Wiring electric baseboard

Page 9 of 10  
On Monday, May 30, 2016 at 6:01:38 PM UTC-4, FromTheRafters wrote:

He's right, you're in way over your head. This silliness over Ohm's law started over the voltage drop over a 100 ft of #12 wire. It's also referred to as voltage loss. Everyone here in the thread at least understands that.
And again, V= IR. With a current of 14 amps, a wire resistance of .16 ohms you get a voltage loss of 2.2 Volts. Now put in a current of zero, and what do you get? Voltage loss of Zero. And it has meaning, with no current flowing the voltage loss is zero, we have the full supply voltage at the far end of wire. Note: No division by zero was done here, no electrons were harmed either.
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On 05/30/2016 06:11 PM, trader_4 wrote:
[snip]

Did you consider that if the heater is 100 feet from the power source, there's 200 feet of wire?
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On Monday, May 30, 2016 at 8:19:32 PM UTC-4, notX wrote:

Yes, go look at my original calcs when we were discussing the actual issue. Now we have sadly descended into lala land because the village idiot doesn't understand basic algebra or Ohm's Law.
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trader_4 used his keyboard to write :

Don't be so hard on yourself, it isn't an easy subject.
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trader_4 presented the following explanation :

They're the same? Tell these guys then, 'cause they have it all wrong.
http://ecmweb.com/electrical-testing/voltage-loss-versus-voltage-drop
https://appauto.wordpress.com/2008/01/04/voltage-loss-or-voltage-drop-that-is-the-question/
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On Monday, May 30, 2016 at 9:26:41 PM UTC-4, FromTheRafters wrote:

OK, let's look at them one at a time. The first guy is just a poor writer who doesn't even define the terms he's using. What he's saying is "voltage drop" is the theoretical voltage drop given by Ohm's Law for a circuit. He then uses the term "voltage loss" for the actual MEASURED voltage difference in that circuit. But there is no such industry standard definition and it has nothing to do with what we are talking about here. That measured "voltage loss" is just the theoretical voltage drops from Ohm's Law, plus other voltage drops, from other resistance in the circuit other than the theoretical resistance of the wire. Each connection, switch, etc will have some resistance, which gets added to the total. So, you wind up with the measured voltage being less than the theoretical from the wire alone, ie the measured voltage drop is larger than the theoretical. That is all he's saying, but he's not very clear about it and it has nothing to do with the example here. Voltage drop and voltage loss here are one and the same, because we are only considering the ideal circuit, without those additional resistances. We could put an additional resistor in to model that, but it would make no difference in the fact that with zero current, the voltage drop or voltage loss is zero.
The second guy is a true village idiot. Note that again, no definition of voltage drop or voltage loss is ever given. He just goes off to diagrams and it's not clear from looking at some of those diagrams, what his point is.
Let me do for you what neither of these two did, nor can you, which is give definitions. Voltage drop is the voltage across a component, which in the case of a resistance load is given by Ohm's Law, V = IR. Voltage loss is typically used to describe unwanted voltage drop in a circuit due to the resistance of the wire, connections, other components, etc.
In the simple case we're talking about the voltage drop across the wire is also the voltage loss. Capiche?
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trader_4 explained on 5/31/2016 :

Which it is.

Yes it is, the voltage across an open circuit like across a blown fuse is not determined by the dissipation of energy through it, but 'voltage drop'is (only if there is current). You or gfretwell or Clare brought the term loss into the discussion, not me. If I had said loss it would have been 'copper loss' not voltage loss or power loss. Somebody (Clare?) said they were the same as 'voltage drop' - which they aren't.

It doesn't matter what resistance is dissipating the energy, it is all 'voltage drop'.

No they are not.

No, it is "undefined".

Maybe they just expect everyone to know about it already? That doesn't make *them* the village idiots now does it?

If you're going to give definitions, give the right ones please.

See above comment regarding correct definitions. IMO the word "typically" does not belong in a proper definition, but maybe that's just me.

See the comment above the comment above for further enlightenment.
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On Tuesday, May 31, 2016 at 11:44:46 AM UTC-4, FromTheRafters wrote:

Then do what neither of your crappy citations did. Define voltage drop and voltage loss for us and then explain how in the context of our 100 ft of #12 wire, the voltage drop across that wire and the voltage loss across that wire are not one and the same. You can't because they are. It's voltage drop following Ohm's LAw and it's also referred to as "voltage loss" because it's voltage that is not available to the load.

Here Johnny, 7th grade math test:
V = IR r=.16 ohms, I = 0
What is the value of V?
All of us say it's ZERO.

Writing a tutorial you don't expect everyone to know about it. If they did there would be no point in his tutorial. Actually there is no point in that tutorial, because he doesn't know WTF he's doing or even trying to do either.

I did. Where are the definitions from those citations? Oh, they have none. Where are yours?

So, tell us your definition. You're the one bringing up this crap about voltage drop vs voltage loss. YOU tell us what you claim the distinction is. The rest of us know that in the case we're talking about, there is no distinction. We have a 100 ft of wire. The voltage drop across it is also the voltage loss.

See the fact that the neither the village idiot nor his cites can define the terms he's trying to argue about.
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trader_4 was thinking very hard :

I can see that you are getting very upset, so I will not continue this.
HAND
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On Tuesday, May 31, 2016 at 3:58:39 PM UTC-4, FromTheRafters wrote:

Yeah, I do tend to get upset when we have village idiots here that are clueless trying to explain to those of us who are degreed electrical engineers, electrical inspectors, etc how ohm's law works and basic algebra.
BTW, did you make that graph yet of V = IR, plotting V vs I? It goes right through the origin, does it not? 0,0 is that "undefined"?
Still waiting for you to explain your new claim too, where you claim that 1 squared can equal 2. I have bets it's only in your special little universe.
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On Monday, May 30, 2016 at 1:57:48 AM UTC-4, Diesel wrote:

Good grief, you need to take electricity 101. Gfre is correct, with no load there is no voltage drop, the full 240V is available and can be measured at the end of the 100 ft of wire. The only way you get voltage drop is with a load and current flowing. Let RL be the load connected to a voltage source V using wires that have a resistance RW. With RL not present the voltage at the open wires where RL would be is equal to V. With the load RL inserted, the voltage across RL is V- RW*V/(RL+RW). You should know this just from common experience. Measure the voltage at a receptacle on a long circuit with no loads on it, you get the same voltage as you do at the panel. Turn all the appliances, lights, etc on and you get some number of volts less, due to the voltage drop.

Resistor is a resistor man. And a heating element looks more like a resistor that an intelligent widget that draws more amps. Here's a though experiment. I take a 3500W heating element that draws 14.6 A at 240V. Now I hook it up instead to a 24V source. Does it draw 146 A?
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Mon, 30 May 2016 15:09:06 GMT in alt.home.repair, wrote:

No, it doesn't. It's about 1.5amps at 24 volts. 35.1 watts. My apologies for the confusion on my end!
used:
ohms=voltage/amperage 240/14.6.4 (16.438)
watts=volts(x2)/ohms 24x24/16.45.1 (35.121)
amps=watts/volts 35.1/24=1.5 (1.4625)
--
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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On Monday, May 30, 2016 at 9:55:51 PM UTC-4, Diesel wrote:

Thank you for admitting your mistake. It's something rarely done around here.
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trader_4 explained on 5/31/2016 :

That's something I really like about Diesel, and you're right - it is a rarity.
--
67654A30666F6E306A654A31666E2F5066654A32676F46306537546B656A3D3D

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On 05/29/2016 09:23 PM, Diesel wrote:
[snip]

I'm often surprised by how little some people know about electricity. Voltage drop depends on current flow. There's no power (wattage) without current flow either.
So you have a load which is not present. How is it drawing power?
[ship]
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Mon, 30 May 2016 03:14:39 GMT in alt.home.repair, wrote:

I sometimes feel the same way.

If you put 120volts AC on a 12/2 wire, and run that wire, 100 ft out, and take a measurement at the end of the run, the available voltage at the end of the run isn't going to match the volts going into the wire. Some is lost during transit due to the wires resistance and the length.
If you try this experiment using DC power, it's even worse. DC really doesn't travel distance well.
https://en.wikipedia.org/wiki/Voltage_drop
For example, an electric space heater may have a resistance of ten ohms, and the wires which supply it may have a resistance of 0.2 ohms, about 2% of the total circuit resistance. This means that approximately 2% of the supplied voltage is lost in the wire itself. Excessive voltage drop may result in unsatisfactory operation of, and damage to, electrical and electronic equipment.
The simplest way to reduce voltage drop is to increase the diameter of the conductor between the source and the load, which lowers the overall resistance. In power distribution systems, a given amount of power can be transmitted with less voltage drop if a higher voltage is used. More sophisticated techniques use active elements to compensate for the undesired voltage drop.

You aren't losing power due to the load, You've already lost it in route to the load via wire resistance and the distance said electricity has to travel on the wire. Thicker wire, less resistance, less voltage drop. Simple concept, really.
Like you wrote though, it does sometimes amaze me how little people know/understand about electricity.
--
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wrote:

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On Monday, May 30, 2016 at 1:57:47 AM UTC-4, Diesel wrote:

BS. There is no "transit", the current flow is zero, the voltage across the wires is the same at both ends. The voltage drop for each wire is V = I * RW, where I is the current, RW is the resistance of the wire. Set I to zero, and what do you get? This is a very simple application of Ohms's Law. And if you think this is wrong, explain what the correct formula for the voltage drop is.

Sigh, more FUD.

More wandering in the wilderness. Just apply Ohm's Law to the actual circuit, like I did above.

Yes, it's simple, unfortunately you have it all fouled up. With no load, there is no loss, because no current is flowing, period.

Even more amazing is how people that don't understand Ohm's Law make comments like that.
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wrote:

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On 05/29/2016 09:23 PM, Diesel wrote:
[snip]

If it says you have voltage drop without load, it SHOULD be dismissed.
[snip]
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