diff --git a/Chapter1.pdf b/Chapter1.pdf index 605907c..401839c 100644 Binary files a/Chapter1.pdf and b/Chapter1.pdf differ diff --git a/Chapter1.tex b/Chapter1.tex index a7cbf1f..7d3794e 100644 --- a/Chapter1.tex +++ b/Chapter1.tex @@ -34,11 +34,12 @@ \item To treat $R_2$ and $R_{load}$ as a single resistor, combine the two resistors which are in parallel to find that the combined (equivalent) resistance is $5\k\Ohm$. Now, we have a simple voltage divider with a $10\k\Ohm$ resistor in series with the $5\k\Ohm$ equivalent resistor. The output voltage is across this equivalent resistance. The output voltage is given by - \[V_{out} = V_{in} \frac{5\k\Ohm}{10\k\Ohm + 5\k\Ohm} = \frac{30\V}{3} = 10\V \] + \[V_{out} = V_{in} \frac{5\k\Ohm}{10\k\Ohm + 5\k\Ohm} = \frac{30\V}{3} = \mans{10\V} \] \item - We can redraw the voltage divider circuit to make the ``port'' clearer. - \begin{circuit} + We can redraw the voltage divider circuit to make the ``port'' clearer. + \begin{circuit}{fig:1.10.1}{Voltage divider with port shown.} + % \label{1.10fig1} (0,2) to[V=$V_{\in}$] (0,0) to[short] (2,0) to[R=$R_2$] (2,2) @@ -50,15 +51,26 @@ We can find $V_{\Th}$ by leaving the ports open (open circuit) and measuring $V_\out$, the voltage across $R_2$. This comes out to be half the input voltage when $R_1 = R_2$, so $V_\out = 15\V$. Thus $V_{\Th} = 15\V$. - To find the Th\'evinen resistance, we need to find the short circuit current, $I_{SC}$. + To find the Th\'evinen resistance, we need to find the short circuit current, $I_{SC}$. We short circuit the port and measure the current flowing through it. + \begin{circuit}{fig:1.10.2}{Voltage divider with short circuit on the output.} + (0,2) to[V=$V_{\in}$] (0,0) + to[short] (2,0) + to[R=$R_2$] (2,2) + to[R=$R_1$](0,2) + (2,0) to[short] (3,0) + (2,2) to[short] (3,2) + (3,0) to[short, i_<=$I_{SC}$] (3,2) + \end{circuit} - \todo{finish} + In this circuit, no current flows through $R_2$, flowing through the short instead. Thus we have $I_{SC} = \dfrac{V_\in }{R_1}$. From this, we can find $R_\Th$ from $R_\Th = \dfrac{V_\Th}{I_{SC}}$. This gives us $R_\Th = \dfrac{15\V \cdot R_1}{V_\in} = \mans{\dfrac{150\k}{V_\in}\Ohm}$. - \begin{circuit} + The Th\'evenin equivalent circuit takes the form shown below. + \begin{circuit}{fig:1.10.3}{Th\'evenin equivalent circuit.} (0,2) to[V=$V_{\Th}$] (0,0) to[short, -o] (3,0) (0,2) to[R=$R_{\Th}$, -o] (3,2) - ; + (3,0) to[open, v_<=$V_\out$] (3,2) \end{circuit} + This circuit is equivalent to the circuit in Figure \ref{fig:1.10.1}. \end{enumerate} \end{document} \ No newline at end of file diff --git a/taoesolutions.sty b/taoesolutions.sty index fef79b8..eb0a8ed 100644 --- a/taoesolutions.sty +++ b/taoesolutions.sty @@ -3,6 +3,12 @@ \usepackage{circuitikz} \usepackage{amsmath} \usepackage{enumitem} +\usepackage{float} +\usepackage[ + margin=1in, + includefoot, + footskip=30pt, +]{geometry} \newcommand{\todo}[1]{\textcolor{red}{\textbf{TODO: #1}}} \renewcommand{\title}[1]{\begin{center}\begin{Huge}#1\end{Huge}\end{center}} @@ -22,13 +28,21 @@ \renewcommand{\in}{\ensuremath{\text{in}}} % In (for V_{In}) \newcommand{\out}{\ensuremath{\text{out}}} -\newenvironment{circuit} - {\begin{center} - \begin{circuitikz}[american]\draw +\newenvironment{circuit}[2]% param1=label, param2=caption + { + \begin{figure}[H] + \caption{#2} + \label{#1} + + \begin{center} + % \centering + \begin{circuitikz}[american]\draw } - {; - \end{circuitikz} - \end{center} + { + ; + \end{circuitikz} + \end{center} + \end{figure} } % Set first level of list to (a) format