1 \begin{problem*}{26.18}
2 Find \Part{a} the equivalent capacitance of the capacitors in
3 Figure~P26.18, \Part{b} the charge on each capacitor, and \Part{c} the
4 potential difference across each capacitor.
17 TwoTerminal B = battery((0,0), ang=180, "$V$", "9.00\U{V}");
18 TwoTerminal C1 = capacitor("$C_1$", "8.00\U{$\mu$F}", draw=false);
19 TwoTerminal C2 = capacitor("$C_2$", "6.00\U{$\mu$F}", draw=false);
20 TwoTerminal C3 = capacitor("$C_3$", "2.00\U{$\mu$F}", draw=false);
21 TwoTerminal C4 = capacitor("$C_4$", "8.00\U{$\mu$F}", draw=false);
23 centerto(B, C3, offset=3u, reverse=true); C3.draw();
24 centerto(C3, C2, offset=2u); C2.draw();
25 pair a = (B.end.x, (C2.beg.y + C3.beg.y)/2);
26 pair b = (B.beg.x, a.y);
27 C1.centerto(B.end, a, offset=2u); C1.draw();
28 C4.centerto(B.beg, b, offset=-2u); C4.draw();
30 a = (a.x - u, a.y); // shift to split x difference between C1.end and B.end
33 wire(B.end, C1.beg, rlsq);
34 wire(C1.end, a, udsq);
35 wire(a, C2.beg, udsq);
36 wire(a, C3.beg, udsq);
38 wire(C2.end, b, rlsq);
39 wire(C3.beg, b, rlsq);
40 wire(b, C4.end, rlsq);
42 wire(C4.beg, B.beg, udsq);
49 Bundling down equivalent capacitances,
56 TwoTerminal B = battery((0,0), ang=180, "$V$");
57 TwoTerminal C1 = capacitor("$C_1$", draw=false);
58 TwoTerminal C23 = capacitor("$C_{23}$", draw=false);
59 TwoTerminal C4 = capacitor("$C_4$", draw=false);
60 centerto(B, C23, offset=2u, reverse=true); C23.draw();
61 pair a = (B.end.x, C23.beg.y);
62 pair b = (B.beg.x, a.y);
63 C1.centerto(B.end, a, offset=u); C1.draw();
64 C4.centerto(B.beg, b, offset=-u); C4.draw();
65 wire(B.end, C1.beg, rlsq);
66 wire(C1.end, C23.beg, udsq);
67 wire(C23.end, C4.end, rlsq);
68 wire(C4.beg, B.beg, udsq);
72 TwoTerminal C1234 = capacitor("$C_{1234}$", draw=false);
73 centerto(B, C1234, offset=2u, reverse=true); C1234.draw();
74 wire(B.end, C1234.beg, rlsq, dist=-u/2);
75 wire(B.beg, C1234.end, rlsq, dist=+u/2);
77 pair c = ((C4.mid.x + B.end.x - u/2)/2, C4.mid.y);
78 draw((c-(u,0)) -- (c+(u,0)), kirchhoff_pen, Arrows);
81 $C_{23}$ comes from $C_2$ and $C_3$ in parallel, so the equivalent
84 C_{23} = C_2 + C_3 = (6.00 + 2.00)\U{$\mu$F} = 8.00\U{$\mu$F} \;.
86 $C_{1234}$ comes from $C_1$, $C_{23}$, and $C_4$ in series, so the
87 equivalent capacitance is
89 C_{1234} = \p({\frac{1}{C_1} + \frac{1}{C_{23}} + \frac{1}{C_4}})^{-1}
90 = \frac{8.00\U{$\mu$F}}{3} = \ans{2.67\U{$\mu$F}} \;.
94 The equivalent capacitor has the total battery voltage across itself,
95 so it carries a charge of
97 Q_{1234} = C_{1234} V = (2.67\U{$\mu$F}) \cdot (9.00\U{V})
100 Because $C_1$, $C_{23}$, and $C_3$ are in series, they each have the
101 same $Q$ as the equivalent capacitor.
103 Q_1 = Q_{23} = Q_3 = Q_{1234} = \ans{24.0\U{$\mu$C}} \;.
105 The voltage across $C_{23}$ is therefore
107 V_{23} = \frac{Q_{23}}{C_{23}} = \frac{24.0\U{$\mu$C}}{8.00\U{$\mu$F}}
110 $C_2$ and $C_3$ are in parallel, so they each have the total $V_{23}$
113 Q_2 &= C_2 V_{23} = 6.00\U{$\mu$F} \cdot 3.00\U{V} = \ans{18.0\U{$\mu$C}} \\
114 Q_3 &= C_3 V_{23} = 2.00\U{$\mu$F} \cdot 3.00\U{V} = \ans{6.00\U{$\mu$C}} \;.
116 Note that $Q_2+Q_3=Q_{23}$, which must be true because charge is conserved.
119 During \Part{b}, we already found $\ans{V_{23}=V_2=V_3=3.00\U{V}}$.
120 Using the charges on $C_1$ and $C_4$, we find
122 V_1 &= \frac{Q_1}{C_1} = \frac{24.0\U{$\mu$C}}{8.00\U{$\mu$F}}
124 V_4 &= \frac{Q_4}{C_4} = \frac{24.0\U{$\mu$C}}{8.00\U{$\mu$F}}
125 = \ans{3.00\U{V}} \;.