Current is conserved, voltage is spent

Follow a single loop: battery, bulb, bulb, back to the battery. Ask how much charge per second crosses each of four points, one in each wire segment. The answer is the same number four times.

The argument is charge conservation, and it is short. If the second bulb passed less current than the first, charge would accumulate somewhere between them at a steady rate forever. That charge would build a field opposing the pile-up in nanoseconds, so a steady state with unequal currents in a series chain does not exist.

Two consequences worth stating as facts:

1. Identical bulbs in series glow equally. Not brighter near the battery, not dimmer at the far end. Same current, same resistance, same power.
2. The wire returning to the battery carries exactly what the wire leaving it carries. A battery is not a charge reservoir that empties; it is a pump in a closed circulation.

What is spent is energy per unit charge. A coulomb leaves the battery's positive terminal with $\varepsilon$ joules of electrical potential energy per coulomb, gives some of it up crossing the first bulb, gives the rest up crossing the second, and arrives back at the negative terminal with nothing left to give. The bookkeeping is in volts, never in amperes. When somebody says "the current is used up", the sentence they meant is "the voltage is used up", and it is a different quantity entirely.

An ideal source of emf holds a fixed potential difference $\varepsilon$ across its terminals no matter what is attached. It does not hold the current fixed. Attach a network of equivalent resistance $R_{\text{eq}}$ and the current follows:

$$I = \frac{\varepsilon}{R_{\text{eq}}}.$$

So the causal order is: the battery clamps $V$, the circuit's resistance is whatever you built, and the current is the consequence. Reverse that order and every prediction downstream is wrong.

Work the two moves that change $R_{\text{eq}}$:

1. Add a resistor in series. $R_{\text{eq}}$ rises, so $I$ falls. A $12$ V battery on $6 \ \Omega$ delivers $2$ A; add another $6 \ \Omega$ in series and it delivers $1$ A.
2. Add a branch in parallel. $R_{\text{eq}}$ falls, so $I$ rises. That same $12$ V battery with two $6 \ \Omega$ resistors side by side delivers $4$ A, two amps down each branch.

A battery labelled "$9$ V, $2$ A" is quoting a voltage it maintains and a current it is rated to survive, not a current it forces. Think of it as a pressure source, not a flow source. Real batteries sag under load because they carry an internal resistance $r$, giving $V_{\text{terminal}} = \varepsilon - Ir$; that refinement belongs to Topic 11.6, and until then treat sources as ideal.

An ideal connecting wire has $R = 0$. Whatever current runs through it, Ohm's law gives

$$\Delta V = IR = I \cdot 0 = 0.$$

So the two ends of any unbroken piece of ideal wire sit at exactly the same potential, however long it is and however it bends. That licenses the single most useful move in circuit analysis: collapse every stretch of plain wire to a single point and give it one potential label. Those points are the nodes of the circuit.

Concretely, for a $12$ V battery driving a lone $6 \ \Omega$ resistor through a meter of lead on each side: every point on the lead from the positive terminal to the resistor is at $12$ V, every point on the lead from the resistor back to the negative terminal is at $0$ V, and the entire $12$ V appears across the resistor. Not $11$ V. Not "some of it lost along the way".

Node thinking also decides what a short circuit does. Lay a plain wire directly across a bulb's two terminals and those terminals become one node at one potential, so the potential difference across the bulb is zero, so its current is zero, so it goes dark while the rest of the circuit carries more current than before. Nothing about that follows from "current still reaches the bulb"; it follows from the two ends being forced to the same potential.

For any single-loop circuit, in this order:

1. Redraw and label nodes. Every stretch of plain wire becomes one labelled point. Choose the negative terminal as $0$ V.
2. Reduce. Combine resistors into $R_{\text{eq}}$.
3. Get the current from the source clamp. $I = \varepsilon / R_{\text{eq}}$. This is one number for the whole loop.
4. Price each element. $\Delta V_k = IR_k$ for each resistor, using the loop current.
5. Check. The drops must sum to $\varepsilon$. If they do not, a node was mislabelled or an element was double-counted.

Two checks are worth running every time. The current you computed should appear unchanged at every point of the loop, and the potential you assign should return to its starting value after one full trip around. If a solution has current shrinking as it goes or voltage disappearing into a wire, one of those two checks has already failed.

Finally, an open switch anywhere in a single loop makes the current zero everywhere, not zero past the break. There is no partial circulation up to the gap; a steady current requires a complete path, and without one the whole loop is at rest.