=9.75\times 10^{-3}s$
$t=9.75ms$
(b) We can find the maximum charge as
$q_{max}=C\mathcal{E}$
We plug in the known values to obtain:
$q_{max}=(55.7\times 10^{-6}F)(12.0V)$
$q_{max}=6.68\times 10^{-4}C$
$q_{max}=668\mu C$
(c) The initial current in the circuit can be determined as
$I_{t}=(\frac{\mathcal{E}}{R})e^{-t/r}$
For initial current $t=0$
$\implies I(0)=(\frac{\mathcal{E}}{R})e^{0}$
$I=\frac{\mathcal{E}}{R}$
We plug in the known values to obtain:
$I=\frac{12.0V}{175\Omega}$
$I=0.06857A$
$I=68.6mA$