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2023-10-08 20:50:15 -07:00
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10
Advanced/Random Walks/parts/3 effective.tex
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@ -38,20 +38,20 @@ 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 $B$.}
\note[Note]{As before, $N_x$ is the set of neighbors of $x$.}
$$
\sum_{b \in B_A} I(S, b) = \sum_{b \in B_B} I(b, B)
\sum_{b \in N_A} I(A, b) = \sum_{b \in N_B} I(b, B)
$$
\begin{solution}
Add Kirchoff's law for all vertices $x \neq A$ to get
$$
\sum_{\forall x} \biggl( ~ \sum_{b \in B_x } I(x, b) \biggr) = 0
\sum_{\forall x} \biggl( ~ \sum_{b \in N_x } I(x, b) \biggr) = 0
$$
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
$$
\sum_{b \in B_A} I(A, b) + \sum_{b \in B_B} I(B, b) = 0
\sum_{b \in N_A} I(A, b) + \sum_{b \in N_B} I(B, b) = 0
$$
\vspace{2mm}
@ -61,7 +61,7 @@ $$
\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,
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.