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This is a very basic question which I'm sure has already been asked and has a simple answer, but I was unable to get a satisfying straightforward explanation via an internet search. Suppose I have a simple circuit composed by a switch and a light controlled by a breaker at the box. When I deactivate the circuit at the breaker, experience shows that it does not really matter whether the switch is open or closed. However, as I attempted to illustrate in this rudimentary sketch, if the switch is open, its line side is at 120V before opening the breaker.
enter image description here All the descriptions I have seen, assume without explanation that the voltage in the part of the circuit "left hanging" in between becomes zero as soon as the breaker is turned off. What is the physical mechanism that makes that happen? Or is it simply the case that the residual charge trapped in that part of the wiring is so small that, as soon as someone touches the wire, it dissipates to ground without producing any current of concern? If that was the case, could there be any concrete situation in which the connected loads make this scenario potentially dangerous, i.e., in which the mid portion of the wiring can act as a large enough capacitor to give a significant shock?

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    Petting a cat or rubbing a balloon on your head will generate more stored electrical energy than what's left on that disconnected 120V line.
    – J...
    Mar 28, 2022 at 14:06

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Yes, you answered it yourself.

Or is it simply the case that the residual charge trapped in that part of the wiring is so small that, as soon as someone touches the wire, it dissipates to ground without producing any current of concern

It is the high-voltage physicists and transmission-line engineers who worry about and safeguard against these effects. These intermediate sections of conductors, when fully insulated but charged, must be de-energized deliberately and gradually, and there are procedures and specialized grounding clamps and rods for this purpose.

This is dangerous if the voltage is high or the capacitance is high, neither of which occur in household wiring.

could there be any concrete situation in which the connected loads make this scenario potentially dangerous

It is not the size of the load, or the current capability of the feeders, panel, breakers and home wiring that would -if at all- create this situation. A circuit with a 15A breaker in this respect is not more dangerous than one with a 50A breaker.

IEC 479-2:1987 states that a discharge with energy greater than 5000 mJ is a direct serious risk to human health. IEC 60065 states that consumer products cannot discharge more than 350 mJ into a person.

Ref: https://en.wikipedia.org/wiki/Static_electricity

If the capacitance of a wire is in the pF range (or perhaps low nF) then the voltage must be 100kV to reach 5000mJ energy.

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  • Thanks @P2000, I tend to think of household circuits as I think of electronic circuits, and as you clarified the models are essentially different
    – MarcoD
    Mar 28, 2022 at 6:14
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    Well, @MarcoD, it's one reason we were not allowed to touch the circuits inside a cathode-ray tube TV even if the device was unplugged. And today, capacitors in power circuits can still hold dangerous charges. But you ask a good question, because people who are afraid of current are sometimes afraid for the wrong reason: it's the heat, not charge, that kills.
    – P2000
    Mar 28, 2022 at 6:18
  • that is a very interesting point you make. If I understand it correctly, the lethality really depends on the scenario. For example in this scenario of basically an exponential discharge, the lowest amount of energy that can "fry" the body is 5J, which is the same amount of energy that flows through the body in 5/6 of a second when one touches a hot 120V wire establishing those 50 mA of current through the body that are considered as the thresold of ventricular fibrillation. I have always thought that the cause of death in this case was the cardiac rhytm disruption, not the heat
    – MarcoD
    Mar 28, 2022 at 7:45
  • @MarcoD I'm not sure, but perhaps P2000 was trying to draw a distinction between "afraid of current" vs "afraid of voltage". High-voltage circuits can electrocute you, whereas high-current circuits are more likely to accidentally catch fire?
    – user253751
    Mar 28, 2022 at 9:04
  • it's the power (W) that kills you, independent of V or I, but higher V can deliver more power to your heart since your body's R is fixed, which is why a car battery that has more amps than a mains outlet isn't dangerous. On the plus side, most power supplies (even the cheapest Chinese one in non-UL devices) nowadays incorporate discharge resistors across the caps, rule of thumb is to drop it to under 1V in under 30secs...
    – dandavis
    Mar 28, 2022 at 12:44

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