Reduce the network by its nodes

An ideal wire has zero resistance, so $V = IR$ drops zero volts across it. Every point you can reach from a given point without crossing a component is therefore at the same potential, and all of those points are one node. The first move in any compound circuit is to put a letter on each node and stop looking at the geometry.

With the nodes labeled, the two definitions become checkable rather than visual:

1. Parallel: two elements share both of their nodes. They therefore have the SAME voltage across them, and the current splits between them.
2. Series: two elements share exactly one node, and nothing else connects to that shared node. They therefore carry the SAME current, and the voltage splits between them.

Redrawing a circuit with a bent wire, a longer lead, or a resistor rotated ninety degrees changes none of this. Two resistors drawn one above the other are not parallel because they look parallel; they are parallel when the wire on each side actually reaches the same two nodes. Two resistors drawn end to end in a single horizontal line are not in series if a third branch taps the point between them, because then they do not carry the same current.

For resistors carrying one common current, the drops add, so

$$R_{\text{series}} = R_1 + R_2 + R_3 + \cdots$$

For resistors across one common voltage, the currents add, so the conductances add:

$$\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots, \qquad R_{\text{parallel}} = \frac{R_1 R_2}{R_1 + R_2} \ \text{for two branches only.}$$

Two habits stop nearly all the arithmetic damage here.

1. Finish the flip. The reciprocal sum gives $1/R_{\text{eq}}$, not $R_{\text{eq}}$. Adding $\tfrac16 + \tfrac13 = \tfrac12$ and writing "$R_{\text{eq}} = 0.5\ \Omega$" is the single commonest slip in the unit.
2. Check the bound. A parallel equivalent is always smaller than the smallest branch. $6\ \Omega$ with $3\ \Omega$ gives $2\ \Omega$, and $2 < 3$. If your answer is $9\ \Omega$, or even $4\ \Omega$, the step is wrong before you look at the algebra.

The bound is not a trick, it is the physics: every additional branch is another road, and roads do not obstruct traffic. Series is where resistance accumulates. Parallel is where conductance accumulates.

Ladder networks are solved from the end farthest from the source, inward. The reason is the series prerequisite: the shared node must connect nothing else. In a ladder, a rung taps the node between two consecutive series-looking resistors, so those two do not carry the same current and cannot be added.

1. Label the nodes. Mark every node that has three or more connections; those are the branch points.
2. Find a combination whose shared node is a plain two-connection node (series) or whose two elements share both nodes (parallel), and collapse only that one.
3. Redraw with the collapsed block in place and relabel. A collapse can merge two nodes, which can create a new legal combination that was not there a moment ago.
4. Repeat until one resistor remains, then work back OUT to recover branch currents and voltages.

Working back out is the half students skip. Once $R_{\text{eq}}$ gives the total current, that current sets the drop across each series block, and each parallel block's drop then divides its own current among its branches by $I_i = V_{\text{block}}/R_i$. Every branch current recovered this way is checkable against the junction rule.

Because series elements share one current $I$, each takes $V_i = IR_i$. The drops are therefore in the same ratio as the resistances:

$$V_i = V_{\text{total}}\,\frac{R_i}{\sum R}.$$

A $2\ \Omega$ and a $10\ \Omega$ resistor across $12$ V carry $1$ A, so they take $2$ V and $10$ V, a ratio of $1:5$. Splitting the source evenly is correct only in the special case of equal resistors, and giving the large resistor the small share has the physics inverted: the large resistor is where the charge spends the most energy.

The mirror statement for parallel is about current, and it runs against intuition. Adding a resistor in parallel lowers $R_{\text{eq}}$, so with an ideal source the total current rises:

$$R_1 = 6\ \Omega \ \rightarrow \ I = \frac{12}{6} = 2 \ \text{A}; \qquad R_1 \parallel R_2 = 3\ \Omega \ \rightarrow \ I = \frac{12}{3} = 4 \ \text{A}.$$

Note what did NOT change: the original $6\ \Omega$ branch still has $12$ V across it and still carries $2$ A. With an ideal source, a parallel addition leaves every existing branch alone and simply draws more from the source. A real source sags a little, which is Topic 11.6.