1 \begin{problem*}{21.32}
2 Four resistors are connected to a battery as shown in Figure P21.32.
3 The current in the battery is $I$, the battery emf is $\epsilon$, and
4 the resistor values are $R_1 = R$, $R_2 = 2R$, $R_3 = 4R$, and $R_4 =
6 \Part{a} Rank the resistors according to the potential difference
7 across them, form largest to smallest. Note any cases of equal
9 \Part{b} Determine the potential difference across each resistor in
11 \Part{c} Rank the resistors according to the current in them, from
12 largest to smallest. Note any cases of equal current.
13 \Part{d} Determine the current in each resistor in terms of $I$.
14 \Part{e} If $R_3$ is increased, what happens to the current in each of
16 \Part{f} In the limit that $R_3 \rightarrow \infty$, what are the new
17 values of the current in each resistor in terms of $I$, the original
18 current in the battery?
22 input makecirc; % circuit drawing functions
28 battery.B(origin, 90, "\mathcal{E}", "");
29 resistor.A(B.B.p, normal, 90, "R_1", "$R$");
30 N[0] := B.B.n + (dx,0);
31 N[1] := R.A.r + (dx,0);
32 centerto.A(R.A.l, R.A.r, dx, res);
33 resistor.D(A, normal, 90, "R_4", "$3R$");
34 resistor.B(N[1]+(dx/2,0), normal, 0, "R_2", "$2R$");
35 resistor.C(N[0]+(dx/2,0), normal, 0, "R_3", "$4R$");
36 centreof.B(R.A.r, N[1], cur);
37 current.A(c.B, phi.B, "I", "");
38 centreof.C(R.B.r, R.C.r, cur);
39 current.B(c.C, phi.C, "I_2", "");
40 centreof.D(R.D.l, N[0], cur);
41 current.C(c.D, phi.D, "I_3", "");
42 wire(R.A.r, R.D.r,rlsq);
43 wire(B.B.n, R.D.l,rlsq);
44 wire(N[1], R.B.l,rlsq);
45 wire(N[0], R.C.l,rlsq);
46 wire(R.B.r, R.C.r,nsq);
47 draw N[0] withpen pencircle scaled 3pt;
48 draw N[1] withpen pencircle scaled 3pt;
56 $R_2$ and $R_3$ both have $I_2$ passing through them, so from Ohm's
57 law we know $V_2=I_2 R_2 < V_3=I_2 R_3$, because $R_2 < R_4$. $R_4$
58 and the equivalent resistance $R_s=R_2+R_3$ are in parallel, so they
59 have the same voltage across them. Because $V_4=V_s=V_2+V_4$, the
60 voltage $V_4$ across $R_4$ is greater than either $V_2$ or $V_3$.
61 Finally, the equivalent resistance of $R_4$ and $R_s$ in parallel is
64 R_p=\p({\frac{1}{R_4}+\frac{1}{R_s}})^{-1}
65 =\p({\frac{1}{3R}+\frac{1}{6R}})^{-1}
66 =3R\p({1+\frac{1}{2}})^{-1}
70 so $R_p > R_1$. Since both $R_p$ and $R_1$ have $I$ going through
71 them, and $V_p=V_4=V_s > V_1$. We still need to place $V_1$ relative
72 to $V_2$ and $V_3$, so we use the formula for voltage across series
75 I=\frac{V_A}{R_A}=\frac{V_B}{R_B} \;.
77 $R_p=2R_1$, so $V_1=V_p/2$, and $R_3=2R_2$, so $V_3=2V_2$.
78 $V_3 + V_2=V_p$, so $V_3=2/3\cdot V_p$ and $V_2=V_p/3$.
79 The final ranking is therefore $\ans{V_4=V_p > V_3=2/3\cdot V_p > V_1=V_p/2 > V_2=V_p/3}$.
82 We've done most of the work in \Part{a}.
84 \mathcal{E}=V_1 + V_p=\frac{3 V_p}{2} \;,
88 V_4 &= V_p =\ans{\frac{2\mathcal{E}}{3}} \\
89 V_1 &= \frac{V_p}{2}=\ans{\frac{\mathcal{E}}{3}} \\
90 V_2 &= \frac{V_p}{3}=\ans{\frac{2\mathcal{E}}{9}} \\
91 V_3 &= \frac{2V_p}{3}=\ans{\frac{4\mathcal{E}}{9}}
95 $I=I_2 + I_3$, and all our currents are positive as we've labled them,
96 so $I$ is greater than $I_2$ and $I_3$. $R_4=3R < R_s=6R$, so $I_3 >
97 I_2$. The final ranking is therefore $I > I_3 > I_2$, with $I_2$
98 passing through both $R_2$ and $R_3$.
101 To be quantitative about \Part{c}, we can use Ohm's law for each current:
103 I &= \frac{V_1}{R_1}=\frac{\mathcal{E}}{3}\cdot\frac{1}{R}=\frac{\mathcal{E}}{3R} \\
104 I_3 &= \frac{V_p}{R_4}=\frac{2\mathcal{E}}{3}\cdot\frac{1}{3R}=\frac{2}{3}\cdot\frac{\mathcal{E}}{3R}=\ans{\frac{2I}{3}}\\
105 I_2 &= \frac{V_p}{R_s}=\frac{2\mathcal{E}}{3}\cdot\frac{1}{6R}=\frac{1}{3}\cdot\frac{\mathcal{R}}{3R}=\ans{\frac{I}{3}}\;.
107 We see that $I=I_2+I_3$, as it should by Kirchhoff's junction rule.
110 If $R_3$ increases, $R_s$ increases and $R_p$ increases, so $I_2$ and
111 $I$ decrease. The change in $I_3$ is a balance of increased flow
112 relative to $I_2$ and decreased overall $I$. We see that less current
113 through $I$ drops $V_1$, but $V_1+V_4=\mathcal{E}$ which doesn't
114 change, so $V_4$ increases, so $I_3$ increases.
117 As $R_3 \rightarrow \infty$, $I_2$ is choked off entirely, so $I=I_3$.
118 So $I$ flows through $R_1$ and $R_4$, and nothing flows through $R_2$