Wiring electric baseboard

[trader_4]

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.

[trader_4]

Thanks Doug. One more step in stopping the insanity! We have a real winner here. He doesn't understand basic math and claims that if you can manipulate any equation so that a zero could possibly be in the denominator, then the whole equation is invalid. That makes pretty much all equations invalid.

He also fails to graph that when an equation produces a result of zero, it's often the most interesting, or defining case. For example, we can apply Newton's Law to give the velocity of a ball shot up in the air. At the peak, the equation give a value of Zero. To you and I, that's a valid answer, it means the velocity is zero, the ball isn't moving, etc. To him, I guess it's "undefined".

[FromTheRafters]

That doesn't mean it is true, as it is an appeal to numbers.

D=RT

Undefined, because not moving (a rate R=D/T of zero) means the time it takes to go *any* distance (even zero distance) is infinite - or you divide by zero to avoid multiplying by infinity. Either way it is undefined.

Do I? Break it down and tell me how you arrived at your conclusion, being careful not to divide by zero or multiply by infinity.

No fair using calculus, because then you have non-zero values which approach zero.

[trader_4]

Doug said it best:

"Complete nonsense."

He and I have explained it to you:

"There's no division in D= RT. That's multiplication."

And now as has been your habit, you want to go off into other area, eg having to use calculus? My God, you don't understand 7th grade math!

BS. No calculus required.

V = IR I = 0, voltage is zero. There is no division by zero.

Did you plot that graph yet of V vs I? It's a straight line right through the origin. At 0 current, there is 0 voltage. Not undefined, no singular event, it's ZERO pure and simple.

[trader_4]

Notice again, there is no definition from you as to the alleged differences and why the voltage drop and voltage loss relative to 100 ft run of wire are not the one and the same. If you're so smart, why is that? I defined them, neither you nor your half-assed cites did.

No answer noted.

So you really are the village idiot, incapable of 7th grade math. There is no division by zero there.

I did, many times. So did Doug.

D = RT

D = 0 M/sec x 60 secs = 0

There was no division by zero, no multiplication by infinity.

You can't even understand math basics, stop already with the false BS about calculus. You're not intimidating us.

[Sam E]

[snip]

How about "What is the weight of 0 lead balls". 0 pounds (or 0 newtons if you prefer, but that doesn't really change anything).

[FromTheRafters]

The term 'voltage drop' refers to a difference in voltage potentially available to other devices in the circuit (read as 'closed circuit') because of the device dissipating some of the available energy because the current through the device causes the device to, for example, emanate heat. Zero current cannot cause this phenomenon, so there is no 'voltage drop'. If there is some other reason for there being a discrepancy between what voltage 'should' be there, it is not due to 'voltage drop' but might be a burnt open fuse or other fault.

I'm saying that Ohm's Law as it pertains to 'voltage drop' requires that the current be non-zero. Algebra is not the same as arithmetic. Algebra is about taking the entities and rearranging them according to certain rules V=IR for instance can be algebraically changed to R=V/I by dividing both sides by I or to I=V/R by dividing both sides by R. This can't be done if the numbers you choose to use are causing you to divide by zero or multiply by infinity. Ohm's Law is valid if you choose the right numbers, not so much if you don't.

Are they teaching algebra in the seventh grade now?

In arithmetic you don't really care about the relations between numbers and I suppose it is okay to overlook the problems which arise by dividing by zero and multiplying by infinity in transformed versions of the same algebraic structure.

The natural number one with an exponent of two can equal two under the right circumstances, but I wouldn't expect a seventh grader to know this or that adding *all* natural numbers together gives -1/12.

[FromTheRafters]

trader_4 was thinking very hard :

I can see that you are getting very upset, so I will not continue this.

HAND

[Mark Lloyd]

Resistance is infinite.

My computer did the calculation right. Some don't. I pressed '1', '2', '0', '/', '0', '=' and the display showed 'inf'.

[trader_4]

You failed again. The assignment was to define voltage drop AND voltage loss and explain why they are not one and the same for the example we're talking about, 100 ft of wire. You defined only voltage drop and even there you have it wrong. A current of zero produces a voltage drop of zero.

A ball thrown up in the air, we can write the equation for it's velocity at any point. At the peak, the equation gives zero for it's velocity, which of course is correct. Following your logic, that wouldn't be right either. We don't say the velocity can't be defined. It is defined, it's zero, just like the voltage across a resistor with no current flowing.

But regardless of that, voltage drop and voltage loss are one and the same in our example of 100 ft of wire going to the heater load. They are both calculated the same, given by Ohm's LAw. I can't believe anyone would argue this.

You just said that with V = IR and a value of .16 for R, 0 for I, it's undefined. Sorry, you failed 7th grade math.

Did you draw that graph yet? Straight line right through the origin? What's up with that?

We are using the right numbers.

V = IR, V= I *.16

I = 0, then V = 0, there is no division by zero. You can take almost any equation, rearrange it so that a variable is in the denominator. Sure if you then try to actually do that, put a zero value there for that variable, then it's undefined or infinity. But no one is doing that in the above equation, it's just multiplication by zero.

You only need to worry about dividing by zero if it actually is happening. No one is dividing by zero, only multiplying by zero.

More diversion into the wilderness. But I'll play along. Explain to us under what circumstance 1 squared is equal to two.

[Scott Lurndal]

Whereas I would not say the ball has an absolute velocity of zero, since it's still moving several hundred miles/hour along several vectors (daily rotation around the earths axis, yearly rotation around the suns axis, longer-term rotation around the center of the galaxy, et cetera, et alia).

If you're refering to relative velocity, then you need to give a referent.

[trader_4]

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.

[Scott Lurndal]

One can postulate conditions where 100 lead balls have a weight of (very close to) zero.

Perhaps you meant mass?

[clare]

EXCEPT there is a rule that the total voltage drop across a circuit and the supply voltage MUST be equal - and when you connect a voltmeter across the open portion of the circuit you WILL read source voltage - so BY DEFINITION it is a "voltage drop" without current flow.

[trader_4]

Oh, please, stop with the nonsense. Apparently you never took high school physics either. That type of question is a typical one found on physics tests covering Newton's Laws. Cannon gets fired, ball goes up in the air, object gets dropped. Calculate the velocity, the height, etc. It's on the physics SATs. They don't preface the whole damn thing with a page of disclaimers about not including the motion of the earth, the universe, realtivity considerations, etc. Did you go to school with Rafters?

[clare]

If there is theoretically a circuit, yes. If all you have is a battery and some wires - which are not connected - then no. As soon as you describe it as a circuit - yes..

[clare]

ferget it!!! It's like a bunch of theologians arguing about how many angels can dance on the point of a pin.

Bend the pin into a circle and it becomes pointless - - - -

[Doug Miller]

FromTheRafters wrote in news:niklfm$kju$1 @news.albasani.net:

That is incorrect. In the equation D = RT, if D and R are both zero, T can -- and must -- be any finite number (e.g. if T = 5 and R = 0, then D = RT = 0 * 5 = 0). The one thing T *cannot* be is infinite: zero times "infinity" is what we call in mathematics an "indeterminate form" that cannot be evaluated algebraically.

Correct, but not relevant to the equation D = RT which is *not* the same thing:

-- D = RT is defined for all real values of all three variables

-- T = D / R is defined for all real values of D and T, and non-zero values of R

Time is not infinite in this equation; as explained above, T must be a finite value.

You can't get around it *with* calculus either.

More nonsense. It does not appear that you ever actually had a course in calculus; at any rate, you certainly didn't *pass* it...

And when either I or R *equals* zero, so does E.

That depends on how it's expressed. As V = IR, zero values are perfectly legitimate.

I'm not getting into *that* argument. I'm just trying to correct the complete and utter nonsense you've been spouting about algebra.

[FromTheRafters]

[e-mail] has brought this to us :

It's not a circuit unless there is continuity "open circuit" is a misnomer. When you put the multimeter across the open (gap) the internal resistance of the meter completes the circuit and you read the 'voltage drop' across that internal resistance which tells you what the supply voltage is, because as you said, they must be equal.

[Doug Miller]

FromTheRafters wrote in news:niklg0$kke$1 @news.albasani.net:

What, you mean your point is that you're wrong? You just stated "you can take any two known quantities [in Ohm's Law] and calculate the third". I just showed that's not true.

Moreover, you're missing the point here rather badly. Given V = IR, with I = 0, it's not possible to calculate any specific value for R precisely because *any finite value* multiplied by zero is still zero. If V = 0 and I = 0, R could be anything at all -- but that doesn't mean that Ohm's Law doesn't work. Rather, it confirms that Ohm's Law *does* work, because the physical representation of 0 = 0R is that no matter what the resistance is, if there's no potential difference no current will flow.

I suppose I'd agree with that statement -- but I'm not sure you realize that there does not have to be current for there to be a potential difference.

It's *your understanding* of Ohm's Law, and of junior high school algebra, that are broken.

False. This is an impossible situation. If current is zero, then voltage *must be* zero; conversely, if voltage is non-zero, then current and resistance must both be non-zero also.

You don't bolster your argument at all by reposting a question asked by someone as ignorant of basic algebra as yourself, especially when you clearly failed to read and understand the many correct explanations in the answers.

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