On Mon, 30 May 2016 06:53:52 -0000 (UTC)
Diesel wrote:


Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.

No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Watts is watts, man. Available voltage determines how many amps
it's going to take to get them, though. Lower voltage=more amps to
do the same job.

If you have a lower voltage, the watts will be lower. In our
example, that 1500 watt heater (at 240v) will be 1407 watts at
232v. Current will actually be less, not more.

Not exactly. You can have high voltage and next to no current behind
it (think of a taser) or low voltage and a considerable amount of
current behind it (think of your car battery) it's no chump and your
starter motor isn't exactly a low drain device...You can also lose a
few volts in our example, and, still be able to pull enough current
to make up the difference.

The easiest way I know of to explain the relationship is this:

It may be useful to consider the image of water in a hose. Voltage is
equivalent to pressure, water flow is equivalent to current and the
diameter of the pipe is equivalent ot the thickness of the wire - or
resistance.

No wonder you are the gopher and not a real electrician.
Stick to fetching the donuts before you burn something
down.

On Monday, May 30, 2016 at 1:57:47 AM UTC-4, Diesel wrote:
Sam E
Mon, 30 May 2016 03:14:39 GMT
in alt.home.repair, wrote:

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

[snip]

1500/232.1=6.5 amps *rounded up*
1500/235=6.39 amps *rounded up*

And that's due entirely to distance with no loads present on the
line yet.

I'm often surprised by how little some people know about
electricity.

I sometimes feel the same way.

Voltage drop depends on current flow. There's no
power (wattage) without current flow either.

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.


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.



If you try this experiment using DC power, it's even worse. DC really
doesn't travel distance well.


Sigh, more FUD.


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.


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




So you have a load which is not present. How is it drawing power?

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.


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




Like you wrote though, it does sometimes amaze me how little people
know/understand about electricity.



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

On Monday, May 30, 2016 at 1:57:48 AM UTC-4, Diesel wrote:
notX Mon,
30 May 2016 03:19:44 GMT in alt.home.repair, wrote:

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

[snip]

Just to clarify, the calculations I provided in this post aren't
FUD, unless you're able to dismiss the yellow Ugly electrician
reference book.

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

http://www.uglys.net/

You have voltage drop due to the length and size of the wire. Because
the wire isn't a super conductor.

At the end of the day, you might have put 120 volts on the line, but,
100ft down that line, you aren't getting 120 volts. Some has been lost
on the wire, due to the wires own resistance.

Heh, smart guy, give us your formula for that voltage drop in the wire.
The rest of us here, have told you that from a very simple application
of Ohm's Law, it's V = I*RW where I is the current flowing and RW is the
resistance of the wire. Set I to zero, what's the voltage drop? ZERO.
So, we can calculate it. Show us YOUR calculation.



Switch from AC to DC with no other changes, and the voltage drop is
more pronounced.


More FUD and diversion. The differences between switching from 240V AC
to DC in this discussion is totally negligible. Another example of an
amateur taking something they heard or saw and misapplying it.

PS: With DC, just like with AC, if the current is zero, the voltage drop
on the wires is zero.

trader_4 brought next idea :
On Monday, May 30, 2016 at 10:37:10 AM UTC-4, FromTheRafters wrote:
on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.
No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero as
well.

That's a good theory, but IMO it is wrong.

No theory, it's correct from electricity 101. With an open circuit, no
current flow, the voltage across the wires is the same at both ends.

That I can agree with. There is no "voltage drop" because there is no
circuit in which current can flow. Voltage drop is defined in terms of
a circuit with current flowing from the source through a load and back
to the source and voltage drop is across all of the resistances in the
circuit including the internal resistance of the source.

Think of it this way:

Distance equals rate times time (D=RT) and you have two trains on a
railroad track. Detroit to Chicago on one end and Chicago to Detroit on
the other. Neither train is moving. Does the distance between them drop
to zero?

No, it doesn't.

Voltage drop has no meaning in an 'open circuit', which isn't actually
a circuit at all, just as 'rate' has no meaning for objects which
aren't moving.

From Wikipedia:

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces through which electric current
can flow."

If current can't flow, it stands to reason that it is not a circuit,
and 'voltage drop' has no meaning.

On Monday, May 30, 2016 at 2:14:05 AM UTC-4, Diesel wrote:
hah
Mon, 30 May 2016 05:25:23 GMT in
alt.home.repair, wrote:

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

[snip]

1500/232.1=6.5 amps *rounded up*
1500/235=6.39 amps *rounded up*

And that's due entirely to distance with no loads present on the
line yet.

[snip]

Where can I get one of these magic heaters, that uses 1500 watts
when it isn't connected?

Hmm? Oh.. ROFL, I see. That's what happens when you don't pay
attention to context and choose to selectively quote.


He's paying attention and his sarcasm is on point.


Those are load calculations based on a 1500watt element running at
two different voltages. It shows the amount of amps required to get
1500watts at those voltages.


Except of course heating elements look like resistors and they don't
magically readjust to pull more amps with a lower voltage to keep
the watts constant. At 240V, a 3500W heating element pulls 14.6A.
Following your logic, if we instead put 24V on it, it will pull 146 A
and continue to put out 3500W. That makes sense to you?



The different voltages are directly related to length and diameter of
wire and have nothing to do with the load from the heater at this
point.


Give us the formula for that voltage drop in the wire. The rest of us
know it's V = I*RW, where I is the current, RW the resistance of the wire.
Ergo, I = 0, V = 0. Capiche? That means that with no load, the full
240V is present at both ends of the long wire. Now put even a tiny load
on and now you have voltage drop. The larger the load, the larger the
voltage drop in the wires, because the current is higher.



Thanks to the wire and the wire alone, some volts have already been
lost (as in they'll never reach the heater) in transit, due again, to
the wires own resistance and length. In this case, the length is the
same; 100ft. But, the wire size or gauge is not. 12ga is on the top,
10ga is on the bottom; for comparison.


Again obvious you don't understand the basics. Volts are not "lost"
If they are and you're so smart, give us the formula for the "lost"
volts with no current flowing.



Might I suggest a simple english comprehension class or two? Muggles
suffers from a similar problem. So, you're in very good company.


Might I suggest a course in electricity 101? Understand how to apply]
Ohm's Law to a simple circuit? Kirchoff's Laws?

On Monday, May 30, 2016 at 12:14:44 AM UTC-4, Diesel wrote:

Sat, 28 May 2016 23:17:05 GMT in alt.home.repair, wrote:

On Sat, 28 May 2016 20:04:33 -0000 (UTC), Diesel
wrote:


Thu, 26 May 2016 05:50:22 GMT in alt.home.repair, wrote:

"Continuous load" means you size the circuit to 125% of total
load (14.6*1.25=18.25a)
so 18.25a is OK on a 12 ga wire with a 20a breaker

You're going over the 80% trade standard load (16amps on a 20amp
circuit) at that point,

Excuse me but the load is 14.6a. The 18.25a IS 125% of the load so
the 80% has been accounted for.

I sincerely do apologize for having missed that in my original reply
to you.

Sadly, that isn't all that you missed and misapplied.

On Monday, May 30, 2016 at 11:46:09 AM UTC-4, FromTheRafters wrote:
trader_4 brought next idea :
On Monday, May 30, 2016 at 10:37:10 AM UTC-4, FromTheRafters wrote:
on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.
No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero as
well.

That's a good theory, but IMO it is wrong.

No theory, it's correct from electricity 101. With an open circuit, no
current flow, the voltage across the wires is the same at both ends.

That I can agree with. There is no "voltage drop" because there is no
circuit in which current can flow. Voltage drop is defined in terms of
a circuit with current flowing from the source through a load and back
to the source and voltage drop is across all of the resistances in the
circuit including the internal resistance of the source.

Think of it this way:

Distance equals rate times time (D=RT) and you have two trains on a
railroad track. Detroit to Chicago on one end and Chicago to Detroit on
the other. Neither train is moving. Does the distance between them drop
to zero?

No, it doesn't.

Voltage drop has no meaning in an 'open circuit', which isn't actually
a circuit at all, just as 'rate' has no meaning for objects which
aren't moving.


From Wikipedia:

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces through which electric current
can flow."

If current can't flow, it stands to reason that it is not a circuit,
and 'voltage drop' has no meaning.

It does have meaning, it's ZERO. Zero in physics has meaning. It can
be measured. The voltage on the ends of that 100 ft of wire is exactly
the same., because the voltage drop across the wire with no current
flowing is zero. It's like saying a velocity of zero has no meaning.

On Mon, 30 May 2016 10:37:01 -0400, FromTheRafters
wrote:

on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.
No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero as
well.

That's a good theory, but IMO it is wrong.

Yeah that Ohm guy was a moron. I suppose we should throw out the rest
of electrical engineering too as just a flawed theory.

After serious thinking trader_4 wrote :
On Monday, May 30, 2016 at 11:46:09 AM UTC-4, FromTheRafters wrote:
trader_4 brought next idea :
On Monday, May 30, 2016 at 10:37:10 AM UTC-4, FromTheRafters wrote:
on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.
No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

That's a good theory, but IMO it is wrong.

No theory, it's correct from electricity 101. With an open circuit, no
current flow, the voltage across the wires is the same at both ends.

That I can agree with. There is no "voltage drop" because there is no
circuit in which current can flow. Voltage drop is defined in terms of
a circuit with current flowing from the source through a load and back
to the source and voltage drop is across all of the resistances in the
circuit including the internal resistance of the source.

Think of it this way:

Distance equals rate times time (D=RT) and you have two trains on a
railroad track. Detroit to Chicago on one end and Chicago to Detroit on
the other. Neither train is moving. Does the distance between them drop
to zero?

No, it doesn't.

Voltage drop has no meaning in an 'open circuit', which isn't actually
a circuit at all, just as 'rate' has no meaning for objects which
aren't moving.


From Wikipedia:

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces through which electric current
can flow."

If current can't flow, it stands to reason that it is not a circuit,
and 'voltage drop' has no meaning.

It does have meaning, it's ZERO. Zero in physics has meaning. It can
be measured. The voltage on the ends of that 100 ft of wire is exactly
the same.,

No argument on that point.

because the voltage drop across the wire with no current
flowing is zero.

There is no such thing as a 'voltage drop' when there is no current
flowing.

https://www.youtube.com/watch?v=ggKnH-95ty0

It's like saying a velocity of zero has no meaning.

D=RT, so when the rate (velocity) is zero then D=zero*T and since
zero*T equals zero then D equals zero, so you are saying that the
distance is zero. This is obviously not the case.

Consider this:

What is the amperage (A) when you put zero ohms (R) resistance across
an ideal 12 volt (V) source?

Answer:

Undefined! Because A=V/R and you can't divide by zero.

This is simple math, why are you struggling with it?

--
55484A76596D466962486B67626D3867633256756332556762 3259676148567462334967
5A576C30614756794943413D

It happens that formulated :
On Mon, 30 May 2016 10:37:01 -0400, FromTheRafters
wrote:

on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.
No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero as
well.

That's a good theory, but IMO it is wrong.

Yeah that Ohm guy was a moron.

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

I suppose we should throw out the rest
of electrical engineering too as just a flawed theory.

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.

On Monday, May 30, 2016 at 2:18:37 PM UTC-4, FromTheRafters wrote:


Think of it this way:

Distance equals rate times time (D=RT) and you have two trains on a
railroad track. Detroit to Chicago on one end and Chicago to Detroit on
the other. Neither train is moving. Does the distance between them drop
to zero?

No, it doesn't.


The distance in your formula is not the distance between the cities,
which of course is constant and never changes. The distance in
your formula is the distance THE TRAIN TRAVELS.




Voltage drop has no meaning in an 'open circuit', which isn't actually
a circuit at all, just as 'rate' has no meaning for objects which
aren't moving.


From Wikipedia:

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces through which electric current
can flow."

If current can't flow, it stands to reason that it is not a circuit,
and 'voltage drop' has no meaning.

It does have meaning, it's ZERO. Zero in physics has meaning. It can
be measured. The voltage on the ends of that 100 ft of wire is exactly
the same.,

No argument on that point.

because the voltage drop across the wire with no current
flowing is zero.

There is no such thing as a 'voltage drop' when there is no current
flowing.


Which is exactly the same thing as saying the voltage drop is zero.
It's like saying if we have 3 apples and take 3 away, how many are
there now? The answer of course is ZERO, but what you're arguing
is that ZERO has no meaning. It does with apples and it does with
regard to Ohm's Law and a voltage drop of zero with no current.



https://www.youtube.com/watch?v=ggKnH-95ty0

It's like saying a velocity of zero has no meaning.

D=RT, so when the rate (velocity) is zero then D=zero*T and since
zero*T equals zero then D equals zero, so you are saying that the
distance is zero. This is obviously not the case.


Say what now? Distance Traveled= Rate x Time. If the rate is zero,
then the distance traveled is zero. Again, clearly zero has meaning.


Consider this:

What is the amperage (A) when you put zero ohms (R) resistance across
an ideal 12 volt (V) source?

Answer:

Undefined! Because A=V/R and you can't divide by zero.


Poor attempt at diversion.
But we're not dividing by zero in the case of Ohm's Law, nor in
your example of motion above. And from a math, physics and engineering perspective what happens in your new example is that as the resistance approaches zero, the current approaches infinity. We deal with infinity
and things approaching limits in engineering and math. It's not a
mystery. But none of this has anything to do with what we are doing
with Ohm's Law, because the voltage drop does not involve dividing by
zero.


This is simple math, why are you struggling with it?


Look, now you're going to start taking cheap shots? Tthe only one
here who was obviously struggling is Diesel. He
doesn't understand Ohm's Law. But now we can add you to the list because
you can't understand that in the simple case of distance = rate * time,
a zero rate gives an answer of ZERO distance traveled and that answer
of zero definitely has meaning, just like zero voltage drop from Ohm's Law
has meaning.

Gfre, WTF has happened to education in America? We now have two idiots here
trying to explain basic math and electricity to us and neither knows
WTF they are talking about.

On 05/30/2016 06:13 AM, Al Gebra wrote:

[snip]

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

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.

trader_4 pretended :
On Monday, May 30, 2016 at 2:18:37 PM UTC-4, FromTheRafters wrote:


Think of it this way:

Distance equals rate times time (D=RT) and you have two trains on a
railroad track. Detroit to Chicago on one end and Chicago to Detroit on
the other. Neither train is moving. Does the distance between them drop
to zero?

No, it doesn't.


The distance in your formula is not the distance between the cities,
which of course is constant and never changes. The distance in
your formula is the distance THE TRAIN TRAVELS.




Voltage drop has no meaning in an 'open circuit', which isn't actually
a circuit at all, just as 'rate' has no meaning for objects which
aren't moving.


From Wikipedia:

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces through which electric current
can flow."

If current can't flow, it stands to reason that it is not a circuit,
and 'voltage drop' has no meaning.

It does have meaning, it's ZERO. Zero in physics has meaning. It can
be measured. The voltage on the ends of that 100 ft of wire is exactly
the same.,

No argument on that point.

because the voltage drop across the wire with no current
flowing is zero.

There is no such thing as a 'voltage drop' when there is no current
flowing.


Which is exactly the same thing as saying the voltage drop is zero.
It's like saying if we have 3 apples and take 3 away, how many are
there now? The answer of course is ZERO, but what you're arguing
is that ZERO has no meaning. It does with apples and it does with
regard to Ohm's Law and a voltage drop of zero with no current.



https://www.youtube.com/watch?v=ggKnH-95ty0

It's like saying a velocity of zero has no meaning.

D=RT, so when the rate (velocity) is zero then D=zero*T and since
zero*T equals zero then D equals zero, so you are saying that the
distance is zero. This is obviously not the case.


Say what now? Distance Traveled= Rate x Time. If the rate is zero,
then the distance traveled is zero. Again, clearly zero has meaning.

Not if you use the D=RT formula, it is 'undefined' when either R or T
is zero (I did *not* say approaching zero).

https://en.wikipedia.org/wiki/Division_by_zero

Consider this:

What is the amperage (A) when you put zero ohms (R) resistance across
an ideal 12 volt (V) source?

Answer:

Undefined! Because A=V/R and you can't divide by zero.


Poor attempt at diversion.
But we're not dividing by zero in the case of Ohm's Law, nor in
your example of motion above. And from a math, physics and engineering
perspective what happens in your new example is that as the resistance
approaches zero, the current approaches infinity. We deal with infinity and
things approaching limits in engineering and math. It's not a mystery. But
none of this has anything to do with what we are doing with Ohm's Law,
because the voltage drop does not involve dividing by zero.

That is true, unless you state that I or R is zero. See above and
below.

This is simple math, why are you struggling with it?


Look, now you're going to start taking cheap shots? Tthe only one
here who was obviously struggling is Diesel. He
doesn't understand Ohm's Law. But now we can add you to the list because
you can't understand that in the simple case of distance = rate * time,
a zero rate gives an answer of ZERO distance traveled and that answer
of zero definitely has meaning, just like zero voltage drop from Ohm's Law
has meaning.

Simple math, you are *not* allowed to divide by zero - it is
*undefined*. Infinity is okay to work with because things in that
formula can be infinitely small or infinitely large.

Gfre, WTF has happened to education in America? We now have two idiots here
trying to explain basic math and electricity to us and neither knows
WTF they are talking about.

E=IR
I=E/R
R=E/I

If you state that either I or R is zero (as is the case with open
circuit, which is not really a circuit, or zero resistance attached to
an ideal source) the formula doesn't work. There was mention of "zero"
current and in the 'superconductor' comment was about zero resistance.
My analogy with the trains was a bit of fun, but still illustrates that
dividing by zero yields the result of 'undefined' *not* 'zero'.

I, of course, agree about 'approaching' these limits as being worthy of
consideration. However, that was *not* what I was replying to.

Also, there is no such thing as 'voltage drop' when there is no current
because 'voltage drop' is defined by there being current. You can have
voltage drop across a fuse just before it blows, but the voltage there
after it blows is not 'voltage drop' because there is no current.

On Monday, May 30, 2016 at 4:19:08 PM UTC-4, Sam E wrote:
On 05/30/2016 06:13 AM, Al Gebra wrote:

[snip]

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

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.

(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.

Sam E expressed precisely :
On 05/30/2016 06:13 AM, Al Gebra wrote:

[snip]

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

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.

Power is E*I and since E=I*R power also equals I*I*R or I squared R.

You are correct that there is *no* 'voltage drop' with no current,
which is not the same as saying that the 'voltage drop' is zero.

trader_4 explained on 5/30/2016 :
On Monday, May 30, 2016 at 4:19:08 PM UTC-4, Sam E wrote:
On 05/30/2016 06:13 AM, Al Gebra wrote:

[snip]

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

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.

(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.

Power drop?

If there is such a thing in this context, which I doubt, it probably
requires current too.

On 05/30/2016 01:18 PM, FromTheRafters wrote:

[snip]

Undefined! Because A=V/R and you can't divide by zero.

This is simple math, why are you struggling with it?

With any 2 nonzero numbers, A/B is the reciprocal of B/A. In other
words, (A/B)*(B/A)=1. If you make A=0, then A/B=0. B/A should then give
the reciprocal of 0 (as in, what do you multiply 0 by to get 1). How
about infinity?

Still doesn't explain 0/0.

If you don't like this, please don't read it :-)

--
Mark Lloyd
http://notstupid.us/

"When the gods wish to punish us, they answer our prayers." [Oscar
Wilde]

On Mon, 30 May 2016 06:53:52 -0000 (UTC), Diesel
wrote:


Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.

No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

You are out of your area of expertise. With no current flow there is
NO voltage loss. Simple Ohm's Law. Nobody has been found crooked
enough to break that law.

Now, TECHNICALLY you are almost right. If you have a very low
impedence volt meter it will draw milliamps. Say it draws 5 milliamps
(0.005 amps) and the resistance of your cable is 100 ft of #10 copper
which means 200 feet of conductor. At 0..999 ohms per 1000 ft trhat
is 0.1998 ohms. Plug that into Ohm's law -E=IxR - and you get a
voltage drop of (.005 x .1998 )= a whopping 0.000999 volt drop across
the wire. Lets jump out on a limb and say you have a really rotten
voltmeter that draws 50 miliams (.050 amp) and do the math -
..050X.1998)=0.00999 volts.

That voltage drop will , in the real world, get "lost in the noise" -
well within the accuracy variance of all but the most expensive lab
type volt meters.

Watts is watts, man. Available voltage determines how many amps
it's going to take to get them, though. Lower voltage=more amps to
do the same job.

If you have a lower voltage, the watts will be lower. In our
example, that 1500 watt heater (at 240v) will be 1407 watts at
232v. Current will actually be less, not more.

Not exactly. You can have high voltage and next to no current behind
it (think of a taser) or low voltage and a considerable amount of
current behind it (think of your car battery) it's no chump and your
starter motor isn't exactly a low drain device...You can also lose a
few volts in our example, and, still be able to pull enough current
to make up the difference.

Yes, exactly. You are forgetting the "R" in E=IxR.
The starter draws a lot of current because it has a very low
resistance. A 12 volt starter drawing 120 amps has an effectice
resistance of 12/120= 0.1 ohms.and consumes 12X120= 1440 watts -
roughly 1 1/2 hp.

The taser is abour 26 watts. The open circuit voltage is about 50,000
vots. That means the current is (26/50,000)=0.00052 amps.. This means
the impedence or resistance of the tazer is something in the
neighbourhood of (50,000/.00052) = 96,153,846 ohms.

The easiest way I know of to explain the relationship is this:

It may be useful to consider the image of water in a hose. Voltage is
equivalent to pressure, water flow is equivalent to current and the
diameter of the pipe is equivalent ot the thickness of the wire - or
resistance.
The water theory of electrical theory is a very simplistic
explanation that "doesn't exactly hold water"

On 05/30/2016 01:25 PM, FromTheRafters wrote:

[snip]

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.

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.

On Mon, 30 May 2016 10:37:01 -0400, FromTheRafters
wrote:

on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.
No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero as
well.

That's a good theory, but IMO it is wrong.
You are not smart enough to brak ohm's law - and the water theory of
electricity doesn't exactly hold water.

On Mon, 30 May 2016 11:45:57 -0400, FromTheRafters
wrote:

trader_4 brought next idea :
On Monday, May 30, 2016 at 10:37:10 AM UTC-4, FromTheRafters wrote:
on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.
No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero as
well.

That's a good theory, but IMO it is wrong.

No theory, it's correct from electricity 101. With an open circuit, no
current flow, the voltage across the wires is the same at both ends.

That I can agree with. There is no "voltage drop" because there is no
circuit in which current can flow. Voltage drop is defined in terms of
a circuit with current flowing from the source through a load and back
to the source and voltage drop is across all of the resistances in the
circuit including the internal resistance of the source.

Think of it this way:

Distance equals rate times time (D=RT) and you have two trains on a
railroad track. Detroit to Chicago on one end and Chicago to Detroit on
the other. Neither train is moving. Does the distance between them drop
to zero?

No, it doesn't.

Voltage drop has no meaning in an 'open circuit', which isn't actually
a circuit at all, just as 'rate' has no meaning for objects which
aren't moving.

From Wikipedia:

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces through which electric current
can flow."

If current can't flow, it stands to reason that it is not a circuit,
and 'voltage drop' has no meaning.
You are burying yourself deeper.
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.

On Mon, 30 May 2016 14:18:28 -0400, FromTheRafters
wrote:

After serious thinking trader_4 wrote :
On Monday, May 30, 2016 at 11:46:09 AM UTC-4, FromTheRafters wrote:
trader_4 brought next idea :
On Monday, May 30, 2016 at 10:37:10 AM UTC-4, FromTheRafters wrote:
on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.
No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

That's a good theory, but IMO it is wrong.

No theory, it's correct from electricity 101. With an open circuit, no
current flow, the voltage across the wires is the same at both ends.

That I can agree with. There is no "voltage drop" because there is no
circuit in which current can flow. Voltage drop is defined in terms of
a circuit with current flowing from the source through a load and back
to the source and voltage drop is across all of the resistances in the
circuit including the internal resistance of the source.

Think of it this way:

Distance equals rate times time (D=RT) and you have two trains on a
railroad track. Detroit to Chicago on one end and Chicago to Detroit on
the other. Neither train is moving. Does the distance between them drop
to zero?

No, it doesn't.

Voltage drop has no meaning in an 'open circuit', which isn't actually
a circuit at all, just as 'rate' has no meaning for objects which
aren't moving.


From Wikipedia:

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces through which electric current
can flow."

If current can't flow, it stands to reason that it is not a circuit,
and 'voltage drop' has no meaning.

It does have meaning, it's ZERO. Zero in physics has meaning. It can
be measured. The voltage on the ends of that 100 ft of wire is exactly
the same.,

No argument on that point.

because the voltage drop across the wire with no current
flowing is zero.

There is no such thing as a 'voltage drop' when there is no current
flowing.

https://www.youtube.com/watch?v=ggKnH-95ty0

It's like saying a velocity of zero has no meaning.

D=RT, so when the rate (velocity) is zero then D=zero*T and since
zero*T equals zero then D equals zero, so you are saying that the
distance is zero. This is obviously not the case.

Consider this:

What is the amperage (A) when you put zero ohms (R) resistance across
an ideal 12 volt (V) source?

Answer:

Undefined! Because A=V/R and you can't divide by zero.

This is simple math, why are you struggling with it?
You are over your head -- WAY over.

On Mon, 30 May 2016 14:25:32 -0400, FromTheRafters
wrote:

It happens that formulated :
On Mon, 30 May 2016 10:37:01 -0400, FromTheRafters
wrote:

on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.
No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero as
well.

That's a good theory, but IMO it is wrong.

Yeah that Ohm guy was a moron.

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

I suppose we should throw out the rest
of electrical engineering too as just a flawed theory.

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.
I've just posted it for you.

Mark Lloyd laid this down on his screen :
On 05/30/2016 01:18 PM, FromTheRafters wrote:

[snip]

Undefined! Because A=V/R and you can't divide by zero.

This is simple math, why are you struggling with it?

With any 2 nonzero numbers, A/B is the reciprocal of B/A. In other words,
(A/B)*(B/A)=1. If you make A=0, then A/B=0. B/A should then give the
reciprocal of 0 (as in, what do you multiply 0 by to get 1). How about
infinity?

Still doesn't explain 0/0.

If you don't like this, please don't read it :-)

Maybe you could multiply it by itself zero times. Anything to the
zeroeth power equals one.

Okay, so multiplicative groups kick out the zero, it was worth a try.

On Mon, 30 May 2016 16:50:22 -0400, FromTheRafters
wrote:

trader_4 explained on 5/30/2016 :
On Monday, May 30, 2016 at 4:19:08 PM UTC-4, Sam E wrote:
On 05/30/2016 06:13 AM, Al Gebra wrote:

[snip]

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

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.

(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.

Power drop?

If there is such a thing in this context, which I doubt, it probably
requires current too.

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.

brought next idea :
On Mon, 30 May 2016 11:45:57 -0400, FromTheRafters
wrote:

trader_4 brought next idea :
On Monday, May 30, 2016 at 10:37:10 AM UTC-4, FromTheRafters wrote:
on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.
No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

That's a good theory, but IMO it is wrong.

No theory, it's correct from electricity 101. With an open circuit, no
current flow, the voltage across the wires is the same at both ends.

That I can agree with. There is no "voltage drop" because there is no
circuit in which current can flow. Voltage drop is defined in terms of
a circuit with current flowing from the source through a load and back
to the source and voltage drop is across all of the resistances in the
circuit including the internal resistance of the source.

Think of it this way:

Distance equals rate times time (D=RT) and you have two trains on a
railroad track. Detroit to Chicago on one end and Chicago to Detroit on
the other. Neither train is moving. Does the distance between them drop
to zero?

No, it doesn't.

Voltage drop has no meaning in an 'open circuit', which isn't actually
a circuit at all, just as 'rate' has no meaning for objects which
aren't moving.

From Wikipedia:

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces through which electric current
can flow."

If current can't flow, it stands to reason that it is not a circuit,
and 'voltage drop' has no meaning.
You are burying yourself deeper.
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.

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.

pretended :
On Mon, 30 May 2016 14:18:28 -0400, FromTheRafters
wrote:

After serious thinking trader_4 wrote :
On Monday, May 30, 2016 at 11:46:09 AM UTC-4, FromTheRafters wrote:
trader_4 brought next idea :
On Monday, May 30, 2016 at 10:37:10 AM UTC-4, FromTheRafters wrote:
on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.
No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

That's a good theory, but IMO it is wrong.

No theory, it's correct from electricity 101. With an open circuit, no
current flow, the voltage across the wires is the same at both ends.

That I can agree with. There is no "voltage drop" because there is no
circuit in which current can flow. Voltage drop is defined in terms of
a circuit with current flowing from the source through a load and back
to the source and voltage drop is across all of the resistances in the
circuit including the internal resistance of the source.

Think of it this way:

Distance equals rate times time (D=RT) and you have two trains on a
railroad track. Detroit to Chicago on one end and Chicago to Detroit on
the other. Neither train is moving. Does the distance between them drop
to zero?

No, it doesn't.

Voltage drop has no meaning in an 'open circuit', which isn't actually
a circuit at all, just as 'rate' has no meaning for objects which
aren't moving.


From Wikipedia:

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces through which electric current
can flow."

If current can't flow, it stands to reason that it is not a circuit,
and 'voltage drop' has no meaning.

It does have meaning, it's ZERO. Zero in physics has meaning. It can
be measured. The voltage on the ends of that 100 ft of wire is exactly
the same.,

No argument on that point.

because the voltage drop across the wire with no current
flowing is zero.

There is no such thing as a 'voltage drop' when there is no current
flowing.

https://www.youtube.com/watch?v=ggKnH-95ty0

It's like saying a velocity of zero has no meaning.

D=RT, so when the rate (velocity) is zero then D=zero*T and since
zero*T equals zero then D equals zero, so you are saying that the
distance is zero. This is obviously not the case.

Consider this:

What is the amperage (A) when you put zero ohms (R) resistance across
an ideal 12 volt (V) source?

Answer:

Undefined! Because A=V/R and you can't divide by zero.

This is simple math, why are you struggling with it?
You are over your head -- WAY over.

Says you. Do you divide by zero for a living?

hah has brought this to us :
On 05/30/2016 01:25 PM, FromTheRafters wrote:

[snip]

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.

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.

Finally, someone who knows what they're talking about. LOL

On Thu, 26 May 2016 06:50:22 +0100, wrote:

On Wed, 25 May 2016 21:59:39 -0700 (PDT), wrote:

Hi,
I'm wiring four new electric baseboards in my basement.
All four are 220Volt units: 500W, 500W, 1000W, 1500W, for a total of 3500W. From the main panel to the farthest unit is less than 100ft.
If I understood the rules, putting all of this on one circuit will require a 20Amp breaker (1.25*3500W/220V), and the wire can be #12AWG.
Did I get that right, and am I missing any other design factor?

Many thanks, especially if you can reference the NEC so I know I'm doing it right.
Theodore

OK first things first, heaters are rated at 240v not 220 so your total
load is 14.6a.

OK lets look at the code (Fixed electric space heating art 424)

Why? Why not just do the sums and work out what's actually needed instead of believing some paper pushing ****wit?

--
Bozone (n.): The substance surrounding stupid people that stops bright ideas from penetrating. The bozone layer, unfortunately, shows little sign of breaking down in the near future.

has brought this to us :
On Mon, 30 May 2016 14:25:32 -0400, FromTheRafters
wrote:

It happens that formulated :
On Mon, 30 May 2016 10:37:01 -0400, FromTheRafters
wrote:

on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.
No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

That's a good theory, but IMO it is wrong.

Yeah that Ohm guy was a moron.

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

I suppose we should throw out the rest
of electrical engineering too as just a flawed theory.

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.
I've just posted it for you.

Where?

On 05/30/2016 03:46 PM, FromTheRafters wrote:
Sam E expressed precisely :
On 05/30/2016 06:13 AM, Al Gebra wrote:

[snip]

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

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.

Power is E*I and since E=I*R power also equals I*I*R or I squared R.

Sorry for the error. I never claimed to be perfect. Although, it'd be
hard to make a mistake like that one (voltage drop with no current).

BTW, as to formulae I always liked P=IE.

You are correct that there is *no* 'voltage drop' with no current, which
is not the same as saying that the 'voltage drop' is zero.

While 'no' and 'zero' are different ideas, either should be correct here.

Voltage drop = I*R = 0v*R = 0v.

wrote on 5/30/2016 :
On Mon, 30 May 2016 06:53:52 -0000 (UTC), Diesel
wrote:


Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.

No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

You are out of your area of expertise. With no current flow there is
NO voltage loss.

It almost looks like you are agreeing with me now, except you said
'voltage loss' instead of 'voltage drop' which are *not* the same
thing.

Also, in another post, you started writing about semiconductors and I
am familiar with forward voltage drop, and it requires a current.

Excerpted from Wikipedia:

"In a small silicon diode operating at its rated currents, the voltage
drop is about 0.6 to 0.7 volts."

Notice the word "currents" in there? There is no 'voltage drop' when no
current is present.

I'm still waiting for "voltage drop" with no reference to current. You
see, devices don't dissipate power when there is no current through
them, so how can there be any 'voltage drop' with no current?

On 05/30/2016 03:41 PM, trader_4 wrote:
On Monday, May 30, 2016 at 4:19:08 PM UTC-4, Sam E wrote:
On 05/30/2016 06:13 AM, Al Gebra wrote:

[snip]

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

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.

(I^2)R is the power drop. The voltage drop is IR

I discovered that mistake soon after posting. It is IR.

And no, it's not about a simple mistake, Diesel doesn't understand
Ohm's Law and electricity 101.


Most likely, considering the multiple errors (including using the
wattage of a heater that wasn't connected).

On 05/30/2016 03:50 PM, FromTheRafters wrote:

[snip]

Power drop?

If there is such a thing in this context, which I doubt, it probably
requires current too.

There's power dissipated by the wire (mostly as heat). And that does
require current.

--
Mark Lloyd
http://notstupid.us/

"When the gods wish to punish us, they answer our prayers." [Oscar
Wilde]

On Mon, 30 May 2016 05:57:44 -0000 (UTC), Diesel
wrote:

Sam E
Mon, 30 May 2016 03:14:39 GMT
in alt.home.repair, wrote:

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

[snip]

1500/232.1=6.5 amps *rounded up*
1500/235=6.39 amps *rounded up*

And that's due entirely to distance with no loads present on the
line yet.

I'm often surprised by how little some people know about
electricity.

I sometimes feel the same way.

Voltage drop depends on current flow. There's no
power (wattage) without current flow either.

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.


So you have a load which is not present. How is it drawing power?

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.
You are WAY in over your head.

On Mon, 30 May 2016 05:57:44 -0000 (UTC), Diesel
wrote:

notX Mon,
30 May 2016 03:19:44 GMT in alt.home.repair, wrote:

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

[snip]

Just to clarify, the calculations I provided in this post aren't
FUD, unless you're able to dismiss the yellow Ugly electrician
reference book.

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

http://www.uglys.net/

You have voltage drop due to the length and size of the wire. Because
the wire isn't a super conductor.

At the end of the day, you might have put 120 volts on the line, but,
100ft down that line, you aren't getting 120 volts. Some has been lost
on the wire, due to the wires own resistance.

Switch from AC to DC with no other changes, and the voltage drop is
more pronounced.
Absolutely totally 110% wrong.. There is NO voltage drop withouit
current flow.

You are WAY in over your head.

On 05/30/2016 04:28 PM, FromTheRafters wrote:

[snip]

Says you. Do you divide by zero for a living?

I seem to remember n/0 = INF (for nonzero values of n). 0/0 = NaN,

I'm not so sure about that last one, I've seen 0, 1, and infinity. All
at the same time. It's a fuzzy number. Very fuzzy.

On Monday, May 30, 2016 at 4:40:19 PM UTC-4, FromTheRafters wrote:

Say what now? Distance Traveled= Rate x Time. If the rate is zero,
then the distance traveled is zero. Again, clearly zero has meaning.

Not if you use the D=RT formula, it is 'undefined' when either R or T
is zero (I did *not* say approaching zero).

https://en.wikipedia.org/wiki/Division_by_zero


Idiot.

There is no division in Distance Traveled = rate x time.
If either the rate or the time is zero, the distance is zero.
And obviously it has meaning, it means the train did not move.


Consider this:

What is the amperage (A) when you put zero ohms (R) resistance across
an ideal 12 volt (V) source?

Answer:

Undefined! Because A=V/R and you can't divide by zero.


Poor attempt at diversion.
But we're not dividing by zero in the case of Ohm's Law, nor in
your example of motion above. And from a math, physics and engineering
perspective what happens in your new example is that as the resistance
approaches zero, the current approaches infinity. We deal with infinity and
things approaching limits in engineering and math. It's not a mystery. But
none of this has anything to do with what we are doing with Ohm's Law,
because the voltage drop does not involve dividing by zero.

That is true, unless you state that I or R is zero. See above and
below.


See this:

V = IR. There is no division by zero. If I or R is zero, V is zero.
And that zero has meaning.



This is simple math, why are you struggling with it?


Look, now you're going to start taking cheap shots? Tthe only one
here who was obviously struggling is Diesel. He
doesn't understand Ohm's Law. But now we can add you to the list because
you can't understand that in the simple case of distance = rate * time,
a zero rate gives an answer of ZERO distance traveled and that answer
of zero definitely has meaning, just like zero voltage drop from Ohm's Law
has meaning.

Simple math, you are *not* allowed to divide by zero - it is
*undefined*. Infinity is okay to work with because things in that
formula can be infinitely small or infinitely large.


We're not dividing by zero. YOU just keep pretending we are.


Gfre, WTF has happened to education in America? We now have two idiots here
trying to explain basic math and electricity to us and neither knows
WTF they are talking about.

E=IR

If either current or resistance is zero, Voltage is zero and contrary to your
BS, it has meaning.


I=E/R

Is there zero resistance in that heater circuit wire?


R=E/I


Anyone here doing that division, with a current of zero, to try to calculate
resistance? No.



If you state that either I or R is zero (as is the case with open
circuit, which is not really a circuit, or zero resistance attached to
an ideal source) the formula doesn't work.

V = IR If either I or R is zero, V is zero. Did you even take algebra?


There was mention of "zero"
current

That right, because with no load, you have zero current. Put zero in for
I above, put a finite value for R and you have zero voltage.


and in the 'superconductor' comment was about zero resistance.

That was brought up by the guy who doesn't even understand Ohm's Law.
No point in going there, it has nothing to do with the current discussion.

My analogy with the trains was a bit of fun, but still illustrates that
dividing by zero yields the result of 'undefined' *not* 'zero'.


I does nothing of the sort. All it demonstrates is that:

Distance Traveled = Rate x Time.

Rate of zero, Distance Traveled is zero. It has meaning the train
didn't move. Idiot.


I, of course, agree about 'approaching' these limits as being worthy of
consideration. However, that was *not* what I was replying to.

Also, there is no such thing as 'voltage drop' when there is no current
because 'voltage drop' is defined by there being current. You can have
voltage drop across a fuse just before it blows, but the voltage there
after it blows is not 'voltage drop' because there is no current.

Idiot. If a fuse blows, the voltage across it after it blows is the full
open circuit voltage. Try using a meter and see.

On Monday, May 30, 2016 at 2:25:40 PM UTC-4, FromTheRafters wrote:
It happens that formulated :
On Mon, 30 May 2016 10:37:01 -0400, FromTheRafters
wrote:

on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.
No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero as
well.

That's a good theory, but IMO it is wrong.

Yeah that Ohm guy was a moron.

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

I suppose we should throw out the rest
of electrical engineering too as just a flawed theory.

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.

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.

On Monday, May 30, 2016 at 5:22:14 PM UTC-4, wrote:
On Mon, 30 May 2016 16:50:22 -0400, FromTheRafters
wrote:

trader_4 explained on 5/30/2016 :
On Monday, May 30, 2016 at 4:19:08 PM UTC-4, Sam E wrote:
On 05/30/2016 06:13 AM, Al Gebra wrote:

[snip]

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

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.

(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.

Power drop?

If there is such a thing in this context, which I doubt, it probably
requires current too.

Lost power in the circuit is a function of the voltage drop.


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.