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While I come across some explanations on why KCL works, it is usually attributed to the Law of Conservation of Charge. But by the statement of KCL, it states that the current entering and leaving a node are equal. For this to hold true, the charge entering and leaving the node must be equal (This is guaranteed by the Law of Conservation of Charge). My question is "Why the time taken by the charges to enter and leave the circuit needs to be equal?"

Could anyone please provide me an intuitive explanation or proof for this? I'm unable to internalize the concept. If I'm clear with this, I could understand about constant current in a series circuit.

P.S.: I'm sorry if my Physics vocabulary is not good because I'm a high schooler.

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  • $\begingroup$ Really thanks guys... Now I could understand this! $\endgroup$ Apr 11, 2023 at 13:59

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Why the time taken by the charges to enter and leave the circuit needs to be equal?

Those do not, in fact, need to be equal. However, if we have a situation where they are at least approximately equal, then we can simplify our analysis a lot.

For some background, circuit theory is a simplification of Maxwell’s equations. It relies on three assumptions. Of these three assumptions, the relevant one here is:

The net charge inside any circuit element is always 0.

So if a charge enters one terminal of a circuit element then the same charge must immediately leave another terminal, otherwise the net charge in the circuit element would be non-zero.

Not all possible circuit elements have that property. For instance, suppose that we wanted to consider each plate of a capacitor as its own separate circuit element. Then current would flow into a plate and not out and it would gain charge. Because of this, in order to use capacitors in circuit theory, we have to consider both plates together as part of one circuit element.

So there is nothing in nature that forces this assumption to be true, but there are many devices where it is true. When we design and build circuits we use those devices so that our circuits are easier to design and understand. So the only reason “why” this is true is because we deliberately construct circuits out of devices where it holds.

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  • $\begingroup$ One plate of a capacitor only has one terminal. $\endgroup$
    – Bob D
    Apr 10, 2023 at 15:25
  • $\begingroup$ @BobD yes. There is no circuit theory assumption on the number of terminals. The assumption above can be used to derive the fact that there must be at least 2 terminals, so if we drop that assumption then a 1 terminal element is no longer excluded. $\endgroup$
    – Dale
    Apr 10, 2023 at 16:00
  • $\begingroup$ Dale, I only mentioned it because of your (correct) statement "So if a charge enters one terminal of a circuit element then the same charge must immediately leave another terminal' $\endgroup$
    – Bob D
    Apr 10, 2023 at 16:37
  • $\begingroup$ @BobD yes, that "leave another terminal" statement is a consequence of the assumption. The "one plate capacitor" statement intentionally violates that assumption precisely to show the consequences of doing so. $\endgroup$
    – Dale
    Apr 10, 2023 at 17:18
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For an electric circuit at steady state, the flow of electrons into and out of a node must occur at the same rate. Electrons obviously can't flow out faster than they arrive. If electrons arrived more often than leaving, they would start to "bunch up" somewhere in the circuit, but that defies our description of the circuit at steady state. There are cases of circuits not in a steady state, like a capacitor that's being charged, where you do have charges accumulate in certain parts of the circuit.

Intuitively, it's like a hose carrying a flow of water. Anything that goes in one end of the hose comes out the other at the same rate. If the inflow is greater than the outflow, you have a backup in the hose and get a cartoonish bulge until the hose bursts - this sort of backup doesn't happen with electrical charge in a wire, you instead always have equal inflow and outflow.

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  • $\begingroup$ Kirchoff's law is very often applied in non-steady state. And in this case, the charges can accumulate and change on the surface of the wires. And, in fact, they do! As explained by @Dale, it is an approximation that neglects parasitic capacitances of the wires, except in the case of capacitor plates. $\endgroup$ Apr 10, 2023 at 16:49

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