1
\$\begingroup\$

I know the question might sound slighly basic, but stay with me here.

Kirchoff's Current Law states that

"The algebraic sum of currents in a network of conductors meeting at a point is zero."

But how do we know that no charge gets "stuck" in the junction? Usually, I see people justifying KCL by pointing to the law of conservation of charge, but that doesn't mean that those charges couldn't get more or less concentrated in a region of a circuit. I understand physically that the electrons would want to spread out as far as possible, so of course they physically shouldn't get stuck in a junction - but how do we know that happens every time, and can we prove it in the lumped element model?

\$\endgroup\$
4
  • \$\begingroup\$ Charges get stuck all the time, but that charge-excess will fill an entire node (fills all the conductors connected to that node.) Like putting excess water pressure on a tree-shaped structure that's made of water-filled pipes. \$\endgroup\$
    – wbeaty
    Commented Feb 13, 2020 at 2:36
  • 3
    \$\begingroup\$ Trytryagain, First off, the electrons can only "get stuck" at a node on the surface -- not in the interior. Second, it only takes a few tens of thousands of electrons difference, one end to the other of a long wire say, to propel \$10^{18}\$ electrons per second through it. You probably have no idea at all just how powerful the electric force is. Electrons pretty much do not accumulate anywhere. If they did, you'd be able to take a tiny "pith ball" hanging on a thread and it would grab some accumulated electrons and then repel itself from the node. But that doesn't happen until some kV or so. \$\endgroup\$
    – jonk
    Commented Feb 13, 2020 at 4:25
  • \$\begingroup\$ Trytryagain, In short, yes. For a very short moment prior to "steady state" after turning on a circuit, there is a moment where various charges distribute themselves around the circuit and at bends in the wiring, etc. All this in order to guide the currents when steady state arrives soon after. One that happens, any change is immediately countered by negative feedback in positioning surface electron charges in the wiring to compensate and re-establish the steady state. From electronics point of view, the numbers of charges are so small compared to the currents that no one cares to count. \$\endgroup\$
    – jonk
    Commented Feb 13, 2020 at 4:29
  • \$\begingroup\$ We know it from experience. \$\endgroup\$
    – Curd
    Commented Feb 13, 2020 at 9:53

3 Answers 3

3
\$\begingroup\$

"The algebraic sum of currents in a network of conductors meeting at a point is zero."

It's a point, so no charge can get 'stuck' there because there is no place for it to get stuck in. This point has no capacitance, inductance, resistance or length, so any current that flows into it must instantly flow out again. You can pump a billion amps into it and it won't store any charge, create a magnetic field, drop voltage or radiate. And how do we know it must do this? Kirchhoff’s Current Law.

Of course in a real circuit this 'point' does not exist. Real wires have capacitance, inductance, and resistance, and they generate magnetic and electric fields and act as antennas. So some charge may be stored there, or the current might take some time to build up, and a voltage drop might occur which causes some current to take a different path. But modelling all the characteristics of every piece of wire in a circuit is usually unnecessary, especially at low frequencies where the 'ideal' model is close enough. If you need to model some of those real-world characteristics you can easily add them to the theoretical circuit as separate components - which also don't exist in reality.

The beauty of electronics is that we work with these 'ideal' components to design and build devices that look nothing like the theoretical circuit, but work just the same no matter how we arrange them physically. We even go to a lot of effort to produce components that are closer to the 'ideal' in order to make designing circuits easier. It's not about describing reality, but being able to easily create a complex mathematical model which can then be 'emulated' with real-word components.

\$\endgroup\$
1
  • \$\begingroup\$ This answer is the only one that really addressed my question. Great, thoughtful response. Thank you! \$\endgroup\$ Commented Feb 13, 2020 at 15:08
5
\$\begingroup\$

The economist George Box once said that "all models are wrong; some are useful". If you asked a physicist they might be able to describe a particular set of circumstances...electric field, magnetic field, quantum effects, whatever...that would cause a temporary nonuniformity in the carrier concentration in a wire.

But we are engineers here. Our models need to be useful. We don't need to prove that a model is correct...because we know that it is not correct. Nevertheless, we know that the model is useful and it works for our purposes.

\$\endgroup\$
5
  • 2
    \$\begingroup\$ You don't need a physicist to come up with that situation. Any wire with parasitic capacitance to some other conductor (i.e. all wires in the real world) will do that. \$\endgroup\$
    – The Photon
    Commented Feb 13, 2020 at 1:48
  • \$\begingroup\$ Also, Box was a statistician who started out studying chemistry, not an economist. \$\endgroup\$
    – The Photon
    Commented Feb 13, 2020 at 1:53
  • \$\begingroup\$ the charges do get concentracted this is part of how radio antennae behave. the effect is ephermeral because the charge spreads like a gas \$\endgroup\$ Commented Feb 13, 2020 at 1:56
  • \$\begingroup\$ @ThePhoton And even charges get stuck in capacitors with their dielectric absorbtion \$\endgroup\$
    – DKNguyen
    Commented Feb 13, 2020 at 2:11
  • \$\begingroup\$ @Photon: seems you do need a physicist, because even in the situation you are describing "current doesn't get stuck"; it's just another kind of current (other than the one consisting of moving charges): displacement current (seems to me many EEs are not aware of) \$\endgroup\$
    – Curd
    Commented Feb 13, 2020 at 10:49
2
\$\begingroup\$

They can accumulate. That tendency gives rise to the phenomenon of capacitance.

With short wires, the resistances, inductances, and capacitances are so small, that the time constants associated with this accumulation and redistribution are also very small.

The redistribution of an accumulation of charge at one end of a wire causes charge to (and I use an evocative but non-scientific description here) slosh from end to end. This is the basis of an antenna. The longer the wire, the lower the frequency of sloshing.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.