# Why does the current stay same everywhere in a series circuit [duplicate]

Current in a series circuit are the same everywhere, but when there is a bulb creating a potential difference, wouldn't the rate of the same number of electrons flowing, coming out from the bulb change as electrons lose energy or voltage to the bulb as voltage is what makes current flow at some rate? Could you explain this without using the analogies?

• But the battery is also "giving" energy to the electrons at the same time.
– user258881
Apr 1, 2020 at 16:47
• But the bulb is always converting the electrical energy into light and heat energy and decreasing the current. So the current will only be back to 'normal' after passing over the battery, and change again after the bulb? Apr 1, 2020 at 17:00
• This might help: physics.stackexchange.com/questions/105954/…
– user258881
Apr 1, 2020 at 17:14
• Does this answer your question? Why current in series circuit is the same?
– user258881
Apr 1, 2020 at 17:15
• @ZiHao That thinking works for voltage, but not current.
– JMac
Apr 1, 2020 at 17:27

Analogy
Think of water running through a pipework: it may run faster where the pipes are narrower, and slower in the wider parts, but the quantity of water which enters is the same as the quantity of water that exits. There may be temporary decompressions here and there, but the water eventually has to come out, unless there is a leak or a reservoir somewhere.

Without analogies
All the charge entering the electric circuit (usually in the form of electrons) should eventually exit it. In narrower wires charge may run faster, but through a smaller cross section, in wider wires it may run slower, but the same quantity of charge will pass through every cross-section... unless there is charge accumulation or a leak somewhere. There may be temporary charge accumulations here and there, but they eventually dissipate.

Note however, that the current density (i.e. charge passing per unit area) will not be the same in the cross-sections of different size.

Mathematically this is described by a so-called continuity equation: $$\frac{dQ}{dt}=I,$$ where $$I$$ is the current entering the circuit, and $$Q$$ is the charge (or the quantity of water) accumulated within the circuit. More precise form of this equation formulated for charge and current densities ($$\rho$$ and $$\vec{j}$$) is $$\frac{d\rho}{dt} = \nabla\cdot \vec{j},$$ but this might be too mathy.

• This answer is perfect! But the OP has explicitly mentioned that they don't want analogies, so you might want to rethink about your answer.
– user258881
Apr 1, 2020 at 17:51
• @FakeMod Thanks, I misunderstood the question. Yet, the analogy is so obvious :) Apr 1, 2020 at 18:28

Current in a series circuit are the same everywhere, but when there is a bulb creating a potential difference, wouldn't the rate of the same number of electrons flowing, coming out from the bulb change...?

Perhaps the following will help clarify things.

The bulb does have an affect on current. The rate of electrons flowing (the current) in the series circuit connected to a fixed voltage source is reduced when you insert the bulb in the series circuit, because when you do you increase the overall resistance of the circuit. But once placed in the circuit the same lower current exists everywhere in the circuit.

If the current coming out of one side of the bulb was less than the current going into the other side, electrons would continue to "build up" inside the bulb. We know that doesn't happen.

Hope this helps.