\begin{problem*}{19.31} % resistor networks
Calculate the currents in each resistor of Fig.~19-49.
\end{problem*}

\begin{nosolution}
\begin{center}
\begin{asy}
import Circ;
real u = 3cm;
MultiTerminal Bc = source(dir=90, type=DC, value="$3.0\U{V}$");
MultiTerminal Rcb = resistor(Bc.terminal[0], dir=-90,
    value=Label("$10\U{\Ohm}$", align=E));
MultiTerminal Rca = resistor(Bc.terminal[1], dir=180, value="$2\U{\Ohm}$");
pair Jtop = Rca.terminal[1], Jbot = (Jtop.x,Rcb.terminal[1].y);
MultiTerminal Rb = resistor(Jtop, dir=-90,
    value=Label("$6\U{\Ohm}$", align=E));
MultiTerminal Ba = source(Jtop, dir=180, type=DC, value="$6.0\U{V}$");
MultiTerminal Rab = resistor(Jbot, dir=180,
    value=Label("$8\U{\Ohm}$", align=S));
MultiTerminal Raa = resistor(Rab.terminal[1], dir=90,
    value=Label("$12\U{\Ohm}$", align=W));
wire(Ba.terminal[1], Raa.terminal[1], rlsq);
wire(Rab.terminal[0], Jbot, nsq);
wire(Jbot, Rb.terminal[1], nsq);
wire(Jbot, Rcb.terminal[1], rlsq);
\end{asy}
\end{center}
\end{nosolution}

\begin{solution}
\begin{center}
\begin{asy}
import Circ;
MultiTerminal Bc = source(dir=90, type=DC, value="$3.0\U{V}$");
MultiTerminal Rcb = resistor(Bc.terminal[0], dir=-90,
    value=Label("$10\U{\Ohm}$", align=E));
MultiTerminal Rca = resistor(Bc.terminal[1], dir=180, value="$2\U{\Ohm}$");
pair Jtop = Rca.terminal[1], Jbot = (Jtop.x,Rcb.terminal[1].y);
MultiTerminal Ic = current((Jbot+Rcb.terminal[1])/2, Label("$I_3$", align=S));
MultiTerminal Rb = resistor(Jtop, dir=-90,
    value=Label("$6\U{\Ohm}$", align=E));
MultiTerminal Ib = current(Rb.terminal[1], dir=-90, "$I_2$");
MultiTerminal Ba = source(Jtop, dir=180, type=DC, value="$6.0\U{V}$");
MultiTerminal Ia = current(Ba.terminal[1], dir=180, "$I_1$");
MultiTerminal Rab = resistor(Jbot, dir=180,
    value=Label("$8\U{\Ohm}$", align=S));
MultiTerminal Raa = resistor(Rab.terminal[1], dir=90,
    value=Label("$12\U{\Ohm}$", align=W));
wire(Ia.terminal[1], Raa.terminal[1], rlsq);
wire(Jbot, Ib.terminal[1], nsq);
wire(Jbot, Ic.terminal[0], nsq);
wire(Ib.terminal[1], Rb.terminal[1], nsq);
wire(Ic.terminal[1], Rcb.terminal[1], rlsq);
dot("a", Jbot, S);
\end{asy}
\end{center}
Label the resistors from left to right: $R_1 = 12\U{\Ohm}$, $R_2 =
8\U{\Ohm}$, $R_3 = 6\U{\Ohm}$, $R_4 = 2\U{\Ohm}$, and $R_5 =
10\U{\Ohm}$.

Label the batteries from left to right: $V_1 = 6.0\U{V}$ and $V_2 =
3.0\U{V}$.

Applying Kirchhoff's junction rule to junction $a$ we have
$$I_1 + I_2 - I_3 = 0$$

Applying Kirchhoff's loop rule to the left-hand loop we have
$$V_1 - I_1 (R_1 + R_2) + R_3 I_2 = 0$$
where we \emph{add} the voltage change over $R_3$ because we cross it
\emph{against} the direction of the current $I_2$.

Applying Kirchhoff's loop rule to the right-hand loop we have
$$V_2 - R_4 I_3 - R_3 I_2 - R_5 I_5 = V_2 - I_3 (R_4 + R_5) - R_3 I_2 = 0$$

We now have three equations for three unknowns (the $I_i$).
Solving the loop rools for $I_1$ and $I_3$ we have
\begin{align*}
  I_1 &= \frac{V_1 + R_3 I_2}{R_1 + R_2} = \frac{V_1 + R_3 I_2}{R_{12}} \\
  I_3 &= \frac{V_2 - R_3 I_2}{R_4 + R_5} = \frac{V_2 - R_3 I_2}{R_{45}}
\end{align*}
where we have used the equivalent resistances $R_{12} \equiv R_1 +
R_2$ and $R_{45} \equiv R_4 + R_5$ to save writing later.  We can then
plug those currents into the junction rule and solve for $I_2$
\begin{align*}
  \frac{V_1 + R_3 I_2}{R_{12}} + I_2 - \frac{V_2 - R_3 I_2}{R_{45}} &= 0 \\
  \frac{V_1}{R_{12}} + \frac{R_3}{R_{12}} I_2 + I_2 
    - \frac{V_2}{R_{45}} + \frac{R_3}{R_{45}}I_2 &= 0 \\
  \p({\frac{R_3}{R_{12}} + 1 + \frac{R_3}{R_{45}}})\cdot I_2
    &= \frac{V_2}{R_{45}} - \frac{V_1}{R_{12}} \\
  I_2 &= \frac{\frac{V_2}{R_{45}} - \frac{V_1}{R_{12}}}{\frac{R_3}{R_{12}}
    + 1 + \frac{R_3}{R_{45}}} \\
  I_2 &= \ans{-28\U{mA}}
\end{align*}
Where the $-$ sign means the true current is in the opposite direction
to the one we have assigned (so the true current flows upward in the
figure).  We can now plug this current in to find $I_1$ and $I_3$.
\begin{align*}
  I_1 &= \frac{V_1 + R_3 I_2}{R_{12}} = \ans{292\U{mA}} \\
  I_3 &= \frac{V_2 - R_3 I_2}{R_{45}} = \ans{264\U{mA}}
\end{align*}

Double-checking our algebra, we see
 $I_1 + I_2 - I_3 = 292 - 27 - 264 = -1\U{mA} \approx 0$
where difference of $1\U{mA}$ is due to rounding errors from forcing
our answers to milli-Volt precision.
\end{solution}
