# Why is current constant in a series circuit?

This is more of a confirmation I require rather than a direct question, please do guide though, since I can be terribly wrong. (I do not have much knowledge of this.)

Current is defined as the directed flow of charge in an electric circuit. According to Ohms Law, I could say that current is directly proportional to voltage and inversely proportional to resistance.

Now, I do know that the electrons cannot pile up since behind and in front of every electron, there exists another. So, if I connect 3 resistors in series, all with different voltages, the current flowing through them has to be constant, right?

Here is my problem. Is it correct if I assume that with varying voltages, the current also fluctuates, but varying voltage is due to presence of resistors (voltage drop, etc), and that current has no part in this?

Let's assume the wire is a river and the current is the water flow. Initially, I have 5 A of current with a bulb only (only 1 resistor). If I add another bulb in the circuit in series with much higher resistance than the first the overall charge would decrease, right? If I measure with an ammeter, so the No. of coulombs per second would decrease as the electrons initially move out of the battery?

If you start out with a circuit that initially contains a battery and a resistor, you will measure a given current: $$I={V\over R}.$$ If you now add another resistor to the circuit in a series configuration as specified, then the current in the circuit will decrease: $$I={V\over(R+R^\prime)}.$$ The larger the net resistance becomes, the smaller the current for a given voltage.

• Thank you for not turning a blind eye to my stupid question. So, i watched some lectures, to help clear my doubts...and so far, ive come to a small conclusion of my own...given the context, (a series circuit) the current is independent (or is it true for all cases? i dont think it does, cuz no voltage would mean no potential difference for the electrons to move) so in the ohms law, I now acts as a constant, and with V = IR, it can be said that V is directly proportional to R... Commented Apr 8 at 16:45
• @Confusedaf The current in such a circuit is dependent on the ratio of voltage to resistance, so that $$I={V\over R}.$$ If the voltage was zero, $V=0$, then the current would also be zero, $I=0$. Commented Apr 8 at 19:23

When we say "the current is constant in a series circuit", we mean that the current through the first resistor is the same as the current through the second resistor.

We don't mean that if you change the resistor values the current won't change. We don't mean that if you change the voltage applied the current won't change. We don't mean that if you add additional resistors or remove one of the resistors (replacing it with a short) the current won't change. Because none of those things are true.

Is it correct if I assume that with varying voltages, the current also fluctuates,

Yes. If you vary the source voltage, the current will change proportionally.

But the current through the first resistor will still be the same as the current through the second resistor.

but varying voltage is due to presence of resistors (voltage drop, etc), and that current has no part in this?

I'm not clear on what you mean here, but it is true that for real-world voltage sources, changing the current drawn will (slightly) change the output voltage.

Regardless of this fact, if you put two resistors in series connected to this source (or to any other circuit), then the current through the first resistor will be equal to the current through the second resistor.

Initially, I have 5 A of current with a bulb only (only 1 resistor). If I add another bulb in the circuit in series with much higher resistance than the first the overall charge would decrease, right?

Yes, increasing the equivalent resistance of the load on the source will decrease the current drawn from the source.

But after adding the second bulb in series with the first one, the current through the first bulb and the current through the second bulb will be equal. Not to the current that was originally going through the first bulb, but to the new current determined by the new configuration of the circuit.