\begin{problem*}{19.24} % internal resistance
Determine the terminal voltage of each battery in Fig.~19-44.
\begin{center}
\begin{asy}
  import Circ;
  real u = 3cm;
  TwoTerminal ra = resistor((0,0), normal, 0, "$r_1 = 1.0\U{\Ohm}$", "");
  TwoTerminal Ba = source(ra.end, DC, 0, "", "$\mathcal{E}_1 = 12\U{V}$");
  TwoTerminal rb = resistor(ra.beg+(0,-u), normal, 0, "$r_2 = 2.0\U{\Ohm}$", "");
  TwoTerminal Bb = source(rb.end, DC, 0, "", "$\mathcal{E}_2 = 18\U{V}$");
  TwoTerminal R = resistor(Bb.end+(0.5u,0.25u), normal, 90, "", "$R = 6.6\U{\Ohm}$");
  wireU(rb.beg, ra.beg, -24pt, rlsq);
  wire(Bb.end, R.beg, rlsq);
  wire(Ba.end, R.end, rlsq);
\end{asy}
\end{center}
\end{problem*}

\begin{solution}
From Kirchhoff's loop rule
\begin{align*}
  \mathcal{E}_1 - IR - \mathcal{E}_2 - Ir_2 - Ir_1 &= 0 \\
  I(R+r_1+R_2) &= \mathcal{E}_1-\mathcal{E}_2 \\
  I &= \frac{\mathcal{E}_1-\mathcal{E}_2}{R+r_1+r_2} = 0.625\U{A}
\end{align*}

So the voltage across the top battery is
$$V_1 = \mathcal{E}_1 - I r_1 = \ans{17\U{V}} \qquad (17.375\U{V})$$
and the voltage across the bottom battery is
$$V_2 = \mathcal{E}_2 - I r_2 = \ans{11\U{V}} \qquad (10.75\U{V})$$
\end{solution}
