Added topic tags to some Giancoli-6 problems + minor typos

[course.git] / latex / problems / Giancoli_6 / problem19.31.tex

diff --git a/latex/problems/Giancoli_6/problem19.31.tex b/latex/problems/Giancoli_6/problem19.31.tex
index 771032caa1788dc4c6b436d0d1c8c949e4a0db94..ff5807769764a1b54e6a29990b0b8f6df7d495df 100644 (file)
--- a/latex/problems/Giancoli_6/problem19.31.tex
+++ b/latex/problems/Giancoli_6/problem19.31.tex
@@ -1,24 +1,24 @@
-\begin{problem*}{31}
-  Calculate the currents in each resistor of Fig.~19-49.
+\begin{problem*}{19.31} % resistor networks
+Calculate the currents in each resistor of Fig.~19-49.
 \end{problem*}
 
 \begin{nosolution}
 \begin{center}
 \begin{asy}
-  import Circ;
-  real u = 3cm;
-  TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$");
-  TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", "");
-  TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$");
-  pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y);
-  TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", "");
-  TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$");
-  TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", "");
-  TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", "");
-  wire(Ba.end, Raa.end, rlsq);
-  wire(Rab.beg, Jbot, nsq);
-  wire(Jbot, Rb.end, nsq);
-  wire(Jbot, Rcb.end, rlsq);
+import Circ;
+real u = 3cm;
+TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$");
+TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", "");
+TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$");
+pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y);
+TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", "");
+TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$");
+TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", "");
+TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", "");
+wire(Ba.end, Raa.end, rlsq);
+wire(Rab.beg, Jbot, nsq);
+wire(Jbot, Rb.end, nsq);
+wire(Jbot, Rcb.end, rlsq);
 \end{asy}
 \end{center}
 \end{nosolution}
@@ -26,24 +26,24 @@
 \begin{solution}
 \begin{center}
 \begin{asy}
-  import Circ;
-  TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$");
-  TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", "");
-  TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$");
-  pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y);
-  TwoTerminal Ic = current((Jbot+Rcb.end)/2, 0, "", "$I_3$");
-  TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", "");
-  TwoTerminal Ib = current(Rb.end, -90, "", "$I_2$");
-  TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$");
-  TwoTerminal Ia = current(Ba.end, 180, "$I_1$", "");
-  TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", "");
-  TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", "");
-  wire(Ia.end, Raa.end, rlsq);
-  wire(Jbot, Ib.end, nsq);
-  wire(Jbot, Ic.beg, nsq);
-  wire(Ib.end, Rb.end, nsq);
-  wire(Ic.end, Rcb.end, rlsq);
-  dot("a", Jbot, S);
+import Circ;
+TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$");
+TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", "");
+TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$");
+pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y);
+TwoTerminal Ic = current((Jbot+Rcb.end)/2, 0, "", "$I_3$");
+TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", "");
+TwoTerminal Ib = current(Rb.end, -90, "", "$I_2$");
+TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$");
+TwoTerminal Ia = current(Ba.end, 180, "$I_1$", "");
+TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", "");
+TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", "");
+wire(Ia.end, Raa.end, rlsq);
+wire(Jbot, Ib.end, nsq);
+wire(Jbot, Ic.beg, nsq);
+wire(Ib.end, Rb.end, nsq);
+wire(Ic.end, Rcb.end, rlsq);
+dot("a", Jbot, S);
 \end{asy}
 \end{center}
 Label the resistors from left to right: $R_1 = 12\U{\Ohm}$, $R_2 =
@@ -75,9 +75,12 @@ R_2$ and $R_{45} \equiv R_4 + R_5$ to save writing later.  We can then
 plug those currents into the junction rule and solve for $I_2$
 \begin{align*}
   \frac{V_1 + R_3 I_2}{R_{12}} + I_2 - \frac{V_2 - R_3 I_2}{R_{45}} &= 0 \\
-  \frac{V_1}{R_{12}} + \frac{R_3}{R_{12}} I_2 + I_2 - \frac{V_2}{R_{45}} + \frac{R_3}{R_{45}}I_2 &= 0 \\
-  \p({\frac{R_3}{R_{12}} + 1 + \frac{R_3}{R_{45}}})\cdot I_2 &= \frac{V_2}{R_{45}} - \frac{V_1}{R_{12}} \\
-  I_2 &= \frac{\frac{V_2}{R_{45}} - \frac{V_1}{R_{12}}}{\frac{R_3}{R_{12}} + 1 + \frac{R_3}{R_{45}}} \\
+  \frac{V_1}{R_{12}} + \frac{R_3}{R_{12}} I_2 + I_2 
+    - \frac{V_2}{R_{45}} + \frac{R_3}{R_{45}}I_2 &= 0 \\
+  \p({\frac{R_3}{R_{12}} + 1 + \frac{R_3}{R_{45}}})\cdot I_2
+    &= \frac{V_2}{R_{45}} - \frac{V_1}{R_{12}} \\
+  I_2 &= \frac{\frac{V_2}{R_{45}} - \frac{V_1}{R_{12}}}{\frac{R_3}{R_{12}}
+    + 1 + \frac{R_3}{R_{45}}} \\
   I_2 &= \ans{-28\U{mA}}
 \end{align*}
 Where the $-$ sign means the true current is in the opposite direction
@@ -88,5 +91,8 @@ figure).  We can now plug this current in to find $I_1$ and $I_3$.
   I_3 &= \frac{V_2 - R_3 I_2}{R_{45}} = \ans{264\U{mA}}
 \end{align*}
 
-Double-checking our algebra, we see $I_1 + I_2 - I_3 = 292 - 27 - 264 = -1\U{mA} \approx 0$ where difference of $1\U{mA}$ is due to rounding errors from forcing our answers to milli-Volt precision.  
+Double-checking our algebra, we see
+ $I_1 + I_2 - I_3 = 292 - 27 - 264 = -1\U{mA} \approx 0$
+where difference of $1\U{mA}$ is due to rounding errors from forcing
+our answers to milli-Volt precision.
 \end{solution}