From e0a4aa29f8ccbc9e86ec202be199756aea2528b6 Mon Sep 17 00:00:00 2001 From: Preston Pan Date: Tue, 28 May 2024 18:37:50 -0700 Subject: change emacs a little; more stuff --- mindmap/LRC circuit.org | 84 +++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 84 insertions(+) create mode 100644 mindmap/LRC circuit.org (limited to 'mindmap/LRC circuit.org') diff --git a/mindmap/LRC circuit.org b/mindmap/LRC circuit.org new file mode 100644 index 0000000..e28d9b0 --- /dev/null +++ b/mindmap/LRC circuit.org @@ -0,0 +1,84 @@ +:PROPERTIES: +:ID: 6dbe2931-cc18-48fc-8cc1-6c71935a6be3 +:END: +#+title: LRC circuit +#+author: Preston Pan +#+html_head: +#+html_head: +#+html_head: +#+options: broken-links:t + +* Introduction +LRC circuits are equivalent to mass-spring oscillation systems in terms of the differential equation generated. In other +words, they are an example of a wave generator. First we introduce the LRC circuit without a voltage source. Later, +another circuit diagram will include a possibly variable voltage source. +#+name: LRC Circuit +#+header: :exports results :file lrc_circuit.png +#+header: :imagemagick yes :iminoptions -density 600 :imoutoptions -geometry 400 +#+header: :fit yes :noweb yes :headers '("\\usepackage{circuitikz}") +#+begin_src latex :exports results :file + \documentclass{article} + \usepackage{circuitikz} + \begin{document} + \begin{center} + \begin{circuitikz} \draw + (0,0) to[resistor, l=\mbox{$R$}] (0,4) + to[inductor, l=\mbox{$L$}] (4,4) + to[capacitor, l=\mbox{$C$}] (4,0) + (4,0) -- (0,0) + (2,0) -- (2,-1) + to (2, -1) node[shape=ground]{}; + \end{circuitikz} + \end{center} + \end{document} +#+end_src + +#+RESULTS: LRC Circuit +#+begin_export latex +#+end_export + +* Mass-Spring Equation Equivalence +We know these relations for the given circuit elements above: +\begin{align} +v(t) = L\frac{di}{dt} \\ +i(t) = C\frac{dv}{dt} \\ +v = iR +\end{align} +if we analyze the current current signal, Kirchhoff's voltage law tells us that the total voltage +drop throughout this circuit is zero. We use the capacitor equation in integral form and sum the voltages: +\begin{align*} +L\frac{di}{dt} + \frac{1}{C}\int i(t)dt + iR = 0 +\end{align*} +We then take a derivative to remove the integral: +\begin{align*} +L\frac{d^{2}i}{dt^{2}} + R\frac{di}{dt} + \frac{1}{C}i = 0 \\ +(LD^{2} + RD + \frac{1}{C}) i(t) = 0 +\end{align*} +it is clear that the characteristic polynomial of this homogeneous linear differential equation is: +\begin{align*} +L\lambda^{2} + R\lambda + \frac{1}{C} = 0 +\end{align*} +which, utilizing the quadratic formula, has the solutions: +\begin{align*} +\lambda_{1} = \frac{-R + \sqrt{R^{2} - \frac{4L}{C}}}{2L}, +\lambda_{2} = \frac{-R - \sqrt{R^{2} - \frac{4L}{C}}}{2L} +\end{align*} +which implies the general solution to this differential equation is: +\begin{align*} +i(t) = \sum_{n=0}^{\infty} A_{n}e^{\lambda_{1} t} + B_{n}e^{\lambda_{2} t} +\end{align*} +We can make this nicer by setting $-\frac{R}{2L} = m$, $\frac{\sqrt{R^{2} - \frac{4L}{C}}}{2L} = p$, +then $\lambda_{1} = m + p$, $\lambda_{2} = m - p$. Then: +\begin{align*} +i(t) = \sum_{n=0}^{\infty} A_{n}e^{(m + p) t} + B_{n}e^{(m - p) t} \\ +i(t) = e^{m}(\sum_{n=0}^{\infty} A_{n}e^{pt} + B_{n}e^{-pt}) +\end{align*} +Then we can just recast our notation for the constants $A_{n}$ and $B_{n}$ to include this $e^{m}$ term: +\begin{align*} +i(t) = \sum_{n=0}^{\infty} A_{n}e^{pt} + B_{n}e^{-pt} +\end{align*} +** Dampened Oscillation +In the case $R^{2} < \frac{4L}{C}$, $p = i\frac{\sqrt{\sigma}}{2L}$ for some $\sigma > 0$. We re-case $\lambda = \frac{\sqrt{\sigma}}{2L}$ so $p = i\lambda$. Then: +\begin{align*} +i(t) = \sum_{n=0}^{\infty} A_{n}e^{i\lambda t} + B_{n}e^{-i\lambda t} +\end{align*} -- cgit