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In the video itself, he stated that electricity propagates from the battery at the speed of light when he describes the diagram.

I think what he has missed describing is that the wire folds carry capacitance (and inductance). The capacitance leads to current flow paths shorter than resistive paths (where resistance is assumed zero).

Some commentors have asked what happens if the wire at the extreme end is cut.

If it is not cut, then intially some current goes via capacitance, making the bulb light up sooner. Ultimately, assuming DC battery output, capacitive current decays down. However, the time-consuming resistive path establishes.

If the wire is cut, capatance makes the bulb light up at first, however, the capacitive current soon decays without establishing of the resistive path, hence the bulb glow would decay.

The 'electrical' model above does not contradict the 'energy flow via fields' description, as the two are one and the same thing at the basic level.



Adding more:

It's a distributed R-C circuit, along with discrete R of the bulb.

C per unit distance depends on the positioning of the wires. How fast current ramps up depends on the distributed C and the bulb's R (as wire's distributed R is ignored in the problem statement).


L too.

Distributed L and C, discrete R.




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