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Im currently reviewing Ohm's law and Fundamentals in Electronics.

There might be similar questions to this, yet my point remains different.

Let's say for example in a DC circuit, a 10 volt battery is able to produce 10 amperes. The power of the 10 volt battery is capable of moving a maximum of 10 coloumbs per second ; The greater the charge the more potential they have to do work. After entering a resistor the volume of charge exiting the resistor is much less compared to entering it. The resistor is allowing far less electrons to pass through per second. I understand that total amount of electrons or charge is conserved. However since the rate of flow has also decreased doesn't that mean the current leaving the resistor is far less? Doesn't that mean we have different values for current across the circuit before interacting with a resistor?

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I understand the voltage drop concept: since the volume of charge has decreased after interacting with a resistor its potential energy has decreased.

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    \$\begingroup\$ What does Kirchhoff's current and voltage laws say about it? Unless you are looking at real world circuits with stray capacitances, inductances, and energy being only able to move only at speed of light, there is no reason to believe why in ideal circuit voltage would be anything else than 10V or 0V and why current would be non-equal in some wire. Ideal voltage source is not limited, it will have 10V even if you draw any amouny of current. \$\endgroup\$
    – Justme
    Commented Nov 5, 2022 at 18:39
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    \$\begingroup\$ In your drawing electrons would leave the resistor forever until none were left (since positive current means loss of electrons). What do you think would happen to a piece of matter if no electrons were left to hold together its chemical bonds? What does the battery do with all the extra electrons you would be charging it up with? \$\endgroup\$ Commented Nov 5, 2022 at 18:53
  • \$\begingroup\$ When you analyze a circuit, you always assume that the same current flows through elements in series. In your circuit, you have two elements in series, so in your circuit the current through both elements is the same. I cannot emphasize this to you enough. If you look at a schematic diagram and think that current can just jump out of any component anywhere and go somewhere else you will never develop any intuition for how circuits work. \$\endgroup\$
    – user57037
    Commented Nov 6, 2022 at 3:27

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Matter is neutral

The greater the charge the more potential they have to do work.

I'd recommend that you set aside this idea of 'greater charge' representing more potential for work.

It's not helpful if it causes you to forget that nature pretty much ensures that everything remains neutral. Even the smallest amounts of charge difference result in almost unimaginable forces.

Here's an example of just how hard it is for us to understand, truly, just how powerful electric forces are:

Suppose you were able to separate 4,000 Coulombs of charge, positive to one side and negative to the other across a distance of one inch. The force that would exist between these would be roughly equal to the force that is holding the moon in orbit around the Earth! This is something like \$2\times 10^{20}\:\text{Newtons}\$.

Electric forces are unimaginably huge. They are far, far beyond our casual understanding. (This link provides some thoughts about just how much more powerful electric force is as compared with gravity.)

So current in equals current out

After entering a resistor the volume of charge exiting the resistor is much less compared to entering it. The resistor is allowing far less electrons to pass through per second. I understand that total amount of electrons or charge is conserved. However since the rate of flow has also decreased doesn't that mean the current leaving the resistor is far less? Doesn't that mean we have different values for current across the circuit before interacting with a resistor?

No, we don't have different values.

Let's take a look at a couple of figures from the 3rd edition of Matter & Interactions by Chabay and Sherwood:

enter image description here

On the left is a description of the circuit before the battery is able to interact with it and set up charges around the loop. On the right is a 'cartoon of sorts' illustrating how the charges set themselves up on the surface of the copper wire leading up to the Nichrome wire 'resistor' (which is intentionally shown thin in order to add some realism to how one might actually make a significant resistor value out of Nichrome.)

Please note that one end of the tiny Nichrome resistor has excess positive charges and the other end has excess negative charges. And in between, along the Nichrome wire itself, there will be a gradual transition (a gradient) that goes between excess positive to excess negative. At the very center of the Nichrome wire, the surface charge will be neutral -- halfway between what's at the two ends.

There are not many such excess charges, mind you.

Let's look at the copper wire itself for a moment. At this Wiki page on resistivity and conductivity we can find that copper has conductivity: \$\sigma=5.96\times 10^7\:\frac{\text{Siemens}}{\text{m}}\$. Let's assume that the cross-section of this copper wire is \$1\:\text{mm}^2\$ and carries a current of \$1\:\text{mA}\$. Then from this we can find the electric field gradient as:

$$\mathscr{E}_\text{copper wire}=\frac{J}{\sigma}=\frac{1\:\text{mA}}{1\:\text{mm}^2\,\cdot\,5.96\times 10^7\:\frac{\text{Siemens}}{\text{m}}}\approx 16.7785 \:\frac{\mu\text{V}}{\text{m}}$$

This is about the same field strength that you'd get from a single electron at a distance of about \$1\:\text{cm}\$!

So we really are not talking about much charge needed to motivate a current in a copper wire.

Now, it's a little bit different for the Nichrome wire.

Here, let's say that the cross-section of our Nichrome wire is one half of the copper wire, so \$0.5\:\text{mm}^2\$. From the same Wiki page above, find: \$\sigma=6.7\times 10^5\:\frac{\text{Siemens}}{\text{m}}\$. Now we'd compute:

$$\mathscr{E}_\text{Nichrome wire}=\frac{1\:\text{mA}}{0.5\:\text{mm}^2\,\cdot\,6.7\times 10^5\:\frac{\text{Siemens}}{\text{m}}}\approx 3 \:\frac{\text{mV}}{\text{m}}$$

The electric field intensity on this Nichrome wire would be 178 times what it is for the copper wires leading up to it! This means that almost all of the electric field gradient takes place at the Nichrome resistor and not in the copper wires leading up to it.

But this is still less than 200 electrons at the same distance of about \$1\:\text{cm}\$!

So it does not take much excess surface charge in order to motivate currents in wires.

Let's now consider those bends shown in the diagram from Matter & Interactions. We have two bends on each side. Let's assume they are \$90^\circ\$ bends to simplify things a bit. The electric field points down the length of the wire until the bend, at which point it must make a sharp turn. So we need to work out what it takes to terminate the electric field at the bend (and restart it around the corner, as well.) Using Gauss's law I get:

$$Q = -\epsilon_{_0} \cdot \mathscr{E} \cdot A= -\epsilon_{_0} \cdot 16.7785 \:\frac{\mu\text{V}}{\text{m}} \cdot 1\:\text{mm}^2\approx 1.5\times 10^{-22}\:\text{Coulomb}$$

This is less than \$\frac1{1000}\$th of the charge represented by one electron.

So there is a slight reshuffling of charge around a bend in a wire. But as you can see it doesn't take very much. If this had been \$1\:\text{A}\$, instead, we would have just one electron's worth of charge redistribution around the kink to re-direct that entire amp so that it takes the corner and moves on.

Summary

There's a good Youtube video called Surface Charge on a High Voltage Circuit. Worth watching.

The main idea you need to take from all this is that for all intents and purposes at our macro-scale of life (and circuitry) the electric current into a 2-terminal device will be exactly equal to the electric current out of it. This includes capacitors, which still must maintain net charge neutrality.

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After entering a resistor the volume of charge exiting the resistor is much less compared to entering it.

No. The current is the same at both terminals. Otherwise there would be an accumulation of charge in the resistor.

The resistor is allowing far less electrons to pass through per second.

Don't worry about electrons. (The mobile charge carriers in some circumstances are positive ions.) Just think of the resistor as restricting the flow of current.

However since the rate of flow has also decreased doesn't that mean the current leaving the resistor is far less? Doesn't that mean we have different values for current across the circuit before interacting with a resistor?

No. The current in your simple circuit will be the same at all points. The current leaving the battery must return to it. Charge is conserved.

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No you have it all wrong.Current in the resistor is equal to current in the wire ,if the currents were different over time there would be a build up of charges at the wire or at the resistor which isnt happening.

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  • \$\begingroup\$ The reason this works is that the excess charges push back on each other all along the wire back to the battery. \$\endgroup\$ Commented Nov 5, 2022 at 19:47
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It is true that the resistor is slowing the rate of flow of charge, but it is wrong to say that more charge enters than leaves. That would imply that the resistor is accumulating charge, which is incorrect. The resistor impedes flow around the entire loop, in the same way the narrowing of a section of a loop of road from 3 lanes to 1 causes cars entering and leaving (and indeed even cars far from the restriction) to slow down. The effect is a slowing everywhere.

It is not the presence of a charge in some medium that is "work". A resistor does not accumulate charge for work to be done. It is a charge's loss of potential energy, and the corresponding conversion of that energy to some other form, that we call work. A charge is required to move from a place of high potential energy to a place of lower potential energy in order for this to occur, and do the corresponding amount of work.

There is no accumulation of charge in the resistor, simply a flow of charge along it, from high to low potential, thus any charge that enters the resistor will result in a an equal charge leaving at the other end.

This is the essence of Kirchhoff's Current Law.

The current around the entire loop is the same at all points. Your claim that their are fewer charges per second leaving the resistor each second, than entering, is incorrect.

For example, If an electron (or any charge, positive or negative) has 10eV more potential energy at one end of the resistor than it would have at the other, then it must travel the entire length of the resistor for it to lose those 10eV of potential energy, and do 10eV of work during the journey.

In other words, if the potential difference across the resistor is 10V, then the only way for one Coulomb of charge to actually do 10J of work (which would be heating in the resistor) is for that Coulomb of charge to make the entire journey from end to end, implying that charge enters the resistor at the same rate that it leaves the other end.

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