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Let's say we have a 5V battery and a ohmic conductor that consumes energy connected in a series circuit with constant resistance. When we increase the voltage, the number of electrons flow through a point per second increases as well. But when the component consumes the energy and creates a potential difference of 5V, the energy per unit of charge decreases, therefore, will the current in a series circuit always changes after the component? Or is it that the change in rate is too insignifant and we ignore it. Of course, current in a series circuit is the same everywhere, if not, there would be a 'traffic jam' of electrons in the wire which is weird. So can I know why?

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  • \$\begingroup\$ when the component consumes the energy and creates a potential difference of 5V, the energy per unit of charge decreases, - wrong way round - it increases. \$\endgroup\$ – Andy aka Apr 4 at 9:00
  • \$\begingroup\$ Be very careful with your language. The velocity (rate) of charge is not the same as current. These are independent concepts. \$\endgroup\$ – Elliot Alderson Apr 4 at 13:35
  • \$\begingroup\$ Since drift velocity is also proportional to voltage and so the drop in voltage would also affect the speed of the electrons and eventually the current because there will be less electrons passing through a point per second? Perhaps I should just accept the fact with my very limited high school physics... \$\endgroup\$ – hao004 Apr 4 at 13:44
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Neil's explanation is a good one but I thought I'd give just another way of looking at it.

You were completely correct saying that the current is constant throughout the circuit. Say we have a 5V source and use wires to connect it to a resistor.

I think where you may be getting confused is thinking all the energy the current contains is in its movement (kinetic energy) whereas it is almost all potential energy (you don't really notice it until connected to a location with a different potential) which is why there are a lot of references to gravity. Here is another image of the problem:

You are holding the marbles at the top of a hill - they have 5V of potential energy. You let them go onto this -almost- perfectly smooth ice which has a very small gradient - this is the electrons passing through the wire - they lose a little bit of potential energy. Then you get to a resistor - this is an extremely rough patch of grass but on a very steep bit of the hill. The marbles maintain their movement but it requires a loss of a lot of their potential energy moving against the grass. Finally there is another section of a small gradient of slightly rough ice of the wire the other side. Overall the marbles maintained their movement throughout but at the cost of different potential energies at different locations.

On a different note about how electrons actually act in the metals which I think you may find interesting. Without a voltage applied - electrons are flying about in both directions of the wire - at about a million km per hour - really fast but in both directions so there is no overall current flow! When a voltage is applied, they are still flying about extremely fast in both directions, but now there is also a small overall trend of movement (this is called the drift velocity). And when I say small, I mean only about a few meters per hour - really small! It is this mass movement we describe as the current flow.

Keep asking these sorts of questions though, it is good thinking!

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Consider a series circuit like a cascade of small waterfalls.

Water arrives at the top at the same flow rate that it leaves at the bottom, flow rate is equivalent to current. Place a cross section at any point in between and you'll still have the same volume of water crossing that plane. You could equate average speed of charge carriers to the speed of the water, higher through a thin pipe, slower in a wide conductor.

The height of the water is equivalent to voltage. It's the potential energy that the water has at that point, its ability to do work. The total drop from top to bottom is equal to the sum of the individual drops on the way down. It doesn't matter whether there's a big drop first and then a small drop, the heights will sum to the total height.

'Why' is somewhat trickier to answer, it can only be answered in the context of what you think is obvious. We observe that it just does. If you chase explanations all the way down, then you end up at quantum mechanics, which isn't obvious to anyone.

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    \$\begingroup\$ I use dodgy water analogies from time to time but a waterfall? The water accelerates as it falls and thins out. \$\endgroup\$ – Transistor Apr 4 at 8:28
  • \$\begingroup\$ @Transistor - But wouldn't it be similar to a circuit with a single 18 awg wire branching off to a whole bundle of 16 awg wires? \$\endgroup\$ – Michael Karas Apr 4 at 8:38
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But when the component consumes the energy and creates a potential difference of 5V, the energy per unit of charge decreases, therefore, will the current in a series circuit always changes after the component?

The current in a circuit will be consistent. What goes in from the source must return to the source.

Or is it that the change in rate is too insignificant and we ignore it.

No, there is no fudging or cheating.

Of course, current in a series circuit is the same everywhere, if not, there would be a 'traffic jam' of electrons in the wire which is weird. So can I know why?

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Equivalent circuits.

Maybe Figure 1 will help. Here your 5 V supply is powering a load R1 and 50 mA is flowing. In the equivalent circuit we have split the load into five equal parts. The voltage drop across each part is now 1/5 × 5 V = 1 V through 1/5 the resistance so the current remains the same, 50 mA. You can divide up R1 as many times as you like and the voltage across each element will decrease proportionally but the current will remain the same through the whole circuit.

enter image description here

Figure 2. A closed circuit of bicycle chain. Image source: Technology Student.

It may help you to think of the charge flow as more like a bicycle chain. The pedal chain wheel is the power source and the rear sprocket is the load. When pedaling the top is in tension and the bottom is slack but no links are consumed or go missing during the ride. The chain flows in a closed circuit and links into the drive = links returned from the drive.

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I decided to give this another try using the drift velocity formula. I = nAvQ, where n is the number of electrons/m^3, A is the cross sectional area of the wire, v is the drift velocity of electrons, Q is the charge of an electron. In order to keep the current same in a series circuit when there is a voltage drop across the resistor, one of the value must increase since drift velocity decreases. This should, I think, cause the material to have higher electron density in order to deliver the same current. But then electron density is a constant. Do correct me so I can delete the answer.

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Yes, the speed of the physical electrons decreases when voltage drops, because it causes a current drop when resistance remains unchanged. Less current (ampere), means less electrons (coulomb) travelling through the wire.

If the wire and paths stay the same, less electrons enter and leave the wire and components of the circuitry. Thus if the density of the electrons and diameter of the wire remain unchanged, they have to move slower travelling the same traject.

Please take a look at the Wikipedia article about Drift Velocity; The numerical example will give you some extra insights.

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  • \$\begingroup\$ I think your first sentence is worded poorly and mixes up cause and effect. Velocity drops when current drops; if the current is through a resistor then the current will drop when the voltage drops. \$\endgroup\$ – Elliot Alderson Apr 5 at 13:29
  • \$\begingroup\$ @ElliotAlderson, I started my first sentence with a direct answer to the question; I prefer doing it that way. \$\endgroup\$ – Sam Van Brussel Apr 13 at 13:53

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