\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, $N_x$ is the set of neighbors of $x$.}
$$
	\sum_{b \in N_A} I(A, b) = \sum_{b \in N_B} I(b, B)
$$

\begin{solution}
	Add Kirchoff's law for all vertices $x \neq A$ to get
	$$
		\sum_{\forall x} \biggl( ~ \sum_{b \in N_x } I(x, b) \biggr) = 0
	$$
	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
	$$
		\sum_{b \in N_A} I(A, b) + \sum_{b \in N_B} I(B, b) = 0
	$$

	\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 $I_A = \sum_{b \in N_A} I(A, b)$, 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}<parallelresistors>
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 parallel 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 parallel.

	\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 
	$\binom{n}{k}(n - k)$ parallel connections from the $k^\text{th}$ layer to the $(k + 1)^\text{th}$
	layer, creating an effective resistance of 
	$$
		\frac{1}{\binom{n}{k}(n - k)}
	$$

	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}