Charge is not consumed, energy is

Put two identical bulbs in a series loop with a battery and place an ammeter before the first bulb and another after it. Both read the same value, and the two bulbs glow equally.

That surprises people because bulbs get hot and batteries run down, so something is clearly being used up. It is, and it is not the charge:

1. Charge per second is the same at every point of a series path. There is no side exit and no place to store it.
2. Energy per charge falls as you go around: each charge arrives at a bulb with more potential energy than it leaves with, and the difference is $\Delta V$ across that bulb.

So an ammeter after the bulb reading less than one before it would mean charge disappeared inside the filament. What actually drops across the bulb is voltage, and a voltmeter is the instrument that shows it.

Mobile charges are spread through every wire and every filament before the switch is closed. Closing it establishes a potential difference around the whole loop essentially at once, and carriers everywhere begin drifting together.

That drift is remarkably slow, often well under a millimetre per second, and the bulb still lights the instant you flip the switch, because the bulb is lit by the carriers already inside its filament. Three predictions of the tank picture fail on this point:

1. A bulb two metres away lights with a noticeable delay. It does not.
2. The wire nearest the battery runs hottest. It does not.
3. A dead battery has run out of electrons. It has run out of chemistry: it can no longer maintain a potential difference.

The useful replacement image is a closed loop of bicycle chain. Push anywhere and the whole chain moves at once; the pedals supply energy and the chain was already all the way round.

An ideal battery holds the potential difference across its terminals at its rated value. It does not deliver a fixed current. The current follows from what is connected:

$$I = \frac{\Delta V}{R_{\text{eq}}}.$$

So a $12$ V battery across a single $6\ \Omega$ resistor drives $2$ A, and adding a second $6\ \Omega$ resistor in series makes $R_{\text{eq}} = 12\ \Omega$ and the current $1$ A. It does not keep supplying $2$ A and share it out.

Any change to the circuit means solving for the current again. Reusing a current computed for the old circuit is the single most common way a correct method produces a wrong answer.

Conventional current is defined as the direction positive charge would flow: out of the positive terminal, around the external circuit, back into the negative terminal. In a metal the actual carriers are electrons, which drift the other way.

That is a convention rather than a mistake, and it costs nothing as long as you commit to one. Every circuit rule in this unit, the loop rule and junction rule included, is written for conventional current, so use it and let the electrons take care of themselves.

Two related quantities are worth naming so they do not get confused with the drift speed. The signal travels near the speed of light, because it is the establishing of the field around the loop. The drift speed is the slow average motion of the carriers. And the individual carriers move fast in random directions between collisions; the drift is only the small bias on top of that.