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