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In a circuit, how do electrons "know" when the circuit is closed? To my understanding, they do not start flowing unless the circuit is closed.

How soon would they start flowing once I closed a switch on that circuit that was one light minute away from the battery?

(This is, assuming that the circuit is a simple DC circuit with 0 resistance and only one on/off switch.) Sorry if this is a stupid question, I am just a high-school senior.

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    $\begingroup$ The electrons "know" about the closing of the switch because an electromagnetic field will travel along the wires at a little less than the speed of light (typically around 60% of c for typical coaxial cables and twisted pairs). These traveling electromagnetic waves on wires are somewhat analogous to traveling mechanical waves on a rope. $\endgroup$
    – CuriousOne
    Commented Oct 28, 2015 at 23:05
  • $\begingroup$ @CuriousOne why do you post answers in the comments? $\endgroup$
    – Asher
    Commented Oct 29, 2015 at 3:14
  • $\begingroup$ @CuriousOne: I am repeating my request again; just post the above comment as answer. I am frustrated repeating the same urge again & again! After all, I'm always being deprived of upvoting these answers:( $\endgroup$
    – user36790
    Commented Oct 29, 2015 at 4:17
  • $\begingroup$ @user36790: Fair. I will try to do better in the future. $\endgroup$
    – CuriousOne
    Commented Oct 29, 2015 at 4:42

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Immediately. They start to flow immediately.

When you connect a resistor to the negative terminal of a battery and a wire to the positive terminal of the battery the whole resistor gets to lower potential and the whole wire gets to a higher potential.

So when you start to connect that wire to the resistor you literally bring a positive voltage wire towards a negative voltage resistor. And current will flow as soon as they get close enough for the potential per meter to be a breakdown field for the air between the wire and the resistor. And once that current starts to flow the resistor will start to develop a voltage difference across its ends until eventually the current is large enough for the potential drop to have the one end of the resistor be at the same potential as the ends of the batteries.

So the resistor started out with both ends of the resistor at the same potential but the two ends of the switch (the part connected to the positive potential wire and the part connected the to the negative potential resistor) started out at different potentials. Eventually that is completely switched, pun not intended.

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  • $\begingroup$ This is the correct answer, but perhaps you should say that it requires that the open circuit had been left in that state for a sufficiently long time (on order of 100 seconds). And the way to analyze it is to treat the switch as a very low capacitance. So while the flow will start immediately it will sag for a while before rising again on a time-scale given by the communication time to the source (i.e. on the order of 100 seconds). $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Oct 29, 2015 at 3:50

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