Signs come from the walk

Potential is a function of position in a circuit: every node has one value. Walk from a node all the way around any closed path and back to that same node, and the accumulated changes must cancel, because you ended where you began:

$$\oint \vec{E}\cdot d\vec{\ell} = 0 \quad \text{(electrostatic)} \qquad \Longrightarrow \qquad \sum_{\text{loop}} \Delta V = 0.$$

Two consequences are worth stating flatly.

1. The sum is zero for every closed walk, in either direction, whether or not the walk follows the current, and whether or not the circuit is in steady state. A nonzero sum is a bookkeeping error, not a discovery.
2. The sum being zero is not the same as "the drops equal the EMF". In a loop with one battery those statements happen to coincide; in a loop with two batteries, or a battery being charged, they do not.

Once a magnetic flux is changing through the loop the right side is no longer zero, and that is Unit 13. In Unit 11 the induced term is absent and the rule is exact.

Fix a walking direction first and commit to it. Then each element is priced locally, by comparing your walk to the labels already on the diagram:

1. Resistor, walked ALONG the labeled current arrow: $-IR$. You are moving downstream, toward lower potential.
2. Resistor, walked AGAINST the labeled current arrow: $+IR$. Moving upstream raises your potential.
3. Battery, entered at the short plate and left at the long plate (minus to plus): $+\varepsilon$, regardless of which way current is flowing through it.
4. Battery, entered at the long plate and left at the short one (plus to minus): $-\varepsilon$, again regardless of the current.

Nothing in that list says "batteries are positive and resistors are negative". That shortcut works for exactly one arrangement, a single battery walked with the current, and it fails the moment a second battery opposes the first, a battery is being charged, or a guessed current arrow points the other way.

Worked example. A $12$ V battery, then $R_1 = 3\ \Omega$, then a $6$ V battery whose plus plate faces the current, then $R_2 = 1\ \Omega$, all in one loop, walked clockwise along the assumed current $I$:

$$+12 - 3I - 6 - 1I = 0 \qquad \Longrightarrow \qquad I = \frac{6}{4} = 1.5 \ \text{A}.$$

The second battery earned a minus sign because the walk crossed it plus to plus-side-first, and that is a fact about the walk, not about it being a battery.

A real source is modeled as an ideal EMF $\varepsilon$ in series with a small internal resistance $r$. Walking from the negative terminal to the positive terminal while a current $I$ leaves the positive terminal, you gain $\varepsilon$ and immediately lose $Ir$:

$$V_{\text{terminal}} = \varepsilon - Ir \quad \text{(discharging)}, \qquad V_{\text{terminal}} = \varepsilon + Ir \quad \text{(being charged)}.$$

The consequences are measurable and they are tested.

1. Open circuit is the only place the label is true. At $I = 0$ the terminals read exactly $\varepsilon$. Any load pulls them below it.
2. The sag grows with current. A $9.0$ V cell with $r = 0.50\ \Omega$ delivering $2.0$ A shows $8.0$ V at its terminals, and the missing volt heats the cell.
3. Charging pushes the terminals ABOVE the EMF. Current now enters the positive terminal, so $Ir$ adds instead of subtracting.
4. Two measurements give $r$. Open circuit gives $\varepsilon$; a loaded reading $V$ at known current gives $r = (\varepsilon - V)/I$.

In a full loop, $r$ is simply one more series resistance: $I = \varepsilon/(R + r)$. The maximum current any source can deliver is $\varepsilon/r$, into a dead short, which is why internal resistance is what keeps a shorted battery from delivering infinite current.

Closing a switch, swapping a resistor, or burning out a bulb changes $R_{\text{eq}}$, which changes the source current, which changes every node potential in the circuit. There is no such thing as a change that stays in its own branch.

The reliable procedure is to re-solve, not to patch:

1. Redraw with the change in place and recompute $R_{\text{eq}}$ from scratch.
2. Get the new source current from $I = \varepsilon/(R_{\text{eq}} + r)$.
3. Work back out to every branch, getting each element's new current and drop.
4. Compare the new per-element power $P = I^2R$ against the old value for that same element. Only now say "brighter" or "dimmer".

The classic result this produces is counterintuitive and appears constantly. Take a bulb $A$ of $2\ \Omega$ in series with bulb $B$ of $4\ \Omega$, across $12$ V, with a switched bulb $C$ of $4\ \Omega$ waiting in parallel with $B$. Switch open: $I = 2$ A, so $P_A = 8$ W and $P_B = 16$ W. Switch closed: $B \parallel C = 2\ \Omega$, $R_{\text{eq}} = 4\ \Omega$, $I = 3$ A, so $P_A = 18$ W and $P_B = 9$ W. Closing the switch made $A$ brighter and $B$ dimmer at the same time. No amount of reasoning confined to one branch produces that.