Wiring electric baseboard

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Mon, 30 May 2016 18:18:28 GMT in alt.home.repair, wrote:

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MID: <nb7u27$crn$ snipped-for-privacy@boaterdave.dont-email.me>
Hmmm. I most certainly don't understand how I can access a copy of a
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Diesel brought next idea :

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My apologies for posting this, but he started it.
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On Mon, 30 May 2016 11:45:57 -0400, FromTheRafters

Quit while you are ahead. If there is no semiconductor in the "open circuit" there will be no voltage drop. As soon as you put a meter on to check the voltage it IS a circuit.. A smiconductor has a "forward voltage drop" that behaves differently than a resistance - but we are not talking about semiconductor physics here.
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snipped-for-privacy@snyder.on.ca brought next idea :

We're not talking about completing a circuit with a meter either are we? Show me how a semiconductor has a voltage drop without any current flowing and maybe I'll take your comments seriously.
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On Mon, 30 May 2016 17:26:52 -0400, FromTheRafters

current - other than requiring the current to be NON zero. - and it is vityually impossible to read voltage anywhere in a circuit without completing the circuit - in a DC circuit requiring a resistive load - an an AC circuit either a resistive, inductive or capacitive load. - which means you can NOT measure voltage in a totally unloaded or "open" circuit.
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snipped-for-privacy@snyder.on.ca explained :

There you go then. Besides, semiconductors are sometimes incompatible with Ohm's Law.
https://www.physicsforums.com/threads/confused-about-ohms-law.310101/

True, but that is not the point. Does voltage drop even exist theoretically when there is theoretically no current?
By definition, you need current to have energy dissipation in the device and thus a 'voltage drop'. If a device is 'heating up' you can be sure there is current through it without attaching a multimeter and completing a previously open circuit (which isn't actually a circuit - since it is open).
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On Tue, 31 May 2016 06:24:56 -0400, FromTheRafters

battery and some wires - which are not connected - then no. As soon as you describe it as a circuit - yes..
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On Monday, May 30, 2016 at 11:11:39 PM UTC-4, snipped-for-privacy@snyder.on.ca wrote:

Sigh. Sadly that is incorrect too. The ideal diode equation for the perfect semiconductor shows the relationship between the current through the diode and the voltage across it. It's exponential. The current is a function of the voltage, it's just that the current goes up a lot for a small change in voltage.
- and it is

Of course you can measure voltage in an open circuit. Stick the test leads of a high impedance meter into a receptacle with no loads on that circuit. For all practical purposes, you're measuring the open circuit voltage, you'll see ~120V. The only load is the insignificant impedance of the meter, in the millions of ohms. That load is so small, it's less than the accuracy of the meter itself. And if you want to use more exotic test equipment, clearly we could measure it without even that insignificant load.
We have a guy here who doesn't understand basic algebra and Ohm's Law. IDK why you're dragging in semiconductor physics, AC circuit dynamics and leading the discussion into rat holes.
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trader_4 expressed precisely :

Thanks for recognizing the fact that you're in way over your head here, brainiac.
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On Tuesday, May 31, 2016 at 11:05:36 AM UTC-4, FromTheRafters wrote:

You're the only one that doesn't understand basic algebra. No need to go into semiconductor physics.
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On Mon, 30 May 2016 10:37:01 -0400, FromTheRafters

Yeah that Ohm guy was a moron. I suppose we should throw out the rest of electrical engineering too as just a flawed theory.
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It happens that snipped-for-privacy@aol.com formulated :

No, *he* knew quite well what he was talking about.

Ah, engineering, no wonder the math skills are so weak. They use calculus all day long but few of them know how or why it works.
Show me a definition of "voltage drop" that doesn't involve current flowing in a circuit. I'll wait, but I haven't got forever so be quick about it.
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On 05/30/2016 01:25 PM, FromTheRafters wrote:
[snip]

Normally, the voltage on a wire will be the same at all points regardless of wire size or current flow. "Voltage drop" is what happens when the copper atoms get tired of being so reasonable, and assert their need for "silly time" (its their version of a "smoke break"). Since they can do this at any time, it needs to be accounted for in electrical circuit design.
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hah has brought this to us :

Finally, someone who knows what they're talking about. LOL
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On Mon, 30 May 2016 14:25:32 -0400, FromTheRafters

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snipped-for-privacy@snyder.on.ca has brought this to us :

Where?
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On Monday, May 30, 2016 at 2:25:40 PM UTC-4, FromTheRafters wrote:

V = IR is the formula for voltage drop. Put in zero for I, any finite resistance for R, you get V = 0. That tells you there is no voltage drop. And yes, contrary to your BS, zero does have meaning. Hell, you can even graph this, V versus I, it's a straight line and it goes right through the origin. At zero current, the voltage is zero and yes, that zero has meaning. PS: I didn't do any division by zero either.
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After serious thinking trader_4 wrote :

Of course there isn't, because there is no current. You can't have a voltage drop when there is no current. Thanks for finally agreeing with me.
--
Getting there was like pulling teeth though

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On Mon, 30 May 2016 20:13:21 -0400, FromTheRafters

the circuit thogh.
You've ( it appears) been arguing both sides of the equation.
To avoid confusion - WHOEVER said there would be a voltage loss in an unloaded wire is WRONG. Also whoever said there is no voltage drop in an open circuit is ALSO WRONG. And to top it all off, in order to measure the voltage across an open circuit, you MUST close the circuit - meaning it is no longer an "open" circuit, AND In the real world there is no such thing as zero ohms. You can get REAL close - but "in the wild" it does not exist. - so you are never REALLY deviding by or multiplying by ZERO when solving ohm's law
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After serious thinking snipped-for-privacy@snyder.on.ca wrote :

I'm not talking about a voltage loss, but a 'voltage drop'.

Then you are calling yourself wrong. There is no current in an open circuit, and you have agreed several times (correctly) that you need current to have a 'voltage drop'.

I'm not talking about measurements at all.

You're wrong about that too.
https://en.wikipedia.org/wiki/Superconductivity
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