Recent content by tedward

All I'm trying to do was explore the implications of the different conventions. But the 'electrical guy' keeps derailing with objections.

My drawing doesn't match up to what I described in 101, true. It actually doesn't matter, but you would measure the same reading if the leads always ran parallel to the resistor wire (see below). The resistor wire has an induced voltage of 6V (shown in blue) and a scalar potential (due to the...

#101

Doesn't mean you're entitled to one. Re-read the Test Lead post where I address this.

Here are the examples I was thinking about. We have flexible resistor wire (12 ohms) in series with conducting wire with zero resistance. In all the cases, total emf, total resistance, and current are the same. The path voltage ##V## is what the voltmeter measures with a flux-free loop (and of...

Thanks for these! I've read Romer's paper on this, it's great - everyone interested in this problem should read it. Loved the analysis in the second link - he sums up all the points of confusion we've been discussing here nicely, and we come to the same conclusions. I also agree with him...

We were going back and forth on this point, that's why I wrote that "test leads" post. Maybe I misinterpreted but I thought you claimed afterwards that the outside voltmeter reading doesn't even measure path voltage correctly as opposed to scalar potential (which it doesn't). Doesn't matter...

My aim right now is simply to compare the conventions (as the scalar potential is new to me, but you've convinced me that it has some legitimacy) to see what they actually correspond to - especially in this distributed case, as the analysis with negligible length resistors hides some glaring...

All I'm trying to do now is see if people, especially EE's, actually use this scalar potential convention. It seems very counter intuitive to me. The arguments back-and-forth have gotten tiresome though, I agree.

I used the voltage 'drop' convention, so ##V_r = \int \vec E \cdot d \vec L##, instead of the potential difference convention of ##V_{ab} = -\int \vec E \cdot d \vec L##. Is that what you're referring to?

Uh... yes(?). Use ##P = I^2 R## or ##P=IV##, which underscores that the true ##V_r## = 12V. What's your point? Can't make heads or tails out of this. There are 12 V (path voltage) across the resistor, 12 ohms of resistance, and current is 1A. ?? Flipping voltmeter leads changes the sign of the...

Now that we're past basic physics, I thought it might be interesting to discuss the two voltage conventions in a bit more detail. One of the things the diagrams that Mabilde or RSD Academy always do is assume the resistors are of negligible length. I think that hides the true nature of what the...

Interesting, I've never seen that visualization before. I always picture the magnetic field vectors as fixed in place, with the magnitude fluctuating. But as you said, it's the induced E-field that actually pushes charge. These are after all just visualizations, and any visualization that leaves...

Here are a couple diagrams, which include both changing magnetic field ##\vec B## and the induced electric field ##\vec E##. This first one shows the changing magnetic field pointing into the page, and the induced electric field circles counter-clockwise WITHIN the flux area. This one...

This is a much better illustration of what's going on - flux is the amount of field that passes through any defined area. Whether the field lines 'bloom' in or out a bit towards the ring is irrelevant. The integral that defines flux (below) only counts the part of the magnetic field ##\vec B##...