From: W. Trevor King <wking@drexel.edu>
Date: Tue, 25 Aug 2009 06:37:28 +0000 (-0400)
Subject: Added topic tags to some Giancoli-6 problems + minor typos
X-Git-Url: http://git.tremily.us/?a=commitdiff_plain;h=862315a1430a21238f3f3015519ba77f1394dfc6;p=course.git

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

diff --git a/latex/problems/Giancoli_6/problem17.40.tex b/latex/problems/Giancoli_6/problem17.40.tex
index e18f780..b5f14eb 100644
--- a/latex/problems/Giancoli_6/problem17.40.tex
+++ b/latex/problems/Giancoli_6/problem17.40.tex
@@ -1,10 +1,10 @@
-\begin{problem*}{40}
+\begin{problem*}{17.40} % capacitor circuits
 A $C_1 = 7.7\U{$\mu$F}$ capacitor is charged by a $V = 125\U{V}$
 battery (Fig. 17-29a) and then is disconnected from the battery.  When
 this capacitor ($C_1$) is then connected (Fig. 17-29b) to a second
 (initially uncharged) capacitor, $C_2$, the final voltage on each
 capacitor is $V_2 = 15\U{V}$.  What is the value of $C_2$?
-[\emph{Hint}: charge is conserved.]
+[\emph{Hint:} charge is conserved.]
 \end{problem*}
 
 \empaddtoprelude{
diff --git a/latex/problems/Giancoli_6/problem19.02.tex b/latex/problems/Giancoli_6/problem19.02.tex
index 05344df..a11b21a 100644
--- a/latex/problems/Giancoli_6/problem19.02.tex
+++ b/latex/problems/Giancoli_6/problem19.02.tex
@@ -1,4 +1,4 @@
-\begin{problem*}{2}
+\begin{problem*}{19.2} % resistor networks
 Four $1.5\U{V}$ cells are connected in series to a $12\U{\Ohm}$
 lightbulb.  If the resulting current is $0.45\U{A}$, what is the
 internal resistance of each cell, assuming they are identical and
@@ -6,11 +6,12 @@ neglecting the wires.
 \end{problem*}
 
 \begin{solution}
-This is simply an application of the procedure outlined in Question 13.
-The external resistance is the lightbulb $R_{ext}=12\U{\Ohm}$.
-The total internal resistance is the sum of all the individual cell resistances $r_{int}=6r$.
-The total voltage is the sum of all the individual cell voltages $V = 6\cdot 1.5\U{V} = 9\U{V}$.
-Putting these together we have
+This is simply an application of the procedure outlined in
+Question~13.  The external resistance is the lightbulb
+$R_{ext}=12\U{\Ohm}$.  The total internal resistance is the sum of all
+the individual cell resistances $r_{int}=6r$.  The total voltage is
+the sum of all the individual cell voltages $V = 6\cdot 1.5\U{V} =
+9\U{V}$.  Putting these together we have
 \begin{align*}
 r_{int} = 6r &= \frac{V}{I} - R_{ext} = \frac{9\U{V}}{0.45\U{A}} - 12\U{\Ohm} \\
  r &= \ans{1.3\U{\Ohm}}
diff --git a/latex/problems/Giancoli_6/problem19.07.tex b/latex/problems/Giancoli_6/problem19.07.tex
index f511132..189a1f2 100644
--- a/latex/problems/Giancoli_6/problem19.07.tex
+++ b/latex/problems/Giancoli_6/problem19.07.tex
@@ -1,4 +1,4 @@
-\begin{problem*}{7}
+\begin{problem*}{19.7} % resistor networks
 A $650\U{\Ohm}$ and a $2200\U{\Ohm}$ resistor are connected in series
 with a $12\U{V}$ battery.  What is the voltage across the
 $2200\U{\Ohm0}$ resistor?
diff --git a/latex/problems/Giancoli_6/problem19.15.tex b/latex/problems/Giancoli_6/problem19.15.tex
index 6dd4418..c636cbe 100644
--- a/latex/problems/Giancoli_6/problem19.15.tex
+++ b/latex/problems/Giancoli_6/problem19.15.tex
@@ -1,4 +1,4 @@
-\begin{problem*}{15}
+\begin{problem*}{19.15} % resistor networks
 Eight $7.0\U{W}$ Christmas tree lights are connected in series to each
 other and to a $110\U{V}$ source.  What is the resistance of each
 bulb.
diff --git a/latex/problems/Giancoli_6/problem19.24.tex b/latex/problems/Giancoli_6/problem19.24.tex
index 38fdf69..27083ee 100644
--- a/latex/problems/Giancoli_6/problem19.24.tex
+++ b/latex/problems/Giancoli_6/problem19.24.tex
@@ -1,4 +1,4 @@
-\begin{problem*}{24}
+\begin{problem*}{19.24} % internal resistance
 Determine the terminal voltage of each battery in Fig.~19-44.
 \begin{center}
 \begin{asy}
diff --git a/latex/problems/Giancoli_6/problem19.31.tex b/latex/problems/Giancoli_6/problem19.31.tex
index 771032c..ff58077 100644
--- 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}
diff --git a/latex/problems/Giancoli_6/problem19.58.tex b/latex/problems/Giancoli_6/problem19.58.tex
index be0fa0f..a3d53c5 100644
--- a/latex/problems/Giancoli_6/problem19.58.tex
+++ b/latex/problems/Giancoli_6/problem19.58.tex
@@ -1,9 +1,9 @@
-\begin{problem*}{58}
-  A $45\U{V}$ battery of negligable internal resistance is connected
-  to a $38\U{k\Ohm}$ and a $27\U{k\Ohm}$ resistor in series.  What
-  reading will a voltmeter, of internal resistance $95\U{k\Ohm}$,
-  give when used to measure the voltage across each resistor?  What is
-  the percent inaccuracy due to meter resistance for each case?
+\begin{problem*}{19.58} % internal resistance
+A $45\U{V}$ battery of negligable internal resistance is connected to
+a $38\U{k\Ohm}$ and a $27\U{k\Ohm}$ resistor in series.  What reading
+will a voltmeter, of internal resistance $95\U{k\Ohm}$, give when used
+to measure the voltage across each resistor?  What is the percent
+inaccuracy due to meter resistance for each case?
 \end{problem*}
 
 \begin{solution}
@@ -11,18 +11,18 @@ Case 1: \\
 The original situation looks like
 \begin{center}
 \begin{asy}
-  import Circ;
-  real u = 0.5cm;
-  TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$");
-  pair a = B.end+(0,u);
-  pair b = B.beg-(0,u);
-  TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$");
-  TwoTerminal Rb = resistor(Ra.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$");
-  TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -90, "", "$I$");
-  wire(Rb.end, I.beg, nsq);
-  wire(I.end, b, udsq);
-  wire(b, B.beg, nsq);
-  wire(a, B.end, nsq);
+import Circ;
+real u = 0.5cm;
+TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$");
+pair a = B.end+(0,u);
+pair b = B.beg-(0,u);
+TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$");
+TwoTerminal Rb = resistor(Ra.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$");
+TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -90, "", "$I$");
+wire(Rb.end, I.beg, nsq);
+wire(I.end, b, udsq);
+wire(b, B.beg, nsq);
+wire(a, B.end, nsq);
 \end{asy}
 \end{center}
 Using Kirchoff's loop rule
@@ -40,23 +40,23 @@ Case 2: \\
 With the voltmeter across $R_1$ we have
 \begin{center}
 \begin{asy}
-  import Circ;
-  real u = 0.5cm;
-  TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$");
-  pair a = B.end+(0,u);
-  pair b = B.beg-(0,u);
-  TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$");
-  TwoTerminal Ia = current(Ra.end, 0, "", "$I_1$");
-  TwoTerminal Rv = resistor(a+(0,4u), normal, 0, "$95\U{k\Ohm}$", "$R_v$");
-  TwoTerminal Iv = current(Rv.end, 0, "", "$I_v$");
-  TwoTerminal Rb = resistor(Ia.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$");
-  TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -90, "", "$I_T$");
-  wire(Rb.end, I.beg, nsq);
-  wire(I.end, b, udsq);
-  wire(b, B.beg, nsq);
-  wire(a, B.end, nsq);
-  wire(a, Rv.beg, nsq);
-  wire(Iv.end, Ia.end, rlsq);
+import Circ;
+real u = 0.5cm;
+TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$");
+pair a = B.end+(0,u);
+pair b = B.beg-(0,u);
+TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$");
+TwoTerminal Ia = current(Ra.end, 0, "", "$I_1$");
+TwoTerminal Rv = resistor(a+(0,4u), normal, 0, "$95\U{k\Ohm}$", "$R_v$");
+TwoTerminal Iv = current(Rv.end, 0, "", "$I_v$");
+TwoTerminal Rb = resistor(Ia.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$");
+TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -90, "", "$I_T$");
+wire(Rb.end, I.beg, nsq);
+wire(I.end, b, udsq);
+wire(b, B.beg, nsq);
+wire(a, B.end, nsq);
+wire(a, Rv.beg, nsq);
+wire(Iv.end, Ia.end, rlsq);
 \end{asy}
 \end{center}
 Using our formula for resistors in parallel, we can bundle $R_v$ and $R_1$ into a single resistor $R_1'$, where
diff --git a/latex/problems/Giancoli_6/question19.04.tex b/latex/problems/Giancoli_6/question19.04.tex
index 2d29fcf..492ae00 100644
--- a/latex/problems/Giancoli_6/question19.04.tex
+++ b/latex/problems/Giancoli_6/question19.04.tex
@@ -1,4 +1,4 @@
-\begin{problem*}{Q4}
+\begin{problem*}{Q19.4} % resistor networks
 Two lightbulbs of resistance $R_1$ and $R_2$ ($R_2 > R_1$) are
 connected in series.  Which is brighter?  What if they are connected
 in parallel?  Explain.
diff --git a/latex/problems/Giancoli_6/question19.07.tex b/latex/problems/Giancoli_6/question19.07.tex
index e395d38..2e679b5 100644
--- a/latex/problems/Giancoli_6/question19.07.tex
+++ b/latex/problems/Giancoli_6/question19.07.tex
@@ -1,4 +1,4 @@
-\begin{problem*}{Q7}
+\begin{problem*}{Q19.7} % resistor networks
 If two identical resistors are connected in series to a battery, does
 the battery have to supply more power or less power than when only one
 of the resistors is connected?  Explain.
diff --git a/latex/problems/Giancoli_6/question19.13.tex b/latex/problems/Giancoli_6/question19.13.tex
index 0143b6f..76338d3 100644
--- a/latex/problems/Giancoli_6/question19.13.tex
+++ b/latex/problems/Giancoli_6/question19.13.tex
@@ -1,4 +1,4 @@
-\begin{problem*}{Q13}
+\begin{problem*}{Q19.13} % internal resistance
 Explain in detail how you could measure the internal resistance of a
 battery.
 \end{problem*}