\subsection{Equivalent Resistance of Series and Parallel Circuits}

This subsection defines equivalent resistance for resistive networks, derives the series and parallel combination rules from Kirchhoff's laws and Ohm's law, and shows how to reduce compound circuits step by step to a single equivalent resistor.

\dfn{Equivalent resistance}{Consider a network of resistors connected to an ideal battery of emf $\mathcal{E}$. The \emph{equivalent resistance} $R_{\mathrm{eq}}$ of the network is defined by
\[
R_{\mathrm{eq}} = \frac{\mathcal{E}}{I_{\mathrm{total}}},
\]
where $I_{\mathrm{total}}$ is the total current delivered by the battery. The equivalent resistance is the resistance of a single resistor that would draw the same current from the same battery as the entire network.}

\nt{Equivalent resistance is a bookkeeping device: it replaces an entire resistive sub-network by a single resistor whose effect on the rest of the circuit is identical. The replacement is always done between two terminals, and it preserves the $I$--$V$ relationship at those terminals.}

\dfn{Series combination}{Two (or more) resistors are in \emph{series} when they share exactly one common node and no other element is connected to that node. Equivalently, the same current $I$ flows through each resistor in a series chain, and the total voltage is the sum of individual voltage drops:
\[
V = V_1 + V_2 + \cdots + V_n.
\]
The equivalent resistance of $n$ resistors in series is
\[
R_{\mathrm{eq}} = R_1 + R_2 + \cdots + R_n = \sum_{i=1}^{n} R_i.
\]}

\thm{Voltage divider rule}{Consider two resistors $R_1$ and $R_2$ in series connected to a potential difference $V_{\mathrm{in}}$. The voltage across $R_2$ alone is
\[
V_{\mathrm{out}} = V_{R_2} = V_{\mathrm{in}}\,\frac{R_2}{R_1 + R_2}.
\]
This relation follows from Ohm's law and the fact that the same current $I = V_{\mathrm{in}}/(R_1+R_2)$ flows through both resistors.}

\nt{The voltage divider distributes the input voltage proportionally to each resistance. A larger resistance drops more voltage. This rule is the electrical analogue of the weighted average: $V_{R_i}$ is the fraction of $V_{\mathrm{in}}$ carried by $R_i$.}

\dfn{Parallel combination}{Two (or more) resistors are in \emph{parallel} when they are connected between the same two nodes, so the potential difference $V$ across each is identical. The total current splits among the branches:
\[
I_{\mathrm{total}} = I_1 + I_2 + \cdots + I_n.
\]
The equivalent resistance of $n$ resistors in parallel is
\[
\frac{1}{R_{\mathrm{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} = \sum_{i=1}^{n} \frac{1}{R_i}.
\]
For two resistors in parallel, this simplifies to
\[
R_{\mathrm{eq}} = \frac{R_1 R_2}{R_1 + R_2}.
\]}

\thm{Current divider rule}{Consider two resistors $R_1$ and $R_2$ in parallel connected to a total current $I_{\mathrm{total}}$. The current through $R_1$ is
\[
I_1 = I_{\mathrm{total}}\,\frac{R_2}{R_1 + R_2}.
\]
This follows from KCL and Ohm's law: the common voltage is $V = I_{\mathrm{total}}/G_{\mathrm{eq}}$ where $G_{\mathrm{eq}} = 1/R_1 + 1/R_2$, and $I_1 = V/R_1$.}

\nt{Current divides inversely to resistance: the smaller resistor carries more current. In the limit $R_1 \ll R_2$, essentially all current flows through $R_1$; we say $R_1$ \emph{shunts} $R_2$.}

\mprop{Series and parallel equivalent resistance}{For $n$ resistors, the equivalent resistance satisfies
\[
\text{Series:}\quad R_{\mathrm{eq}} = \sum_{i=1}^{n} R_i
\qquad\text{and}\qquad
\text{Parallel:}\quad \frac{1}{R_{\mathrm{eq}}} = \sum_{i=1}^{n} \frac{1}{R_i}.
\]
The current through and voltage across each individual resistor are obtained by applying Ohm's law $V_i = I_i R_i$ once $I_i$ or $V_i$ is known from the reduction analysis.}

\pf{Series combination}{Kirchhoff's voltage law (the loop rule) states that the sum of potential differences around any closed loop is zero. For two resistors in series with a battery of emf $\mathcal{E}$,
\[
\mathcal{E} - V_1 - V_2 = 0,
\]
so $V_1 + V_2 = \mathcal{E}$. Since the same current $I$ flows through both resistors, $V_1 = IR_1$ and $V_2 = IR_2$. Substituting gives $I(R_1 + R_2) = \mathcal{E}$, or $R_{\mathrm{eq}} = \mathcal{E}/I = R_1 + R_2$. The result extends immediately to $n$ resistors by induction.}

\pf{Parallel combination}{Kirchhoff's current law (the node rule) states that the sum of currents entering a node equals the sum of currents leaving. At the top node of the parallel connection, $I_{\mathrm{total}} = I_1 + I_2 + \cdots + I_n$. Since each branch has the same voltage $V$, we have $I_i = V/R_i$. Thus
\[
I_{\mathrm{total}} = V\sum_{i=1}^{n} \frac{1}{R_i}.
\]
But $I_{\mathrm{total}} = V/R_{\mathrm{eq}}$, so $1/R_{\mathrm{eq}} = \sum 1/R_i$.}

\cor{Bounds on equivalent resistance}{For any combination of series and parallel resistors, the equivalent resistance of a parallel block is always less than the smallest resistance in that block:
\[
R_{\mathrm{parallel}} < \min_i(R_i).
\]
Conversely, the equivalent resistance of a series chain is always greater than the largest individual resistance. These bounds follow from the positivity of all resistances and provide a useful sanity check on computed results.}

\ex{Illustrative example}{When resistors are identical, the formulas simplify. $n$ identical resistors of resistance $R$ in series give $R_{\mathrm{eq}} = nR$. In parallel, $R_{\mathrm{eq}} = R/n$. Thus three $12\,\Omega$ resistors in parallel give $R_{\mathrm{eq}} = 4\,\Omega$, while in series they give $36\,\Omega$.}

\nt{The strategy for analysing a mixed (compound) circuit is: (1) identify the innermost series or parallel groups. (2) Replace each group by its equivalent resistance. (3) Repeat until the entire network is reduced to a single resistor $R_{\mathrm{eq}}$. (4) Use $I_{\mathrm{total}} = \mathcal{E}/R_{\mathrm{eq}}$ to find the battery current, then work backwards through the reduction steps to find individual branch currents and voltages.}

\nt{On the AP Physics C E\&M exam, equivalent resistance problems test your ability to correctly identify series versus parallel connections and to apply Kirchhoff's laws systematically. The most common errors are misidentifying which elements share the same current (series) and which share the same voltage (parallel). Always trace the current paths and label the nodes to verify your classification.}

\qs{Worked example}{An ideal battery of emf
\[
\mathcal{E} = 24\,\mathrm{V}
\]
is connected to a resistive network consisting of four resistors arranged as follows. Resistors $R_1 = 2.0\,\Omega$ and $R_2 = 4.0\,\Omega$ are connected in series. This series combination is connected in parallel with resistor $R_3 = 6.0\,\Omega$. Finally, resistor $R_4 = 3.0\,\Omega$ is connected in series with this entire parallel block, and the combination is connected to the battery.

Find:
\begin{enumerate}[label=(\alph*)]
\item the total equivalent resistance $R_{\mathrm{eq}}$ of the network,
\item the total current $I_{\mathrm{total}}$ delivered by the battery,
\item the voltage drop across each resistor, and
\item the current through each resistor.
\end{enumerate}}

\sol \textbf{Part (a).} Begin by reducing the network from the inside out. Resistors $R_1$ and $R_2$ are in series, so their combined resistance is
\[
R_{12} = R_1 + R_2 = 2.0\,\Omega + 4.0\,\Omega = 6.0\,\Omega.
\]
This series combination is in parallel with $R_3 = 6.0\,\Omega$. The equivalent resistance of the parallel block is
\[
\frac{1}{R_p} = \frac{1}{R_{12}} + \frac{1}{R_3} = \frac{1}{6.0\,\Omega} + \frac{1}{6.0\,\Omega} = \frac{2}{6.0\,\Omega} = \frac{1}{3.0\,\Omega}.
\]
Hence $R_p = 3.0\,\Omega$.

The parallel block is in series with $R_4$, so the total equivalent resistance is
\[
R_{\mathrm{eq}} = R_p + R_4 = 3.0\,\Omega + 3.0\,\Omega = 6.0\,\Omega.
\]

\textbf{Part (b).} Ohm's law gives the total current from the battery:
\[
I_{\mathrm{total}} = \frac{\mathcal{E}}{R_{\mathrm{eq}}} = \frac{24\,\mathrm{V}}{6.0\,\Omega} = 4.0\,\mathrm{A}.
\]

\textbf{Parts (c) and (d).} Now work backwards through the reduction. The current through $R_4$ equals the total current, since $R_4$ is in series with the battery:
\[
I_4 = I_{\mathrm{total}} = 4.0\,\mathrm{A}.
\]
The voltage drop across $R_4$ is
\[
V_4 = I_4 R_4 = (4.0\,\mathrm{A})(3.0\,\Omega) = 12\,\mathrm{V}.
\]
The remaining voltage appears across the parallel block:
\[
V_p = \mathcal{E} - V_4 = 24\,\mathrm{V} - 12\,\mathrm{V} = 12\,\mathrm{V}.
\]
Since the resistors $R_1$, $R_2$, and $R_3$ are all connected to the parallel block, the voltage across each is $V_p = 12\,\mathrm{V}$ (for $R_3$ directly) and $V_p = 12\,\mathrm{V}$ (for the series pair $R_1$--$R_2$).

The current through $R_3$ is
\[
I_3 = \frac{V_p}{R_3} = \frac{12\,\mathrm{V}}{6.0\,\Omega} = 2.0\,\mathrm{A}.
\]
The current through the $R_1$--$R_2$ branch is
\[
I_{12} = \frac{V_p}{R_{12}} = \frac{12\,\mathrm{V}}{6.0\,\Omega} = 2.0\,\mathrm{A}.
\]
Since $R_1$ and $R_2$ are in series, the same current flows through both:
\[
I_1 = I_2 = I_{12} = 2.0\,\mathrm{A}.
\]
The individual voltage drops are
\[
V_1 = I_1 R_1 = (2.0\,\mathrm{A})(2.0\,\Omega) = 4.0\,\mathrm{V},
\]
\[
V_2 = I_2 R_2 = (2.0\,\mathrm{A})(4.0\,\Omega) = 8.0\,\mathrm{V}.
\]

\noindent\textbf{Checks:}
\begin{itemize}
\item KVL on the left loop: $V_1 + V_2 = 4.0\,\mathrm{V} + 8.0\,\mathrm{V} = 12\,\mathrm{V} = V_p$.
\item KCL at the junction: $I_{12} + I_3 = 2.0\,\mathrm{A} + 2.0\,\mathrm{A} = 4.0\,\mathrm{A} = I_{\mathrm{total}}$.
\item KVL on the full loop: $V_p + V_4 = 12\,\mathrm{V} + 12\,\mathrm{V} = 24\,\mathrm{V} = \mathcal{E}$.
\end{itemize}

Therefore,
\begin{align*}
R_{\mathrm{eq}} &= 6.0\,\Omega,\\
I_{\mathrm{total}} &= 4.0\,\mathrm{A},\\
(V_1, V_2, V_3, V_4) &= (4.0\,\mathrm{V}, 8.0\,\mathrm{V}, 12\,\mathrm{V}, 12\,\mathrm{V}),\\
(I_1, I_2, I_3, I_4) &= (2.0\,\mathrm{A}, 2.0\,\mathrm{A}, 2.0\,\mathrm{A}, 4.0\,\mathrm{A}).
\end{align*}
