Regarding the diagrams in Volume I DC - Basics - Conventional vs Electron Flow, if the diode in the LAST diagram was flipped, would the light still turn on? (Thanks to another post, I was able to download TinyCAD and make a quick drawing - not a very big picture...).

My first guess was "yes" since the electrons would cause friction along the filament before reaching the diode. However, after thinking it over a few times, I'm changing the answer to "no." The reason is that current is never induced due to the switched diode. No current, hence, no light.

If "no" is the correct anwser, then it seems almost like a paradox since the diode comes "after" the lightbulb when considering the correct flow of electrons. It makes perfect sense if the diode came before the lightbulb, at least in terms of sequencing.

I'm having a lot of fun reading these chapters and brushing up on my physics.

Thanks for any input here,

Chris

 

Its a circuit, a loop. So the diode is always before the light, or it is never before the light, depending on the point of view one wishes to take.

Current in a series circuit is equal at all points in the circuit. If it does not flow at one point, it will not flow at any point. If we open a switch, current won't flow to the switch and then stop - it will simply not flow at all. In order for a charge carrier to move, another charge carrier has to be getting out of the way.

 

I dislike the idea of electricity to be thought of as a single direction of flow, since both the electron flow and conventional theory are correct. The common misunderstanding is that electricity is entirely just the flow of electrons, although its also about the flow of the absence of electrons(a positive charge). Try to think of electricity as more of a presence, with 2 simultaneous flows happening, with each having to flow in order for the other to flow, thus negating the "this comes first and this comes second" idea.

 

My first guess was "yes" since the electrons would cause friction along the filament before reaching the diode.

The reason this guess would be incorrect is because this condition would violate Kirchoff's Current Law.

Dave

 

If you cut the diode out and left the wires unconnected would it work?
If you left that doide in and cut the wires before the bulb would it work?
If you do not trust what you are reading then an experiment is mandatory.

That picture is too small.