# Wiring electric baseboard

Page 7 of 10
• posted on June 1, 2016, 5:07 am
On Tuesday, May 31, 2016 at 6:36:49 PM UTC-4, FromTheRafters wrote:

OMG, more stupidity.
You posted:
"The natural number one with an exponent of two can equal two under the right circumstances."
An exponent of two is a number raised to the second power. It's the number times itself. Now explain to us how 1 x 1 = 2.
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<%-name%>
• posted on June 1, 2016, 11:38 am
trader_4 presented the following explanation :

I didn't say 1x1=2, and you would not understand what I meant if I told you, so I won't even attempt to do it.
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<%-name%>
• posted on May 31, 2016, 7:45 pm
On 05/31/2016 08:49 AM, trader_4 wrote:
[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).
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<%-name%>
• posted on May 31, 2016, 8:35 pm

One can postulate conditions where 100 lead balls have a weight of (very close to) zero.
Perhaps you meant mass?
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<%-name%>
• posted on May 31, 2016, 10:41 pm
Scott Lurndal expressed precisely :

Nice! :) Watch out though, E=MC^2 can take on all sorts of funny stuff when it is just an equation where C can be set to zero and solved by seventh graders.
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<%-name%>
• posted on May 30, 2016, 9:06 pm
On Mon, 30 May 2016 10:37:01 -0400, FromTheRafters

You are not smart enough to brak ohm's law - and the water theory of electricity doesn't exactly hold water.
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<%-name%>
• posted on May 30, 2016, 3:12 pm
On Monday, May 30, 2016 at 7:13:00 AM UTC-4, Al Gebra wrote:

Seems to me Mr. Ohm would be proud, like most of us, you have it right.
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<%-name%>
• posted on May 30, 2016, 8:19 pm
On 05/30/2016 06:13 AM, Al Gebra wrote:
[snip]

IIRC, it's I squared R. Of course, it's still no voltage drop with no current. Is it possible there's a confused poster here, who has R and thinks it's E.
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• posted on May 30, 2016, 8:41 pm
On Monday, May 30, 2016 at 4:19:08 PM UTC-4, Sam E wrote:

(I^2)R is the power drop. The voltage drop is IR
And no, it's not about a simple mistake, Diesel doesn't understand Ohm's Law and electricity 101.
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<%-name%>
• posted on May 30, 2016, 8:50 pm

Power drop?
If there is such a thing in this context, which I doubt, it probably requires current too.
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<%-name%>
• posted on May 30, 2016, 9:22 pm
On Mon, 30 May 2016 16:50:22 -0400, FromTheRafters

Lost power in the circuit is a function of the voltage drop.
It is all moot in the case of the resistance heater in the example that started this nonsense thread. All of the power lost to voltage drop will still be returned to the home in the form of heat and that was the object of the exercise in the first place.
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<%-name%>
• posted on May 30, 2016, 11:00 pm
On Monday, May 30, 2016 at 5:22:14 PM UTC-4, snipped-for-privacy@aol.com wrote:

Yeah, here's the Rafters guy trying to explain electricity basics and algebra to us, and he can't even grasp the idea of the power drop that corresponds to the voltage drop.
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<%-name%>
• posted on May 31, 2016, 12:23 am
trader_4 was thinking very hard :

Sorry to have to tell you this, but there can be no power without any current either.
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<%-name%>
• posted on May 31, 2016, 1:56 pm
On Monday, May 30, 2016 at 8:23:53 PM UTC-4, FromTheRafters wrote:

Yes, and that's why P = R*I^2 gives a value of ZERO, when I = 0. That zero is the value of the power and it too has meaning.
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<%-name%>
• posted on May 30, 2016, 10:07 pm
On 05/30/2016 03:50 PM, FromTheRafters wrote:
[snip]

There's power dissipated by the wire (mostly as heat). And that does require current.
--
Mark Lloyd
http://notstupid.us/
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• posted on May 31, 2016, 12:40 am
Mark Lloyd formulated on Monday :

Yes, it sure does. I always heard of that as power *loss* or copper *loss* though. Power Drop was a sort of tap which went from the pole to the customer.
The term 'voltage drop' was only used in circuits with current flowing through them. All of the deflections aside, I'm still not ready to believe that a blown fuse has a voltage drop across it no matter what these brainiacs say.
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<%-name%>
• posted on May 31, 2016, 3:50 am
On Mon, 30 May 2016 20:40:56 -0400, FromTheRafters

It HAS tyo have, because with zero(or as close to zero as the real world can produce) current flow in the conductors and an ipressed voltage of 120, or whatever volts across the cirduit, the voltage HAS to drop across something - in this case the "infinite" or "undrfined" resistance across the blown fuse.
Simple physics.
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• posted on May 31, 2016, 10:50 am
snipped-for-privacy@snyder.on.ca has brought this to us :

It's not a 'voltage drop' when there is no current. Voltage drop exists because of the dissipation of energy across the device under consideration.

There's not much simple about physics.
From Wikipedia:
"Zero electrical DC resistance
[...]
The simplest method to measure the electrical resistance of a sample of some material is to place it in an electrical circuit in series with a current source I and measure the resulting voltage V across the sample. The resistance of the sample is given by Ohm's law as R = V / I. If the voltage is zero, this means that the resistance is zero."
If there was no voltage (a voltage drop actually) measured, you would be fooled into believing that there is zero resistance if you leave out the part about the need to have current flowing through the device.
With no current, you can't trust Ohm's Law to give a meaningful result. See above where R = V / I. If I is zero, R is *undefined* not "zero" or "infinity".
The above example uses a "circuit" (not an open circuit which isn't actually a circuit at all) and a "current source" with current flowing through the device.
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• posted on May 31, 2016, 1:54 pm
On Monday, May 30, 2016 at 8:41:04 PM UTC-4, FromTheRafters wrote:

A blown fuse that's in a circuit has the full open circuit voltage across it. Use a meter and see. It's one of the ways we find blown fuses in a circuit.
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<%-name%>
• posted on May 31, 2016, 3:16 pm
trader_4 presented the following explanation :

Yes, but it is not a 'voltage drop' because there is no current across the open fuse and no energy dissipation taking place.