Answer
(a) $6.7\times 10^{-4}s$
(b) $1.4A$
(c) increased
Work Step by Step
(a) We can find the time constant as follows:
$R_{eq}=(\frac{1}{24\Omega}+\frac{1}{(6.5\Omega+13\Omega)})^{-1}=10.76\Omega$
Now $t=R_{eq}C$
We plug in the known values to obtain:
$t=10.76\Omega(62\times 10^{-6}F)$
$t=6.7\times 10^{-4}s$
(b) We can find the initial current as
$I=\frac{\mathcal{E}}{R_{eq}}$
We plug in the known values to obtain:
$I=\frac{15V}{10.76\Omega}$
$I=1.4A$
(c) We know that the time constant and equivalent resistance $R_{eq}$ are directly proportional. Thus, if the time constant is required to increase, then $R_{eq}$ will increase and as a result, the resistance ($6.5\Omega$) of the resistor should be increased.
