Read the nodes, not the drawing

Two tests, and they are about connections rather than layout:

1. Parallel: both ends of the two elements connect to the same pair of nodes.
2. Series: the two elements share one node that nothing else touches, so the same current has to pass through both.

Redrawing a circuit must never change the answer. If it does, the classification was made from the drawing.

The combining rules run opposite ways, and the parallel one has a step that is easy to drop:

$$R_{\text{series}} = R_1 + R_2 + \cdots, \qquad \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$$

The last operation in the parallel calculation is a flip. Computing $1/R_{\text{eq}} = 1/12 + 1/4 = 1/3$ and reporting $0.33\ \Omega$ is the same slip as adding $12$ and $4$ to get $16\ \Omega$: both fail the one check that catches everything here. A parallel combination is always smaller than its smallest branch, because each branch is one more route available. Here $R_{\text{eq}} = 3\ \Omega$, below the $4\ \Omega$ branch.

More components feels like more obstruction, and for a series addition that is right. For a parallel addition it is exactly backward:

1. The new branch lowers $R_{\text{eq}}$ below every existing branch.
2. So the current the battery delivers rises.
3. With an ideal battery, every existing branch keeps its full potential difference and its brightness.

Household outlets are the everyday version: each appliance plugged in raises the total current drawn, which is what a circuit breaker is watching for. Reserve the phrase "adding resistance" for series connections; a parallel addition adds paths.

Series elements share one current, so each potential difference is $IR$ and the drops divide in the same ratio as the resistances. An even split is the special case of equal resistances, and it is not the general rule.

A $2\ \Omega$ and a $10\ \Omega$ resistor in series across $12$ V:

$$I = \frac{12}{12} = 1\ \text{A}, \qquad \Delta V_1 = 2\ \text{V}, \qquad \Delta V_2 = 10\ \text{V},$$

the same $1:5$ ratio as the resistances. The larger resistance always takes the larger share. Handing the larger share to the smaller resistor, on the grounds that current passes through it more easily, gets it backward: the current is the same in both, so only the resistance can differ the drops.

A real battery behaves as an emf $\varepsilon$ in series with an internal resistance $r$, so its terminal potential difference is

$$\Delta V_{\text{terminal}} = \varepsilon - Ir.$$

It equals the printed value only at zero current. A $9$ V battery with $r = 0.4\ \Omega$ delivering $1.5$ A reads $9 - 0.6 = 8.4$ V at its terminals, and those $0.6$ V are dissipated inside the battery. That is why five bulbs in parallel each glow a little less brightly than one bulb alone, and why a voltmeter across the terminals drops when the circuit closes. Plot terminal potential difference against current and the slope is $-r$ with intercept $\varepsilon$.

Meters are placed by what they have to sample, and the ideal values explain the rule:

1. An ammeter measures charge per second passing through itself, so it goes in the path, and ideally has $0\ \Omega$ so it changes nothing.
2. A voltmeter compares two points, so it bridges them, and ideally has infinite resistance so it draws no current.

Swapped, each destroys the circuit it was measuring: an ammeter clipped across an element shorts it out, and a voltmeter placed in a path opens it.