Problem 21.41

21.41. a)
After a ``long'' time, the capacitor is fully charged and the capacitor acts like an open circuit. Then the current on the left is given by
$I_{L} = \frac{V}{R}$
$I_{L} = \frac{10 V}{1 \Omega + 4 \Omega}$
I_L = 2 A
Then, the voltage on the left side of the capacitor is given by
$V_{L} = 10 V - (2A)(1 \Omega)$
V_L = 8 V
Now, looking at the right side,
$I_{R} = \frac{V}{R}$
$I_{R} = \frac{10 V}{8 \Omega + 2 \Omega}$
I_R = 1 A
Then, the voltage on the left side of the capacitor is given by
$V_{R} = 10 V - (1A)(8 \Omega)$
V_R = 2 V
Therefore, the potential difference across the resistor is given by
$\Delta V = V_{L} - V_{R}$
$\Delta V = 8 - 2 ~=~ 6 V$

b)
We know that the charge on the capacitor, given as a function of time is
q(t) = Q₀ e^-t/RC
So, if we want
$q(t) = \frac{Q_{0}}{10}$
Then we need to find when
$e^{-t/RC} = \frac{1}{10}$
or, when $t = RC \ln 10$
We need to find what R is. Now, since the battery is disconnected, the $1 \Omega$ and $8 \Omega$ resistors are in series, and so are the $4 \Omega$ and $2 \Omega$ resistors. And the resultant resistances are in parallel with each other. In other words,
$R_{eq} = \frac{1}{\left(\frac{1}{1+8} + \frac{1}{2+4}\right)}$
$R_{eq} = 3.6 \Omega$
So, substituting in to find the time,
$t = (3.6 \Omega)(1.00 \mu F) \ln 10$
$t = 8.29 \times 10^{-6}$ seconds.