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I understand the algebra, and I can also use Kirchoff's Law in calculation.

But I struggle to understand how it makes sense logically.

The law states, that the sum of the currents entering a junction must also be the same sum for the currents leaving that junction.

Now, this would make sense in a circuit without resistors, but if I am to apply resistors, the current would be slowed, and therefore, the amount of coulombs passing through per second (amps) would be lowered, and the current on the other side would surely be lower, than before the resistor?

How does this make sense with Kirchhoff's current law?

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    \$\begingroup\$ Current is electrons. Do you think electrons are turning into heat? \$\endgroup\$
    – DKNguyen
    Feb 25, 2020 at 15:14
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    \$\begingroup\$ but if I am to apply resistors, some of the current will be transformed into heat energy Uhm, Nope. Current is the amount of electrons passing per second. So you're saying that electrons are turned into heat? Are electrons "escaping" or converted into heat? Does that make sense? \$\endgroup\$ Feb 25, 2020 at 15:15
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    \$\begingroup\$ Back to the old water analogy- Suppose you have water doing work by flowing over a paddle wheel. The amount of water flowing before and after the paddle wheel is exactly the same, yet the paddle wheel is turning and doing useful work. \$\endgroup\$
    – John D
    Feb 25, 2020 at 15:19
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    \$\begingroup\$ @GeorgeWTrump, Re, "their electrical charge declines with a resistor." No. Sticking with that water-over-the-wheel analogy, after the water has gone over the wheel, it's still the same amount of water, and its weight (the weight of the water is what turns the wheel) has not changed either. But the water now is closer to the Earth's center of gravity. Being closer to the Earth's center of gravity means that the water now has less potential energy. Electrons still have the same charge after the resistor as before, but they are closer (in Voltage) to the negative terminal of the power supply. \$\endgroup\$ Feb 25, 2020 at 15:22
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    \$\begingroup\$ do the resistors not limit the flow of electrons going through them? They do but no electrons "get lost", all the electrons that go in must come out at the other side. That means that the current stays the same. \$\endgroup\$ Feb 25, 2020 at 15:28

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Current is the amount of electric charge passing through a cross-sectional area per unit time. A voltage can be thought of (but isn't!) a force. Electric charge is carried by electrons. Voltage provides an energy differential to cause the electrons to move in one direction overall.

With that picture, we can now see that what is being converted into work isn't the electrons, it is the energy potential given to it by the voltage that results in the electron flow to occur. Electrons are a conserved physical quantity, thus what enters a junction must be equally conserved by what exists a junction.

What is being converted to heat is the potential, the voltage. Hence why electrons are not converted to heat, still exist in equal number, and exit the nodes in the same quantity. However, you do get a voltage drop by the resistance.

To give a mechanical interpretation. If you roll a big rock up a hill and then push it down the hill toward a wall. The wall will provide resistance, the rock will push through the wall and lose energy in the process, but the rock is still a rock, same physical quantity, same physical properties...etc. What has changed is the energy potential given by some outside entity (you pushing it up the hill) was converted to heat and kinetic energy. The rock wasn't converted to heat energy, it remains a rock, the energy potential given to it was. Now, if you rolled multiple rocks down a hill one after the other, despite that resistance and energy loss that occurred, the rocks behind it haven't experienced this loss and will push the first rock forward, and the one behind that rock will have itself pushed forward...etc. Since current is a connected chain of such rocks, your flow over the entry and exit will be equal. What comes in must come out. You simply must keep presenting an energy differential to keep propelling them forward (voltage or gravity in the rock example).

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Yes, a resistor will slow down the rate of current flow (for a given potential). But that will cause a "traffic backup" which will also slow the rate of current flow into or out of any junction feeding that resistor to that same slow rate. Basically, in steady state, electrons can't "bunch up", because the increased charge concentration will repel them away from the traffic jam location.

And electrons don't "fly away" due to heating unless the resistor is an emitter, such as the heated filament in a vacuum tube, with an attractive charged anode nearby.

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  • \$\begingroup\$ This is why the water analogies break down: water needs to be in contact with other water to apply pressure to equalize its distribution. As soon as you lose contact via voids, you lose equal distribution which means you can have uneven local rates. Electrons don't have this issue due to charge repulsion. \$\endgroup\$
    – DKNguyen
    Feb 25, 2020 at 15:42
  • \$\begingroup\$ Wait, so the current will be the same on either side of the resistor, and therefore the sum of the beginning of the resistor will be a "slowered" version compared to before adding the resistor, and therefore the sum at the end of the resistor will also be a "slowered" version, so that they match? I think I am beginning to understand. My logic of thought was that, if the pre-resistor current is fast, and the after-resistor current is slow, the sum of those two would be different, so that the pre-resistor current would be higher. \$\endgroup\$ Feb 25, 2020 at 15:45
  • \$\begingroup\$ @GeorgeWTrump Instead of slowered/fast which carries false mechanical connotations that couple flow rate with energy, think more or less energy before and after the resistor. \$\endgroup\$
    – DKNguyen
    Feb 25, 2020 at 15:46
  • \$\begingroup\$ @DKNguyen so... Coulomb is a measure of charge, and not particles? But wouldn't it still mean, that there will pass less energy through a certain point after the resistor than before the resistor, unless the amount of energy moving through the whole circuit is lowered to balance out this effect? \$\endgroup\$ Feb 25, 2020 at 16:00
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    \$\begingroup\$ @GeorgeWTrump This is going to sound weird, but an electron's charge does not represent how much energy it carries Yes, energy will be less after the resistor, but not charge. An electron's charge doesn't change. \$\endgroup\$
    – DKNguyen
    Feb 25, 2020 at 16:01
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if the sum of electrons entering a node is not longterm zero, then eventually there will be arcing.

Does your phone arc, even if you leave it on?

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