diff --git a/Advanced/Random Walks/main.tex b/Advanced/Random Walks/main.tex index 7d18ccc..fc1560a 100755 --- a/Advanced/Random Walks/main.tex +++ b/Advanced/Random Walks/main.tex @@ -24,5 +24,6 @@ \input{parts/0 random.tex} \input{parts/1 circuits.tex} \input{parts/2 equivalence.tex} + \input{parts/3 effective.tex} \end{document} \ No newline at end of file diff --git a/Advanced/Random Walks/parts/3 effective.tex b/Advanced/Random Walks/parts/3 effective.tex new file mode 100644 index 0000000..4a6915c --- /dev/null +++ b/Advanced/Random Walks/parts/3 effective.tex @@ -0,0 +1,281 @@ +\section{Effective Resistance} + +As we have seen, calculating the properties of a circuit by creating an equation for each vertex is +a fairly time-consuming ordeal. Fortunately, there is a better strategy we can use. + +\vspace{2mm} + +Consider a graph (or a circuit) with source and ground vertices. All parts of the circuit that aren't these +two vertices are hidden inside a box, as shown below: + + +\begin{center} +\begin{circuitikz}[american voltages] + \draw + (0,0) node[above left] {$A$ (source)} + to[short, *-] (1,0) + (5, 0) to[short, -*] (6,0) node[above right] {$B$ (ground)} + to[short] (6, -2) node[below] {$-$} + to[battery,invert,l={1 Volt}] (0, -2) node[below] {$+$} + to[short] (0, 0) + ; + + \node[ + draw, + minimum width = 4cm, + minimum height = 2cm, + anchor = south west + ] at (1, -1) {Unknown circuit}; + +\end{circuitikz} +\end{center} + + +What do we know about this box? If this was a physical system, we'd expect that the current flowing +out of $A$ is equal to the current flowing into $B$. + + +\problem{} +Using Kirchoff's law, show that the following equality holds. \par +Remember that we assumed Kirchoff's law holds only at nodes other than $A$ and $B$. \par +\note[Note]{As before, $B_x$ is the set of neighbors of $x$. Naturally, $B_B$ is the set of neighbors of $B$.} +$$ + \sum_{b \in B_A} I(S, b) = \sum_{b \in B_B} I(b, B) +$$ + +\begin{solution} + Add Kirchoff's law for all vertices $x \neq A$ to get + $$ + \sum_{\forall x} \biggl( ~ \sum_{b \in B_x } I(x, b) \biggr) = 0 + $$ + This sum counts both $I(x, y)$ and $I(x, y)$ for all edges $x, y$, except $I(x, y)$ when $x$ is + $A$ or $B$. Since $I(a, b) + I(b, a) = 0$, these cancel out, leaving us with + $$ + \sum_{b \in B_A} I(A, b) + \sum_{b \in B_B} I(B, b) = 0 + $$ + + \vspace{2mm} + + Rearrange and use the fact that $I(a, b) = -I(b, a)$ to get the final equation. +\end{solution} + +\vfill + +If we call this current $I_A = \sum_{b \in B_A} I(A, b)$, we can pretend that the box contains only one resistor, +carrying $I_A$ units of current. Using this information and Ohm's law, we can calculate the +\textit{effective resistance} of the box. + +\pagebreak + + +\problem{Resistors in parallel} +Using Ohm's law and Kirchoff's law, calculate the effective resistance $R_\text{eff}$ of the circuit below. + +\begin{center} +\begin{circuitikz}[american voltages] + \draw + (0,0) node[above left] {$A$ (source)} + to[short, *-*] (1, 0) + + (1, 0) to[short] (2, 1) + to[R, l=$R_1$, o-o] (4, 1) + to[short] (5, 0) + + (1, 0) to[short] (2, -1) + to[R, l=$R_n$, o-o] (4, -1) + to[short] (5, 0) + + (1, 0) to[short, -o] (2, 0.5) + (2, 0.5) to (2.3, 0.5) + (4, 0.5) to (3.7, 0.5) + (4, 0.5) to[short, o-] (5, 0) + + (1, 0) to[short, -o] (2, -0.5) + (2, -0.5) to (2.3, -0.5) + (4, -0.5) to (3.7, -0.5) + (4, -0.5) to[short, o-] (5, 0) + + (1, 0) to[short] (1.7, 0) + (4.3, 0) to[short] (5, 0) + + (5, 0) to[short, *-*] (6, 0) node[above right] {$B$ (ground)} + to[short] (6, -2) node[below] {$-$} + to[battery,invert,l={1 Volt}] (0, -2) node[below] {$+$} + to[short] (0, 0) + ; + + \node at (3, 0) {$...$ a few more $...$}; +\end{circuitikz} +\end{center} + +\begin{solution} + Let $I_i$ be the current across resistor $R_i$, from left to right. \par + By Ohm's law, $I_i = \frac{V}{R_i}$ (Note that $V = 1$ in this problem). \par + + \vspace{2mm} + + The source current is then $I_A = \sum_{i=1}^n = \Bigl( V \Bigr) \Bigl( \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n} \Bigr)$. + Applying Ohm's law again, we find that + $$ + R_\text{eff} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}} + $$ +\end{solution} + +\vfill + + +\problem{Resistors in series} +Using Ohm's law and Kirchoff's law, calculate the effective resistance $R_{\text{total}}$ of the circuit below. + +\begin{center} +\begin{circuitikz}[american voltages] + \draw + (0,0) node[above left] {$A$ (source)} + to[R, l=$R_1$, *-*] (2,0) + to[short] (2.5, 0) + + + (5.5, 0) to[short] (6, 0) + to[R, l=$R_n$, *-*] (8,0) node[above right] {$B$ (ground)} + to[short] (8, -1) node[below] {$-$} + to[battery,invert,l={1 Volt}] (0, -1) node[below] {$+$} + to[short] (0, 0) + ; + + \node at (4, 0) {$...$ a few more $...$}; +\end{circuitikz} +\end{center} + +\begin{solution} + This solution uses the same notation as the solution for \ref{parallelresistors}. + + \vspace{2mm} + + By Kirchoff's law, all $I_i$ are equal in this circuit. Let's say $I = I_i$. \par + Let $V_i$ denote the voltage at the node to the left of $R_i$. \par + By Ohm's law, $V_i - V_{i+1} = IR_i$. + + \vspace{2mm} + + The sum of this over all $i$ telescopes, and we get $V(A) - V(B) = I(R_1 + R_2 + ... + R_n)$. \par + Dividing, we find that + $$ + R_\text{eff} = R_1 + R_2 + ... + R_n + $$ +\end{solution} + +\vfill +\pagebreak + +We can now use effective resistance to simplify complicated circuits. Whenever we see the above constructions +(resistors in parellel or in series) in a graph, we can replace them with a single resistor of appropriate value. + + +\problem{} +Consider the following circuits. Show that the triangle has the same effective resistance as the star if +\begin{itemize} + \item $x = R_1R_2 + R_1R_3 + R_2R_3$ + \item $S_1 = \nicefrac{x}R_3$ + \item $S_2 = \nicefrac{x}R_1$ + \item $S_3 = \nicefrac{x}R_2$ +\end{itemize} + +\vspace{2mm} + +\hfill +\begin{minipage}{0.48\textwidth} + \begin{center} + \textbf{The star:}\par + \begin{circuitikz}[american voltages] + \draw + (-1, 1.732) to[R, l=$R_1$, *-*] (0, 0) + (2, 0) to[R, l=$R_2$, *-*] (0, 0) + (-1, -1.732) to[R, l=$R_3$, *-*] (0, 0) + + (-1, 1.732) to[short, *-o] (-1.5, 1.732) + (-1, -1.732) to[short, *-o] (-1.5, -1.732) + (2, 0) to[short, *-o] (2.5, 0) + ; + + \node[above] at (-1, 1.732) {a}; + \node[above] at (2, 0) {b}; + \node[below] at (-1, -1.732) {c}; + \end{circuitikz} + \end{center} +\end{minipage} +\hfill +\begin{minipage}{0.48\textwidth} + \begin{center} + \textbf{The triangle:}\par + \begin{circuitikz}[american voltages] + \draw + (-1, 1.732) + to[R, l=$S_1$, *-*] (2, 0) + to[R, l=$S_2$, *-*] (-1, -1.732) + to[R, l=$S_3$, *-*] (-1, 1.732) + + (-1, 1.732) to[short, *-o] (-1.5, 1.732) + (-1, -1.732) to[short, *-o] (-1.5, -1.732) + (2, 0) to[short, *-o] (2.5, 0) + ; + + \node[above] at (-1, 1.732) {a}; + \node[above] at (2, 0) {b}; + \node[below] at (-1, -1.732) {c}; + \end{circuitikz} + \end{center} +\end{minipage} +\hfill + + + +\vfill +\pagebreak + +\problem{} +Suppose we construct a circuit by connecting the $2^n$ vertices of an $n$-dimensional cube with $1\Omega$ resistors. +If we place $A$ and $B$ at opposing vertices, what is the effective resistance of this circuit? \par +\textbf{Bonus:} As $n \rightarrow \infty$, what happens to $R_\text{eff}$? \par +\note[Note]{Leave your answer as a sum.} + +\begin{solution} + Think of the vertices of the $n$-dimensional cube as $n$-bit binary strings, with $A$ at \texttt{000...0} + and $B$ at \texttt{111...1}. We can divide our cube into $n+1$ layers based on how many ones are in each + node's binary string, with the $k^\text{th}$ layer having $k$ ones. By symmetry, all the nodes in each layer + have the same voltage. This means we can think of the layers as connected in series, with the resistors + inside each layer connected in parellel. + + \vspace{2mm} + + There are $\binom{n}{k}$ nodes in the $k^\text{th}$ layer. Each node in this layer has $k$ ones, so + there are $n - k$ ways to flip a zero to get to the $(k + 1)^\text{th}$ layer. In total, there are + $\binom{n}{k}(n - k)$ parellel connections from the $k^\text{th}$ layer to the $(k + 1)^\text{th}$ + layer, creating an effective resistance of $(\binom{n}{k}(n - k))^{-1}$. + + \vspace{2mm} + + The total effective resistance is therefore + $$ + \sum_{k = 0}^{n-1} \frac{1}{\binom{n}{k}(n - k)} + $$ + + \linehack{} + + To calculate the limit as $n \rightarrow \infty$, note that + $$ + \binom{n}{k}(n - k) = \frac{n!}{(n - k - 1)! \times k!} = n \binom{n-1}{k}. + $$ + So, the sum is + $$ + \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{\binom{n - 1}{k}} + $$ + + \vspace{8mm} + + Note that for $n \geq 4$, $\binom{n}{k} \geq \binom{n}{2}$ for $2 \leq k \leq n-2$, so + $$ + \sum_{k = 0}^{n} \frac{1}{\binom{n}{k}} \leq 2 + \frac{2}{n} + \frac{n - 3}{\binom{n}{2}} + $$ + which approaches $2$ as $n \rightarrow \infty$. + So, $R_\text{eff} \rightarrow 0$ as $n \rightarrow \infty$. +\end{solution} \ No newline at end of file