diff options
author | Holden Rohrer <hr@hrhr.dev> | 2020-04-04 02:21:04 -0400 |
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committer | Holden Rohrer <hr@hrhr.dev> | 2020-04-04 02:21:04 -0400 |
commit | 609afd96f4069c69945fedcdfa7787ae4f7de967 (patch) | |
tree | cbeae89d61b88007a2ece9bdcafc3cf692f229d1 /com.tex | |
parent | 2c18d2f1f2888eb2eb8779457db70f2f8615c2b9 (diff) |
attempted (but wrong) start of solution
Diffstat (limited to 'com.tex')
-rw-r--r-- | com.tex | 34 |
1 files changed, 34 insertions, 0 deletions
@@ -0,0 +1,34 @@ +\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. + |