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authorPreston Pan <ret2pop@nullring.xyz>2026-03-16 15:22:07 -0700
committerPreston Pan <ret2pop@nullring.xyz>2026-03-16 15:22:07 -0700
commitc0384ae720e2b274bfeebbc8a5303fb8a85d66d5 (patch)
tree28a74f6b9fc18e2afbfa586a5762e76f9f4b6c32 /mindmap/LRC circuit.org
parent0547cba1e38e1cdde62c885e1b198170ef9e72bc (diff)
test my website more deeply; add more CSS
Diffstat (limited to 'mindmap/LRC circuit.org')
-rw-r--r--mindmap/LRC circuit.org20
1 files changed, 10 insertions, 10 deletions
diff --git a/mindmap/LRC circuit.org b/mindmap/LRC circuit.org
index 51afe3c..950d57b 100644
--- a/mindmap/LRC circuit.org
+++ b/mindmap/LRC circuit.org
@@ -43,21 +43,21 @@ another circuit diagram will include a possibly variable voltage source.
* Mass-Spring Equation Equivalence
We know these relations for the given circuit elements above:
-\begin{equation}
+\begin{align}
v(t) = L\frac{di}{dt} \\
i(t) = C\frac{dv}{dt} \\
v = iR
-\end{equation}
+\end{align}
if we analyze the 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{equation}
L\frac{di}{dt} + \frac{1}{C}\int i(t)dt + iR = 0
\end{equation}
We then take a derivative to remove the integral:
-\begin{equation}
+\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{equation}
+\end{align}
it is clear that the characteristic polynomial of this homogeneous linear [[id:4be41e2e-52b9-4cd1-ac4c-7ecb57106692][differential equation]] is:
\begin{align*}
L\lambda^{2} + R\lambda + \frac{1}{C} = 0
@@ -143,7 +143,7 @@ i(t) = V_{0}e^{i\phi}\mathcal{L}^{-1}\{\frac{1}{(s - z_{1})(s - z_{2})(s - z_{3}
where $z_{2}$ and $z_{3}$ are the two roots we found for the homogeneous case. We then use partial fraction decomposition:
\begin{align}
\label{}
-\frac{1}{(s - z_{1})(s - z_{2})(s - z_{3})} &= \frac{A}{s - z_{1}} + \frac{B}{s - z_{2}} + \frac{C}{s - z_{3}} \\
+\frac{1}{(s - z_{1})(s - z_{2})(s - z_{3})} = \frac{A}{s - z_{1}} + \frac{B}{s - z_{2}} + \frac{C}{s - z_{3}} \\
A(s - z_{2})(s - z_{3}) + B(s - z_{1})(s - z_{3}) + C(s - z_{1})(s - z_{2}) &= 1
\end{align}
from this we know:
@@ -185,9 +185,9 @@ C = \frac{z_{2} - z_{1}}{z_{2}z_{3}^{2} + z_{1}z_{2}^{2} + z_{1}^{2}z_{3} - z_{1
\end{align}
So we have the three coefficients:
\begin{align}
-A &= \frac{z_{3} - z_{2}}{z_{2}z_{3}^{2} + z_{1}z_{2}^{2} + z_{1}^{2}z_{3} - z_{1}z_{3}^{2} - z_{1}^{2}z_{2} - z_{2}^{2}z_{3}} \\
-B &= \frac{z_{1} - z_{3}}{z_{2}z_{3}^{2} + z_{1}z_{2}^{2} + z_{1}^{2}z_{3} - z_{1}z_{3}^{2} - z_{1}^{2}z_{2} - z_{2}^{2}z_{3}} \\
-C &= \frac{z_{2} - z_{1}}{z_{2}z_{3}^{2} + z_{1}z_{2}^{2} + z_{1}^{2}z_{3} - z_{1}z_{3}^{2} - z_{1}^{2}z_{2} - z_{2}^{2}z_{3}}
+A = \frac{z_{3} - z_{2}}{z_{2}z_{3}^{2} + z_{1}z_{2}^{2} + z_{1}^{2}z_{3} - z_{1}z_{3}^{2} - z_{1}^{2}z_{2} - z_{2}^{2}z_{3}} \\
+B = \frac{z_{1} - z_{3}}{z_{2}z_{3}^{2} + z_{1}z_{2}^{2} + z_{1}^{2}z_{3} - z_{1}z_{3}^{2} - z_{1}^{2}z_{2} - z_{2}^{2}z_{3}} \\
+C = \frac{z_{2} - z_{1}}{z_{2}z_{3}^{2} + z_{1}z_{2}^{2} + z_{1}^{2}z_{3} - z_{1}z_{3}^{2} - z_{1}^{2}z_{2} - z_{2}^{2}z_{3}}
\end{align}
The resulting solution looks like this:
\begin{align}
@@ -224,10 +224,10 @@ so the full solution including the terms used for the [[id:bc7e9e01-9721-4b3e-a8
i(t) = V_{0}e^{i\phi}(Ae^{z_{1}t} + Be^{z_{2}t} + Ce^{z_{3}t}) + (i'(0) + i(0))(De^{z_{2}t} + Ee^{z_{3}t})
\end{align}
the sinusoidal part of the solution looks like this:
-\begin{equation}
+\begin{align}
\label{}
\frac{(z_{3} - z_{2})V_{0}e^{i\phi}e^{2\pi i\omega t}}{z_{2}z_{3}^{2} + z_{1}z_{2}^{2} + z_{1}^{2}z_{3} - z_{1}z_{3}^{2} - z_{1}^{2}z_{2} - z_{2}^{2}z_{3}}
-\end{equation}
+\end{align}
* Mass-Spring System
Starting from [[id:6e2a9d7b-7010-41da-bd41-f5b2dba576d3][Newtonian mechanics]] in a single dimension: