1 \begin{problem*}{28.11}
2 A battery with $\EMF=6.00\U{V}$ and no internal resistance supplies
3 current to the circuit shown in Figure~P28.11. When the double-throw
4 switch $S$ is open as shown in the figure, the current in the battery
5 is $1.00\U{mA}$. When the switch is closed in position $a$, the
6 current in the battery is $1.20\U{mA}$. When the switch is closed in
7 position $b$, the current in the battery is $2.00\U{mA}$. Find the
8 resistances \Part{a} $R_1$, \Part{b} $R_2$, and \Part{c} $R_3$.
10 % +---R1---+----R2----+
17 % +--------+----------+
21 MultiTerminal S = switchSPDT(dir=180, label=Label("$S$", align=dir(-70)));
22 label("$a$", S.terminal[2], align=W);
23 label("$b$", S.terminal[1], align=W);
24 MultiTerminal R2v = resistor(S.terminal[2], dir=90, "$R_2$");
25 MultiTerminal R1 = resistor(
26 R2v.terminal[1], dir=180, label=Label("$R_1$", align=N));
27 MultiTerminal R2h = resistor(R2v.terminal[1], "$R_2$");
28 MultiTerminal R3 = resistor(
29 (R2h.terminal[1].x, S.terminal[0].y), dir=-90, "$R_3$");
30 MultiTerminal B = battery(label=Label("$\EMF$", align=W), draw=false);
31 B.centerto(R1.terminal[1], (R1.terminal[1].x, R3.terminal[1].y));
33 wire(R1.terminal[1], B.terminal[0]);
34 wire(B.terminal[1], R3.terminal[1], udsq);
35 pair bx = (S.terminal[1].x, R3.terminal[1].y);
36 wire(S.terminal[1], bx);
38 wire(S.terminal[0], R3.terminal[0]);
39 wire(R3.terminal[0], R2h.terminal[1]);
47 When the switch is in position $b$, the vertical $R_2$ resistor
48 recieves no current. Applying Kirchhoff's loop rule gives
50 0 &= \EMF - I R_3 - I R_2 - I R_1 \\
51 \frac{\EMF}{I} &= R_1 + R_2 + R_3 \;.
54 When the switch is in position $a$, the two $R_2$ resistors are in
55 parallel, so they can be replaced by an equivalent resistance
57 R_2' = (1/R_2 + 1/R_2)^{-1} = R_2/2 \;.
59 After you've made this replacement, there is only a single loop in the
60 circuit. Applying Kirchhoff's loop rule gives
62 0 &= \EMF - I_a R_3 - I_a R_2' - I_a R_1 \\
63 \frac{\EMF}{I_a} &= R_1 + \frac{R_2}{2} + R_3 \;.
66 When the switch is in position $b$, the vertical $R_2$ resistor
67 recieves no current and $R_3$ is shorted. Kirchoff's loop rule gives
69 0 &= \EMF - I_b R_2 - I_b R_1 \\
70 \frac{\EMF}{I_b} &= R_1 + R_2 \;.
73 This gives three equations with three unknowns. Solve however you
77 \frac{6.00\U{V}}{1.00\E{-3}\U{A}} \\
78 \frac{6.00\U{V}}{1.20\E{-3}\U{A}} \\
79 \frac{6.00\U{V}}{2.00\E{-3}\U{A}}
84 1 & \frac{1}{2} & 1 \\
100 1 & \frac{1}{2} & 1 \\