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

[course.git] / latex / problems / Giancoli_6 / problem19.58.tex

diff --git a/latex/problems/Giancoli_6/problem19.58.tex b/latex/problems/Giancoli_6/problem19.58.tex
index be0fa0ff6ab56d90b95afaa78594e6641a8442ea..a3d53c5fbbe5b488e7bb4cc14b5f7e1b1ec23fe7 100644 (file)
--- 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