# How does current flow in the given case?

Suppose each Coulomb of charge in a circuit has 6J of energy. When these electrons pass through the bulb (which is like a resistor) they give away all the energy to the bulb. In that case, the electrons are now left with 0J of energy. How can the electrons make it to the positive terminal if they are left with no energy?

The electrons flowing through the battery, and the wire, and then....your bulb, are many, many in number. ALL of them do not give up their entire energies to heat the filament of the bulb. Some might lose more energy and some less, on colliding with the atoms in the filament. These electrons are always under the force of electric field, generated by the battery driving the circuit. So they are always pulled by the positive terminal of the battery and keep on flowing, hence, completing the circuit, for which the bulb lights up.

Even if you consider your case, where an electron is losing its entire energy in a collision, then so what? It might get kicked by other flowing electrons, and hence gain energy, Or it might feel the field force of the battery and get accelerated again.

I'll only add that "0 joules" is absolutely irrelevant. Saying that the negative pole is 0V is absolutely arbitrary. We don't care about the real "energy", which is unknown. We can only measure differences of energy, and that is work (and heat). Energy is free, work isn't.

A simplification of the situation is to consider the gravitational analogy of the passage of an electron through the light bulb.

Imagine a block of wood (the electron) at the top of a slope (the bulb).
The slope has a number strips across it some of which present no friction to the passage of the block (gaps between ions in the conductor) and some horizontal stripes which result in friction between the block and the strips (electron colliding with an ion).

The inclination of the slope is so arranged that the speed of the block at the top of the slope is the same as the speed of the block at the bottom of the slope.

At the top of the slope the block has some gravitational potential energy and some kinetic energy (electron leaves the negative terminal of the power supply and enters the light bulb with some kinetic energy and electrical potential energy).

The block accelerates on a friction less stripe gaining kinetic energy whilst losing some gravitational potential energy (electron loses some electric potential energy and gains kinetic energy between collisions with ions).

On traversing a horizontal stripe which provides a frictional force the block loses kinetic energy ie slows down and heat is produced (the electron interacts with an ion and make the ion vibrate more whilst at the same time the electron loses some kinetic energy).

The sequence repeats itself each time the block losing some gravitational potential energy and gaining kinetic energy and then losing the kinetic energy it gained to produce heat and at the bottom of the incline the block has the same kinetic energy as it had at the top of the incline.
(The electron gains kinetic energy whilst losing some electric potential between collisions with the ions and then loses the kinetic energy it gained on collision with an ion. After passing through the bulb the electron has as much kinetic energy as it had when it entered the light bulb).

So it is not an all or nothing situation the electron loses electrical potential energy as it passes through the light bulb and that energy ends up as heat (and light).
When the electron leaves the light bulb it still has enough kinetic energy and electrical potential energy to travel through the connecting wire and arrive at the positive terminal of the power sourc

It is ok to define the potential of the positive terminal as zero and measure the potential or voltage everywhere else in the circuit relative to that reference point.

By that definition, an electron that was pulled from the positive terminal and moved to the negative terminal by the battery, gaining some energy in the process, will lose all its energy, as it moves back to the positive terminal through the circuit, only when it gets back to the positive terminal. Not sooner and not later, because in our reference frame that is the point where electron's potential energy was defined as zero.

This balancing act happens automatically through the adjustment of the current in the circuit.

As an example, in the case you are describing, where each Coulomb of negative charge passing through the circuit, starts up at the negative terminal with 6 Joules of energy, which just corresponds to 6V (V=U/Q=6J/1C). If the total effective resistance of the circuit, including the bulb, the wires and possibly other bulbs, is equivalent to, say, 10ohm, the current in the circuit will automatically become 6V/10ohm=0.6A, which will guarantee that every Joule of energy will be spent exactly at the time the charge will arrive to the positive terminal.

Of course, in reality (as far as we know), the electrons in the circuit move very slowly and create current by pushing each other, therefore, it is likely that 6J of energy will be spent before the electrons starting at the negative terminal reach the bulb, but there will be another group of electrons, closer to the positive terminal, that will reach it in their stead and their energy at that moment will be exactly zero.