Add solutions and graphics to Serway and Jewett v8's chapter 28 problems.

[course.git] / latex / problems / Serway_and_Jewett_8 / problem28.26.tex

diff --git a/latex/problems/Serway_and_Jewett_8/problem28.26.tex b/latex/problems/Serway_and_Jewett_8/problem28.26.tex
index 0d33b3cc4a7040b28b05b24453a1da5c276239a2..81162d1ff05ec816206c27469ba9b3b1da78dfd8 100644 (file)
--- a/latex/problems/Serway_and_Jewett_8/problem28.26.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem28.26.tex
@@ -6,9 +6,116 @@ The following equations describe an electric circuit:
   I_1 + I_3 - I_2 &= 0
 \end{align}
 \Part{a} Draw a [possible] diagram of the circuit.  \Part{b}
-Calculate the unknowns sand identify the physical meaning of each
+Calculate the unknowns and identify the physical meaning of each
 unknown.
 \end{problem*}
 
 \begin{solution}
+\Part{a}
+The last equation looks like a junction rule, for a junction like
+this:
+\begin{center}
+\begin{asy}
+import Circ;
+
+real dx = 6pt;
+
+MultiTerminal I1 = current((0, 0), label="$I_1$");
+real ilen = I1.terminal[1].x - I1.terminal[0].x;
+pair P = I1.terminal[1] + (dx, 0);
+MultiTerminal I2 = current(P, dir=-90, label=Label("$I_2$", align=dir(-20)));
+MultiTerminal I3 = current(
+    P + (dx, 0), dir=180, label=Label("$I_3$", align=N), draw=false);
+I3.shift((ilen, 0));
+I3.draw();
+
+wire(I1.terminal[1], I3.terminal[1]);
+wire(P, I2.terminal[1]);
+dot(P);
+\end{asy}
+\end{center}
+
+The first equation looks like a loop rule traveling over the $I_1$ and
+$I_2$ branches, and the second equation looks like a loop rule
+traveling over the $I_2$ and $I_3$ branches.  Luckily, the $I_2$
+portions match between these equations, and we can fill in the circuit
+elements.
+\begin{center}
+\begin{asy}
+import Circ;
+
+real u = 2cm;
+
+MultiTerminal R1 = resistor(dir=90, value="$220\U{\Ohm}$");
+MultiTerminal V1 = battery(R1.terminal[1], dir=90, value="$5.80\U{V}$");
+MultiTerminal R2 = resistor(value="$370\U{\Ohm}$", draw=false);
+R2.centerto(R1.terminal[0], V1.terminal[1], offset=-u);  R2.draw();
+pair a = (R2.center.x, V1.terminal[1].y);
+pair b = (a.x, R1.terminal[0].y);
+MultiTerminal R3 = resistor(
+    a+(u,0), dir=-90, value=Label("$150\U{\Ohm}$", align=E));
+MultiTerminal V3 = battery(b+(u,0), dir=90, value="$3.10\U{V}$");
+MultiTerminal I1 = current(label="$I_1$", draw=false);
+I1.centerto(V1.terminal[1], a);  I1.draw();
+MultiTerminal I2 = current(label="$I_2$", draw=false);
+I2.centerto(a, R2.terminal[1]);  I2.draw();
+MultiTerminal I3 = current(label=Label("$I_3$", align=N), draw=false);
+I3.centerto(R3.terminal[0], a);  I3.draw();
+wire(V1.terminal[1], I1.terminal[0]);
+wire(I1.terminal[1], I2.terminal[0], rlsq);
+wire(I2.terminal[1], R2.terminal[1]);
+wire(a, I3.terminal[1]);
+wire(I3.terminal[0], R3.terminal[0]);
+dot(a);
+wire(R1.terminal[0], R2.terminal[0], rlsq);
+wire(b, V3.terminal[0]);
+dot(b);
+\end{asy}
+\end{center}
+
+\Part{b}
+The unknowns are the currents through each branch of the circuit.  You
+can solve for them however you like.
+\begin{align}
+ \begin{pmatrix}
+  5.80\U{V} \\
+  3.10\U{V} \\
+  0
+ \end{pmatrix}
+  &=
+ \begin{pmatrix}
+  220\U{\Ohm} & 370\U{\Ohm} & 0 \\
+  0 & 370\U{\Ohm} & 150\U{\Ohm} \\
+  1\U{\Ohm} & -1\U{\Ohm} & 1\U{\Ohm}
+ \end{pmatrix}
+ \begin{pmatrix}
+  I_1 \\
+  I_2 \\
+  I_3
+ \end{pmatrix} \\
+ \begin{pmatrix}
+  I_1 \\
+  I_2 \\
+  I_3
+ \end{pmatrix}
+  =&
+ \begin{pmatrix}
+  220\U{\Ohm} & 370\U{\Ohm} & 0 \\
+  0 & 370\U{\Ohm} & 150\U{\Ohm} \\
+  1\U{\Ohm} & -1\U{\Ohm} & 1\U{\Ohm}
+ \end{pmatrix}^{-1}
+ \begin{pmatrix}
+  5.80\U{V} \\
+  3.10\U{V} \\
+  0
+ \end{pmatrix}
+  =
+ \ans{
+  \begin{pmatrix}
+    11.0 \\
+    9.10 \\
+    -1.87
+  \end{pmatrix} \U{mA}
+ } \;.
+\end{align}
 \end{solution}