1 \begin{problem*}{19.24} % internal resistance
2 Determine the terminal voltage of each battery in Fig.~19-44.
7 TwoTerminal ra = resistor((0,0), normal, 0, "$r_1 = 1.0\U{\Ohm}$", "");
8 TwoTerminal Ba = source(ra.end, DC, 0, "", "$\mathcal{E}_1 = 12\U{V}$");
9 TwoTerminal rb = resistor(ra.beg+(0,-u), normal, 0, "$r_2 = 2.0\U{\Ohm}$", "");
10 TwoTerminal Bb = source(rb.end, DC, 0, "", "$\mathcal{E}_2 = 18\U{V}$");
11 TwoTerminal R = resistor(Bb.end+(0.5u,0.25u), normal, 90, "", "$R = 6.6\U{\Ohm}$");
12 wireU(rb.beg, ra.beg, -24pt, rlsq);
13 wire(Bb.end, R.beg, rlsq);
14 wire(Ba.end, R.end, rlsq);
20 From Kirchoff's loop rule
22 \mathcal{E}_1 - IR - \mathcal{E}_2 - Ir_2 - Ir_1 &= 0 \\
23 I(R+r_1+R_2) &= \mathcal{E}_1-\mathcal{E}_2 \\
24 I &= \frac{\mathcal{E}_1-\mathcal{E}_2}{R+r_1+r_2} = 0.625\U{A}
27 So the voltage across the top battery is
28 $$V_1 = \mathcal{E}_1 - I r_1 = \ans{17\U{V}} \qquad (17.375\U{V})$$
29 and the voltage across the bottom battery is
30 $$V_2 = \mathcal{E}_2 - I r_2 = \ans{11\U{V}} \qquad (10.75\U{V})$$