Why does the same current flow through resistors in series?

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SUMMARY

The discussion centers on the behavior of electric current through resistors in series, emphasizing that the current remains constant across all resistors due to the principles of Ohm's Law (V = IR). Resistors do not slow down the charge but instead reduce the potential difference, affecting the entire circuit simultaneously. The analogy of hydraulic systems is suggested for better understanding, highlighting that changes in resistance influence the overall circuit dynamics rather than just localized effects. The conversation clarifies that while resistors discourage charge movement, they do not create or lose charge, maintaining a steady state in the circuit.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Basic knowledge of electric circuits and current flow
  • Familiarity with the concept of potential difference
  • Awareness of transient states in electrical circuits
NEXT STEPS
  • Study the principles of Kirchhoff's laws in circuit analysis
  • Explore the concept of transient response in electrical circuits
  • Learn about the relationship between voltage, current, and resistance in series circuits
  • Investigate hydraulic analogies in electrical systems for better conceptual understanding
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Students of physics, electrical engineers, educators, and anyone seeking to deepen their understanding of electric circuits and the behavior of current through resistors.

Andrea Perry
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This is something I have been trying to figure out lately, but I can't seem to wrap my head around it.
So, according to a definition in my Physics textbook, while the electrical potential difference established between two terminals encourages the movement of charge, it is resistance that discourages it.
If current is the rate of the flow of charge over a given amount of time, and it is the same through all resistors, then what exactly are the resistors doing to the charge if they're not slowing it down (decreasing the potential energy?)?
 
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Andrea Perry said:
If current is the rate of the flow of charge over a given amount of time, and it is the same through all resistors, then what exactly are the resistors doing to the charge if they're not slowing it down (decreasing the potential energy?)?

If you are thinking of an "energy of motion", don't you mean "kinetic energy"?

If you try to use a model of current as particles under action of a net constant force, the particles would accelerate. They wouldn't have a constant velocity. So a very basic question is why the idea of current, as a constant flow is a good model in the first place. I don't think you can wrap your mind around the phenomena of electric current by thinking of simple mechanical models.

Electrical phenomena are often presented by analogy to hydraulic phenomena. Perhaps you could try that route.
 
This was discussed many times on this forums. Maybe you can look it up.

The idea is that the charge is not entering the resistor (or any portion of circuit) with a larger velocity and leave with a lover velocity.
The average velocity of a charge carrier is the same everywhere in a homogeneous resistor.

At this level is better to think of the effect of the resistor in terms of current and voltage and forget the microscopic picture.
The resistor "discouragement" of charge movement translate into reducing the current through the circuit. The whole circuit, not just through the resistor. And this means that less charge flows through any cross section in a given time. It does not mean something about the speed of this charge.

The current is the same everywhere simply because otherwise there will be accumulation of charge in some portions of circuit and this will create extra fields which will very quickly will take care to equalize the flow.
 
Andrea Perry said:
If current is the rate of the flow of charge over a given amount of time, and it is the same through all resistors, then what exactly are the resistors doing to the charge if they're not slowing it down (decreasing the potential energy?)?

That's exactly what each resistor is doing when a current flows thru it: the potential difference before and after the resistor drops a bit according to Ohm's Law, V = IR.

There is no additional charge created by passage thru a resistor, nor is any charge lost. The potential difference, i.e. the voltage, however does change.
 
Andrea Perry said:
if they're not slowing it down (decreasing the potential energy?)?

Speed determines kinetic energy, not potential energy.
 
SteamKing said:
That's exactly what each resistor is doing when a current flows thru it: the potential difference before and after the resistor drops a bit according to Ohm's Law, V = IR.

.
This may be confusing. It sounds like there is a potential difference before and another one after the resistor. Difference between what?
Actually the potential is lower at one end than at the other end. And the difference between the potentials at the two ends is given by Ohm's law.
 
nasu said:
The resistor "discouragement" of charge movement translate into reducing the current through the circuit. The whole circuit, not just through the resistor. And this means that less charge flows through any cross section in a given time.

This seems to make more sense; however, how is it possible that the resistor affects the entire circuit at once?
 
Andrea Perry said:
This seems to make more sense; however, how is it possible that the resistor affects the entire circuit at once?
It's not clear what you mean here.

To clarify, potential differences are measured with respect to ground, which is taken to be V = 0. It's analogous to mechanical potential energy, PE = mgh, which is measured w.r.t. some convenient datum.
 
If you squeeze the garden hose at some point, won't the debit of water decrease everywhere, including at the end of the hose?
Te water circuit is not a very good analogy, but it may give some image.

What actually happens is that inserting an extra resistor in series with an existing one, the potential energy that initially was used only to push charge through one resistor will be divided between the two resistors (not equally, in general).

The "at once" does not make much sense here. For any change in circuit, there is some time before a new steady state is reached. For example, when you close the switch, the current increases from zero to some value (given by ohm's law). All the discussion is about what happens after this transitory regime.
When we say that the current is the same for the entire circuit, we mean in this steady state. If in a circuit with one resistor and a current I1 we cut the wire and introduce a second one, it will again take some time (very-very short) to reach the new steady state, with a current I2. This I2 is less than I1.
 
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  • #10
Andrea Perry said:
What I meant is: How is it that a resistor discourages movement of charge throughout the entire circuit, instead of just through the resistor?

The charge is still moving thru the network. It's just that the potential difference, i.e. the voltage, decreases a little bit every time the current flows thru a resistor.

Terms like "discourages movement" are pretty imprecise when describing the behavior of electricity and electric circuits.
 
  • #11
Andrea Perry said:
This seems to make more sense; however, how is it possible that the resistor affects the entire circuit at once?

That's like asking how does a tap/faucet effect the flow of water in the entire pipe?
 
  • #12
Andrea Perry said:
how is it possible that the resistor affects the entire circuit at once?
This is similar to the "how does a resistor know what to do in its circuit?" question. The fact is that, at switch on, the rules about how circuits work do not apply in the simple way that we learn for starters. There is always a 'settling in' time, after a circuit is first connected up. Charges flow and electric fields are set up, the effects traveling around a circuit at near the speed of light. Simple rules like Kirchoff and Ohm's law only apply in their simple form after the EM wave has rippled around the circuit and then everything 'knows' what to do.
 

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