1 \begin{problem*}{19.24} % internal resistance
2 Determine the terminal voltage of each battery in Fig.~19-44.
7 MultiTerminal ra = resistor(Label("$r_1 = 1.0\U{\Ohm}$", align=S));
8 MultiTerminal Ba = source(
9 ra.terminal[1], type=DC, label="$\mathcal{E}_1 = 12\U{V}$");
10 MultiTerminal rb = resistor(ra.terminal[0]+(0,-u),
11 Label("$r_2 = 2.0\U{\Ohm}$", align=S));
12 MultiTerminal Bb = source(
13 rb.terminal[1], type=DC, label="$\mathcal{E}_2 = 18\U{V}$");
14 MultiTerminal R = resistor(
15 Bb.terminal[1]+(0.5u,0.25u), dir=90, label="$R = 6.6\U{\Ohm}$");
16 wire(rb.terminal[0], ra.terminal[0], rlsq, dist=-24pt);
17 wire(Bb.terminal[1], R.terminal[0], rlsq);
18 wire(Ba.terminal[1], R.terminal[1], rlsq);
24 From Kirchhoff's loop rule
26 \mathcal{E}_1 - IR - \mathcal{E}_2 - Ir_2 - Ir_1 &= 0 \\
27 I(R+r_1+R_2) &= \mathcal{E}_1-\mathcal{E}_2 \\
28 I &= \frac{\mathcal{E}_1-\mathcal{E}_2}{R+r_1+r_2} = 0.625\U{A}
31 So the voltage across the top battery is
32 $$V_1 = \mathcal{E}_1 - I r_1 = \ans{17\U{V}} \qquad (17.375\U{V})$$
33 and the voltage across the bottom battery is
34 $$V_2 = \mathcal{E}_2 - I r_2 = \ans{11\U{V}} \qquad (10.75\U{V})$$