\begin{problem*}{21.45}
The circuit in Figure P21.45 has been connected for a long time.
\Part{a} What is the voltage $V_c$ across the capacitor?
\Part{b} If the battery is disconnected, how long does it take the
capacitor to discharge to $V_c'=1/10\cdot V$?
\begin{center}
\begin{empfile}[4]
\begin{emp}(0,0)
  pair N[];
  numeric dx, Lres;
  dx := 3cm; 
  % calculate the length of a resistor
  centreof.R((1,0), (-1,0), res);
  Lres := 2*(xpart c.R);
  % draw the bridge
  N[0] := origin;
  N[3] := N[0] - (0,Lres*sqrt(2));
  N[1] := N[0] - ((1,1)*Lres/sqrt(2));
  N[2] := N[0] + ((1,-1)*Lres/sqrt(2));
  resistor.A(N[1], normal, 45, "R_1=1.00\ohm", "");
  resistor.B(N[3], normal, 135, "R_2=4.00\ohm", "");
  resistor.C(N[3], normal, 45, "", "$R_3=2.00\ohm$");
  resistor.D(N[2], normal, 135, "", "$R_4=8.00\ohm$");
  centreof.C(N[1], N[2], cap);
  capacitor.C(c.C, normal, phi.C, "C", "");
  %centerto.B(N[3], N[1], -dx, bat);  % I don't know why this way doesn't work, but it appears to not fully define B
  %battery.B(B, 90, "\mathcal{E}", "10.0 V");
  centreof.B(N[3]-(dx,0), N[0]-(dx,0), bat);
  battery.B(c.B, phi.B, "V", "10.0 V");
  % draw the wires
  wire(N[1], C.C.l, nsq);
  wire(N[2], C.C.r, nsq);
  wireU(N[0], B.B.p, 3pt, udsq);
  wireU(N[3], B.B.n, -3pt, udsq);
  % draw the nodes
  draw N[0] withpen pencircle scaled 3pt;
  draw N[1] withpen pencircle scaled 3pt;
  draw N[2] withpen pencircle scaled 3pt;
  draw N[3] withpen pencircle scaled 3pt;
\end{emp}
\end{empfile}
\end{center}
\end{problem*} % problem 21.45

\begin{solution}
Labeling the resistors counterclockwise from the upper left we have
 $R_1 = 1.00\Omega$,
 $R_2 = 4.00\Omega$,
 $R_3 = 2.00\Omega$, and
 $R_4 = 8.00\Omega$.
Let $V = 10.0\U{V}$ be the voltage on the battery
 and $C = 1.00\U{$\mu$F}$ be the capacitance of the capacitor.

\Part{a}
Because the system has been running for a long time, the system must
be close to equilibrium.  Therefore, the current through the capacitor
must be zero (otherwise the voltage across the capacitor would be
changing, and you wouldn't be at equilibrium).  The resistor bridge
then reduces to two parallel circuits, and we can apply Ohm's law to
determine $V_c$

Starting with the left side of the bridge (calling the current $I_L$), 
\begin{align}
 V &= I_L (R_1 + R_2)  &
 I_L &= \frac{V}{R_1 + R_2}
\end{align}
And on the right calling the current $I_R$
\begin{equation}
 I_R = \frac{V}{R_3 + R_4}
\end{equation}

So using Ohm's law to compute the voltage across the capacitor, we
call the voltage on the bottom wire $0$ and have the voltage $V_L$ on the left at
\begin{equation}
 V_L = I_L R_2 = \frac{V R_2}{R_1+R_2} = 8\U{V}
\end{equation}
And the voltage $V_R$ to the right of the capacitor is
\begin{equation}
 V_R = I_R R_3 = \frac{V R_3}{R_3+R_4} = 2\U{V}
\end{equation}
So the voltage across the capacitor is 
\begin{equation}
 V_c = V_L - V_R = \ans{6\U{V}}
\end{equation}

\Part{b}
Once we remove the battery, we see that the capacitor discharges
through two paths in parallel, $R_1 \rightarrow R_4$ and $R_2
\rightarrow R_3$.
The eqivalent resistances of these two parallel branches (top and
bottom) are
\begin{align}
 R_T &= R_1 + R_4  &
 R_B &= R_2 + R_3
\end{align}
So the total equivalent resistance is
\begin{equation}
 R = \left(\frac{1}{R_T}+\frac{1}{R_B}\right)^{-1}
   = 3.60\Omega
\end{equation}

The voltage of a discharging capacitor depends on time according to
\begin{equation}
 V_c' = V_c e^{-t/RC}
\end{equation}
So using $V_c' = V_c/10$ we have
\begin{align}
 10 &= \frac{V_c}{V_c'} = \frac{V_c}{V_c e^{-t/RC}} = e^{t/RC} \\
 \ln(10) &= \frac{t}{RC} \\
 t &= RC\ln(10) = \ans{8.29\U{$\mu$s}}
\end{align}
\ens{solution}
