From a9ac7ef3dbb1e3e9f6a09d162c277c4ec86ab6eb Mon Sep 17 00:00:00 2001
From: "W. Trevor King" <wking@drexel.edu>
Date: Wed, 11 Apr 2012 13:07:27 -0400
Subject: [PATCH] Update MetaPost to Asymptote Giancoli v6's problem 17.40.

---
 latex/problems/Giancoli_6/problem17.40.tex | 124 +++++++++------------
 1 file changed, 55 insertions(+), 69 deletions(-)

diff --git a/latex/problems/Giancoli_6/problem17.40.tex b/latex/problems/Giancoli_6/problem17.40.tex
index b5f14eb..3a579a6 100644
--- a/latex/problems/Giancoli_6/problem17.40.tex
+++ b/latex/problems/Giancoli_6/problem17.40.tex
@@ -5,81 +5,67 @@ 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.]
-\end{problem*}
+\begin{center}
+\begin{asy}
+%    C1         C1
+% +--||--+   +--||--+
+% |      |   |      |
+% +--i|--+   +--||--+
+%    V          C2
+%
+%   (a)        (b)
+import Circ;
 
-\empaddtoprelude{
-  input makecirc; % circuit drawing functions
-  initlatex("");
-  pair A, B, Cl, Cr, Dl, Dr, El, Er;
-  numeric a;
-  a := 3cm;
-  A := (-3a/4,0);
-  B := (3a/4, 0);
-  def drawA = 
-    % add elements
-    centreof.b(A+(1,0), A-(1,0), bat);
-    battery.B(c.b, phi.b, "V", "");
-    centreof.c(B.B.n, B.B.p, cap);
-    capacitor.C(c.c+(0,a), normal, phi.c, "C_1", "");
-    % add wiring along the bottom
-    wire(B.B.n, A+(a/2,0), nsq);
-    wire(B.B.p, A-(a/2,0), nsq);
-    % add wiring along the sides and top
-    wire(A+(a/2,0), C.C.l, udsq);
-    wire(A-(a/2,0), C.C.r, udsq);
-    puttext.bot("($a$)", A-(0,24pt));
-    Cl = C.C.r;
-    Cr = C.C.l;
-  enddef;
-  def drawB = 
-    % add elements
-    centreof.d(B+(1,0), B-(1,0), cap);
-    capacitor.D(c.d, normal, phi.d, "C_2", "");
-    centreof.e(C.D.l, C.D.r, cap);
-    capacitor.E(c.e+(0,a), normal, phi.e, "C_1", "");
-    % add wiring along the bottom
-    wire(C.D.l, B+(a/2,0), nsq);
-    wire(C.D.r, B-(a/2,0), nsq);
-    % add wiring along the sides and top
-    wire(B+(a/2,0), C.E.l, udsq);
-    wire(B-(a/2,0), C.E.r, udsq);
-    puttext.bot("($b$)", B-(0,24pt));
-    Dl = C.D.r;
-    Dr = C.D.l;
-    El = C.E.r;
-    Er = C.E.l;
-  enddef;
-}
+real u = 1cm;
+real dx = u;
+real dy = 1.5u;
 
-\begin{nosolution}
-\begin{center}
-\begin{empfile}[5p]
-\begin{emp}(0cm, 0cm)
-  drawA;
-  drawB;
-\end{emp}
-\end{empfile}
+TwoTerminal bat = battery("$V$");
+TwoTerminal c1 = capacitor("$C_1$", draw=false);
+centerto(bat, c1, dy); c1.draw();
+wire(bat.end, c1.end, rlsq, dx/2);
+wire(bat.beg, c1.beg, rlsq, -dx/2);
+label("(a)", bat.mid + (0, -dy/2));
+
+c1.shift(3*dx); c1.draw();
+TwoTerminal c2 = capacitor("$C_2$", draw=false);
+centerto(c1, c2, -dy); c2.draw();
+wire(c2.end, c1.end, rlsq, dx/2);
+wire(c2.beg, c1.beg, rlsq, -dx/2);
+label("(b)", c2.mid + (0, -dy/2));
+\end{asy}
 \end{center}
-\end{nosolution}
+\end{problem*}
 
 \begin{solution}
 \begin{center}
-\begin{empfile}[5]
-\begin{emp}(0cm, 0cm)
-  drawA;
-  drawB;
-  puttext.ulft("$Q_{1a}$", Cl);
-  puttext.urt("$-Q_{1a}$", Cr);
-  puttext.ulft("$Q_{2b}$", Dl);
-  puttext.urt("$-Q_{2b}$", Dr);
-  puttext.ulft("$Q_{1b}$", El);
-  puttext.urt("$-Q_{1b}$", Er);
-  labeloffset := 9pt;
-  puttext.top("$V$", (Cl+Cr)/2);
-  puttext.top("$V_2$", (Dl+Dr)/2);
-  puttext.top("$V_2$", (El+Er)/2);
-\end{emp}
-\end{empfile}
+\begin{asy}
+import Circ;
+
+real u = 1cm;
+real dx = u;
+real dy = 1.5u;
+
+TwoTerminal bat = battery("$V$");
+TwoTerminal c1 = capacitor("$C_1$", draw=false);
+centerto(bat, c1, dy); c1.draw();
+wire(bat.end, c1.end, rlsq, dx/2);
+wire(bat.beg, c1.beg, rlsq, -dx/2);
+label("(a)", bat.mid + (0, -dy/2));
+label("$Q_{1a}$", c1.end, NE);
+label("$-Q_{1a}$", c1.beg, NW);
+
+c1.shift(3*dx); c1.draw();
+TwoTerminal c2 = capacitor("$C_2$", draw=false);
+centerto(c1, c2, -dy); c2.draw();
+wire(c2.end, c1.end, rlsq, dx/2);
+wire(c2.beg, c1.beg, rlsq, -dx/2);
+label("(b)", c2.mid + (0, -dy/2));
+label("$Q_{1b}$", c1.end, NE);
+label("$-Q_{1b}$", c1.beg, NW);
+label("$Q_{2b}$", c2.end, SE);
+label("$-Q_{2b}$", c2.beg, SW);
+\end{asy}
 \end{center}
 
 Because the voltage drop across $C_1$ in situation $a$ is the same as
-- 
2.26.2