\begin{problem*}{21.40}
A dead battery is charged by connecting it to the live battery of
another car with jumper cables (Fig.~P21.40).  Determine the current
in the starter and in the dead battery.
\end{problem*} % problem 21.40

\begin{solution}
Let $V_L = 12\U{V}$ and $R_L = 0.01\Omega$ be the parameters of the
live battery, $V_D = 10\U{V}$ and $R_D = 1.00\Omega$ be the parameters
of the dead battery, and $R_S = 0.06\Omega$ be the resistance of the
starter.  Let $I_L$ be the current going upward in the left branch,
$I_D$ be the current going upward in the middle branch, and $I_S$ be
the current going downward in the right branch.

Applying Kirchhoff's junction rule to the top node, we have
\begin{equation}
 I_L + I_D - I_S = 0
\end{equation}

Applying Kirchhoff's loop rule to the outer and right loops
respectively, we have
\begin{align}
 V_L - I_L R_L - I_S R_S &= 0 \\
 V_D - I_D R_D - I_S R_S &= 0
\end{align}

Solving these using linear algebra (or your method of choice)
\begin{align}
 \begin{pmatrix}
  0 \\
  V_L \\
  V_D
 \end{pmatrix}
    &=
      \begin{pmatrix}
       1 & 1 & -1 \\
       R_L & 0 & R_S \\
       0 & R_D & R_S
      \end{pmatrix}
      \begin{pmatrix}
       I_L \\
       I_D \\
       I_S
      \end{pmatrix}
    =
      \begin{pmatrix}
       1 & 1 & -1 \\
       0.01 & 0 & 0.06 \\
       0 & 1.00 & 0.06
      \end{pmatrix}
      \begin{pmatrix}
       I_L \\
       I_D \\
       I_S
      \end{pmatrix} \\
 \begin{pmatrix}
  I_L \\
  I_D \\
  I_S
 \end{pmatrix}
    &=
      \begin{pmatrix}
       0.850 & 15.0 & -0.850 \\
       0.0085 & -0.850 & 0.992 \\
       -0.142 & 14.2 & 0.142
      \end{pmatrix}
      \begin{pmatrix}
       0 \\
       12 \\
       10
      \end{pmatrix}
    =
     \ans{
      \begin{pmatrix}
       172 \\
       -0.283 \\
       171
      \end{pmatrix}
      \U{A}
     }
\end{align}
Where $I_2 < 0$ indicates that the current in the middle branch
actually flows downward, recharging the dead battery.
\end{solution}
