From 609afd96f4069c69945fedcdfa7787ae4f7de967 Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Sat, 4 Apr 2020 02:21:04 -0400
Subject: attempted (but wrong) start of solution

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 create mode 100644 com.tex

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+\section{Project Topic}
+
+Our group will be working on the \link{Mystery Circuit Modelling
+Scenario from SIMIODE}{https://simiode.org/resources/3187/download/4-23%
+-S-MysteryCircuit-StudentVersion.pdf}. This applies Kirchhoff's Voltage
+and Current Laws to the given circuit, which describe, respectively,
+that the sum of all voltages in a closed loop is zero and the sum of all
+currents at a node is zero. The circuit we're examining is an RLC
+(resistor, inductor, capacitor) circuit, with zeroed initial conditions.
+The specific circuit has two linked loops of resistors and capacitors,
+in which ``gain,'' the ratio between chosen voltage differentials in the
+circuit can be modeled mathematically. Because there are two connected
+loops, there are three different currents. There is the current coming
+off of the battery $x(t)$, the current split at the middle node becoming
+$y(t)$ and $z(t)$. We are examining ${E(t)\over z(t)\rload}$ as the
+``gain'' in the system. The first part uses $\omega = 100$ and the
+entire problem uses $E(t) = \sin(\omega t)$.
+
+\section{Progress}
+
+From Kirchhoff's Voltage law over the first (xy) loop,
+$$E(t) = \sin(\omega t) = x(t)R_1 + {1\over C_1}\int y(t)dt.$$
+Kirchhoff's Voltage law also applies to the second (yz) loop:
+$${1\over C_1}\int y(t)dt = {1\over C_2}\int z(t)dt + z(t)\rload.$$
+Differentiating and rearranging gives:
+$$x'(t) = -{y(t) \over R_1C_1} + {\omega\cos(\omega t) \over R_1},$$
+$$z'(t) = {y(t) \over C_1\rload} - {z(t) \over C_2\rload}$$
+
+Kirchhoff's current law tells us that $y(t) + z(t) = x(t)$, so
+$$y'(t) = x'(t) - z'(t) = -{y(t)\over R_1C_1} +
+{\omega\cos(\omega t) \over R_1} - {y(t)\over C_1\rload}
++ {z(t) \over C_2\rload},$$
+giving a system of differential equations to solve.
+
-- 
cgit