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Probably the answer should be yes, it's like there would be no electricity source at all.

But when I ask what's happening after a switch is turned on, I get the answer that the electricity then goes from the switch to the device. That makes me think that an open circuit is not completely dead, because if yes, then before the switch was turned on, there should be no electricity in the part of the wire between the power source and the switch - any more than in the part of the wire between the switch and the device. So the path that the electricity has to travel after being switched on, should be from the power source to the device.

So what's going on really?

Edit: It looks like people understand my saying "completely dead" as I'm asking if there is some amount of current there. But really my question is if there is (in an open circuit) some type of life for the full amount of current that will flow in the wire after the circuit will be closed. Because my question is based on the path electricity have to travel after a circuit is completed until it reaches the device and powers it, and I believe that a smaller amount of current doesn't have any effect on that. In case I'm wrong with this, please clarify that for me.

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  • $\begingroup$ You could say that there is no true on or off state. Rather there is only more or less current flow that is asymptotic on both ends. Especially towards the "off" end. Once the materials in your circuit or battery degrade faster than it takes for your battery to drain even one iota, you probably stop caring about that off current flow. $\endgroup$
    – DKNguyen
    Commented May 4, 2022 at 18:48
  • $\begingroup$ Maybe of interest: When you close an open switch, the "news" that the circuit is complete and current can flow travels as a wave that starts at the switch, and spreads out in both directions to the rest of the circuit. $\endgroup$
    – Solomon Slow
    Commented May 4, 2022 at 19:01
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    $\begingroup$ The terminals of an outlet receptacle are an open circuit with nothing plugged into the outlet. Do you really think that means the receptacle is "dead"? $\endgroup$
    – Bob D
    Commented May 4, 2022 at 20:22
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    $\begingroup$ There's this thing called "voltage". $\endgroup$
    – Hot Licks
    Commented May 4, 2022 at 20:42
  • $\begingroup$ @SolomonSlow The "news" and "current" both refer to the same thing, right? And my question is just "why". $\endgroup$
    – George Lee
    Commented May 5, 2022 at 0:28

5 Answers 5

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It depends on what you mean by "dead". In an open circuit there is usually leakage current from the switch. This means that some current actual passes through the switch even though it's off. This occurs since the air is slightly conductive. There could also be currents in an open circuit due to induced electric fields. Another possibility is the one you mentioned. It takes time for the fields to dissipate when an closed circuit becomes an open circuit.

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  • $\begingroup$ Self understood that I'm not talking about minor leaks. I'm talking about what causes the device to get power so fast. $\endgroup$
    – George Lee
    Commented May 5, 2022 at 0:21
  • $\begingroup$ Please see my new edit on the question. $\endgroup$
    – George Lee
    Commented May 12, 2022 at 18:49
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Yes, the open end connected to a voltage source is not "dead", but it has the same electric potential as the voltage source terminal it is connected to. This is because electric charges continuously keep redistributing themselves on surface of conductors (including wires) in such a way that the whole conductor has the same potential.

When the voltage source is an alternating voltage source, this redistribution has oscillatory character so there is a very small alternating current in the conductor due to surface currents. Even when the circuit is not closed.

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  • $\begingroup$ This "distributing" you mention is not what we call "current"? $\endgroup$
    – George Lee
    Commented May 5, 2022 at 0:30
  • $\begingroup$ It is a special kind of current, that is only on conductor surface. Ordinary dynamic or low frequency AC current runs through the whole cross-section of wire. $\endgroup$
    – Ján Lalinský
    Commented May 5, 2022 at 1:47
  • $\begingroup$ So there is current, but less than after the circuit is complete. So where does the additional current come from once the circuit is completed? $\endgroup$
    – George Lee
    Commented May 5, 2022 at 1:52
  • $\begingroup$ Current is due to voltage source, and increase in current is due to connection that creates a conductive loop of small resistance. While the ends are open, charge redistributes mostly on its surface and this current is low because it takes very little charge to create potential on the wires. But when the ends are connected, then much higher current can flow due to closed low resistance path connecting the voltage source terminals. $\endgroup$
    – Ján Lalinský
    Commented May 5, 2022 at 2:04
  • $\begingroup$ Does this answer my question why after the circuit is completed the electricity is ready at the switch? When I say electricity I mean the full amount of current that the device requires, smaller amounts probably shouldn't have any effect on the time the device gets powered. $\endgroup$
    – George Lee
    Commented May 5, 2022 at 14:12
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what's happening after a switch is turned on, I get the answer that the electricity then goes from the switch to the device.
Electricity is related to electrical charges and their separation. If a current needs to flow, there must be a potential difference for it to do so.
To illustrate what happens when an open circuit is connected, think of two wires connected to the power source and a capacitor between the ends of the wires. This is an open circuit. Now, excess electrons flow up to the capacitor, displacing electrons on the opposite side of the capacitor, and these electrons in turn flow to the other pole of the power source. This also happens without a capacitor, only in insignificant quantities.

So the electricity has to travel after being switched on, should be from the power source to the device.
There is also some potential difference in an open circuit, and a certain amount of electrons will flow up to an open switch after being connected to a current source, and some electrons will flow from the other side to the other pole of the source.

Is an open circuit completely dead? As stated above, no. At the moment of connection, electrons move for a moment.
Furthermore, in the case of an AC source, electrons are constantly moving. Antenna rods are an extreme and important case. In this case, you can clearly see that nothing is dead, because a lot of power is emitted from the antenna rod to the environment.

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  • $\begingroup$ "This also happens without a capacitor, only in insignificant quantities" — Those insignificant quantities of charge probably are not capable of powering a device. So in order for a device to get powered do we have to wait longer than the time electricity takes to travel from the switch to the device? $\endgroup$
    – George Lee
    Commented May 9, 2022 at 0:02
  • $\begingroup$ Please just ask a new question $\endgroup$
    – HolgerFiedler
    Commented May 9, 2022 at 3:29
  • $\begingroup$ It's not a new question, it's just asking for clarification. In fact I've asked this question in a different way before, but it was closed for no good reason. Again, that's the same question. physics.stackexchange.com/questions/700673/… $\endgroup$
    – George Lee
    Commented May 9, 2022 at 18:47
  • $\begingroup$ When you carefully read my question you will see that I'm concerned about useful current that actually powers the device, not about insignificant quantities (if you really mean insignificant quantities). $\endgroup$
    – George Lee
    Commented May 9, 2022 at 18:51
  • $\begingroup$ Please see my new edit on the question. $\endgroup$
    – George Lee
    Commented May 12, 2022 at 18:51
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Try the water pipe analogy, which is sometimes useful though not correct.

Imagine that you have water flowing through a pipe and the water does some work at some position along the pipe before the water then flows back to the pump through the return pipe.

Imagine that after you turn off a valve, the water drains out of the pipe.

Then when you open the valve, the water has to fill up the pipe before it can do much work, and you can predict and measure how long it will take for that to happen.

Now imagine that the pipe does not get empty but is full the whole time. When you open the valve it doesn't take as long to do work. The pressure builds up much faster. The water is ready to move wherever the pressure gradient sends it.

Kind of like that, if you open a switch, and you have a full positive charge everywhere on one side of the switch and a negative charge everywhere on the other side. When you close the switch again, you have a great big gradient at the switch and the current can move. If the gradient had to start at both ends of the battery and gradually build up from nothing, it would be slightly slower.

Does that make sense? Talking about whether the current is completely dead might lead to confusion, but if you have an idea about what's going on then you can figure out how to relate your language to that. Of course, your picture of what's going on might also be wrong.

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  • $\begingroup$ Which part of the circuit is analogous to a pipe full of water which needs pressure? The part from the source to the switch, or from the switch to the device? My question is about the first part. $\endgroup$
    – George Lee
    Commented Aug 15, 2022 at 0:48
  • $\begingroup$ The whole circuit is like the pipe full of water. Open the valve and the water starts flowing through the pipe at almost the same time everywhere. If the pipe emptied out, then you'd have to fill it up before it could do much work. You'd at least have to fill it from the pump to the place the work is done. $\endgroup$
    – J Thomas
    Commented Aug 15, 2022 at 4:17
  • $\begingroup$ Pipe full of water is analogous to a dead wire, which is full of atoms. My question is if there is part of the wire which is full of "pressurized water" which means that there is current there. $\endgroup$
    – George Lee
    Commented Aug 15, 2022 at 19:08
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    $\begingroup$ If it's full of pressurized water that isn't flowing, that isn't current. but it's different from having no pressure there too. $\endgroup$
    – J Thomas
    Commented Aug 15, 2022 at 19:11
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If a circuit is truly open then no current will flow.

However, an open circuit can still have large potential difference which have the “potential” to drive energy if the circuit is closed.

In fact, in building wiring the high voltage wires are literally referred to as hot or “live” (the opposite of dead) because if you accidentally do connect then to something they can drive energy.

So to the extent that an open circuit has non zero voltage we might say the circuit is not “completely” dead. But again, if the circuit is open then zero current will flow.

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