-"Elektrizität für Anfänger"!

[DIYSwede]

Surprisingly, the light will illuminate at much less voltage. There are capacitive and inductive effects at work. Think of the wires as two plates of a parallel plate capacitor, or the loop as two tightly coupled dipole antennas. Even if you cut the wires (at the farthest points) the bulb would still illuminate at perhaps a few hundred, or a few thousand volts. I suspect it would also oscillate because that's essentially a giant RLC circuit with a 0 ohm termination at each end.

Sure, a 12 VDC LED lamp will dimly turn on at app 8 V.
I can sorta buy that these "nearly infinitesimally long" wires could acts as dipoles and/or as a capacitance (with a pretty big serial inductance to boot).
But then - I simply can't wrap my head around getting DC to flow continously thru a cap, transformer (or an antenna),
other than for an initial, single LED blink (and then perhaps another when disconnecting the battery) - but then that's only me.
AC is a whole different ball game, though.

2 cents
Johan

Positively irrelevant to the thread, but still funny:



Waste 10 (+22 in Pt 2) more minutes of your life watching a solution to another hypothetical reasoning/ unprovable hypothesis:

[nova_robotics]

But then - I simply can't wrap my head around getting DC to flow continously thru a cap, transformer (or an antenna),

To be honest, I can't either. I would have been right there with you. The system is one light second in width, so we'd have at least a second of dv/dt to keep current flowing through the capacitor. And it's a fairly complex RLC network, so it would probably ring a bit... that's where my knowledge ends. I haven't taken any courses in transmission line theory and it just all turns into magic at that point. But then I was reading through the comments and and there was a bunch of conversation that was way above my pay grade. Someone seemed to come to the conclusion that just the inductance and capacitance gave 800 ohms of DC resistance. I have absolute no idea. Maybe? I dunno. I didn't even know that was possible. I will have to defer to people smarter than myself on this one. All I know is at least some current will flow as soon as you close the switch, but beyond that I really can't speak to it.