On Monday, May 30, 2016 at 6:01:38 PM UTC-4, FromTheRafters wrote:
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.


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.

On Monday, May 30, 2016 at 6:11:29 PM UTC-4, wrote:
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.

We should also point out that it's more BS that the voltage drop would
be more pronounced if we switched from AC to DC. It would be exactly
the same, excluding negligible second order effects like skin effect,
inductance, etc. And those produce the reverse effect, ie more resistance
to AC current than DC.

On Monday, May 30, 2016 at 5:59:52 PM UTC-4, Sam E wrote:
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.

Bingo!
Sam, maybe you can explain it to math challenged Rafters. He says
that you can't do what you just did with that simplest of equations
because with a current of zero, you divided by zero. I don't see
any division there, neither do you. Go figure.

trader_4 brought next idea :
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.

D=RT is a relationship and can be written as T=D/R or R=D/T and it is
still the same relationship.

http://mathforum.org/dr.math/faq/faq.distance.html

See this:

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

Okay, so if I is zero, what is R? Can you show that the relationship
still holds?

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

Well duh! The thing is that it is *not* "voltage drop" because an open
fuse does not dissipate energy.

"Ohm's law states that the *current through a conductor* . . ."

Where's the current through an open fuse, brainiac?

"Voltage drop describes how the supplied energy of a voltage source is
reduced *as electric current moves through the passive elements* . . ."

Show me how Ohm's law holds when the current is zero, brainiac.

notX has brought this to us :
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.

You might be correct in that particular context.

https://en.wikipedia.org/wiki/Divisi...ter_arithmetic

If you are happy with the 'voltage drop' being somewhere between zero
and a very large integer (or infinity) then that's a plus for you.

Still, it can't really be a 'voltage drop' without energy being
dissipated by the device under consideration.

After serious thinking trader_4 wrote :
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.

Of course there isn't, because there is no current. You can't have a
voltage drop when there is no current. Thanks for finally agreeing with
me.

--
Getting there was like pulling teeth though

On 05/30/2016 06:11 PM, trader_4 wrote:

[snip]

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.

Did you consider that if the heater is 100 feet from the power source,
there's 200 feet of wire?

trader_4 was thinking very hard :
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.

Sorry to have to tell you this, but there can be no power without any
current either.

Mark Lloyd formulated on Monday :
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.

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.

Sam E brought next idea :
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.

I have zero problems with dividing by no.

But with 'voltage drop' you can't have zero current, because 'voltage
drop' is all about the energy delivered to and dissipated by the
device, not the capability of the source to deliver voltage.

A stepper contactor relay's burnt contacts do not have a voltage drop
unless there is current flowing through them no matter how much voltage
the source can deliver to those closed contacts. There is no voltage
drop between the poles of a car battery, unless there is current
through it and some internal resistance to dissipate some of the
energy. Voltage drop is not a static thing like voltage is.

trader_4 presented the following explanation :
On Monday, May 30, 2016 at 6:01:38 PM UTC-4, FromTheRafters wrote:
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.


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.

They're the same? Tell these guys then, 'cause they have it all wrong.

http://ecmweb.com/electrical-testing...s-voltage-drop

https://appauto.wordpress.com/2008/0...-the-question/

pretended :
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.

That's what *I* said, and you said I was wrong. Sheeesh, make up your
mind.

Al Gebra
Mon, 30 May 2016
11:13:02 GMT in alt.home.repair, wrote:

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.

*sigh* you are correct. I wasted a significant amount of time by
ignoring the fact that the circuit wasn't completed yet. I will eat a
large amount of crow.


--
MID:
Hmmm. I most certainly don't understand how I can access a copy of a
zip file but then not be able to unzip it so I can watch it. That
seems VERY clever!
http://al.howardknight.net/msgid.cgi?ID=145716711400

Mon, 30 May 2016 15:07:39 GMT in alt.home.repair, wrote:

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.


Bull****. If there is no current flowing, there is no voltage
drop.

Volts (dropped) = Amps x Resistance. If there are no Amps, it
doesn't matter what the resistance is, volts (dropped) is still
zero.

You might argue that when you actually measure the volts at the
far end, you are loading the circuit with your meter but a digital
meter has in impedance in the meg ohms so the current is still
virtually zero and you will still see the full voltage within the
accuracy of the meter.

You are correct. I wasted a significant amount of time by ignoring
the fact that the circuit wasn't completed yet. I will eat a large
amount of crow.

As a sanity check, notice voltage drop charts always base the
number on the load in amps times the resistance of the wire.

Yes, I've noticed.



--
MID:
Hmmm. I most certainly don't understand how I can access a copy of a
zip file but then not be able to unzip it so I can watch it. That
seems VERY clever!
http://al.howardknight.net/msgid.cgi?ID=145716711400

trader_4
Mon, 30
May 2016 15:21:54 GMT in alt.home.repair, wrote:

The thinner and longer the length, the
more is lost in transit.

That's true, but with no load, there is no current, no pushing, no
transit. Capiche?

Aye.

You are correct. I wasted a significant amount of time by ignoring the
fact that the circuit wasn't completed yet. I will eat a large amount
of crow.



--
MID:
Hmmm. I most certainly don't understand how I can access a copy of a
zip file but then not be able to unzip it so I can watch it. That
seems VERY clever!
http://al.howardknight.net/msgid.cgi?ID=145716711400

trader_4
Mon, 30
May 2016 15:09:06 GMT in alt.home.repair, wrote:

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?

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=16.4 (16.438)

watts=volts(x2)/ohms
24x24/16.4=35.1 (35.121)

amps=watts/volts
35.1/24=1.5 (1.4625)


--
MID:
Hmmm. I most certainly don't understand how I can access a copy of a
zip file but then not be able to unzip it so I can watch it. That
seems VERY clever!
http://al.howardknight.net/msgid.cgi?ID=145716711400

On Mon, 30 May 2016 16:40:12 -0400, FromTheRafters
wrote:


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)

An open circuit does not have 0 ohms. and the only time you device by
current is to determine resistance knowing only voltage and amperage -
which only works if you have current flow - which means I does not
equal 0. - and an "ideal source" feeding 0 ohms gives you an answer of
infinite current.

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'.

Undefined or infinite - the current drawn from an "ideal supply" into
0 ohms would be undefined or infinite for a split second untill the
resistance would change due to the heating effect of the current and a
split second later the resistance would become infinite and the
current zero as the "fuse" opened.

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.

Technically trhe voltage drop across an open fuse is considered to be
supply voltage. If you have a series circuit with zero current on all
the defined resistances and therefore no voltage drop across any of
them the total supply voltage is dropped across the "infinite" or
"undefined" resistance element (talking DC) In an AC circuit there
will be a capacitance between the 2 terminals that can be measured,
and the capacitive reactance will cause a miniscule but measurable
current flow

On Mon, 30 May 2016 16:46:39 -0400, 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.

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.


Correct - the voltage drop across the open circuit is the supply
voltage.. There is no voltage drop across the conductors, but the sum
of all voltage drops in a circuit MUST equal the supply voltage.
Putting a voltmeter across any segment of a circuit will give the
voltage drop across it. In an open circuit you will read zero except
across the source and across the infinite resistance of an "open
switch" - and both of those will be identical on a DC circuit - and
close enough to identical as to be virtually impossible to measure the
difference on an AC circuit below radio frequencies, where the
capacitance of the pen switch: starts to have a small but
measureable effect.

On Mon, 30 May 2016 17:26:52 -0400, FromTheRafters
wrote:

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.
The voltage drop across a perfect semiconductor is not affected by
current - other than requiring the current to be NON zero. - and it is
vityually impossible to read voltage anywhere in a circuit without
completing the circuit - in a DC circuit requiring a resistive load -
an an AC circuit either a resistive, inductive or capacitive load. -
which means you can NOT measure voltage in a totally unloaded or
"open" circuit.

On Mon, 30 May 2016 17:28:23 -0400, FromTheRafters
wrote:

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?
Deviding by zero is not involved because, as stated before, by
measuring the voltage you are introducing a non-zero value to both
resistance and current. - Resistance WAY off from zero - approaching
(but never COMPLETELY reaching infinite (by the very definition of
infinity) meaning current- for all practical reasons being ZERO - but
in reality just being infinitesimally small, Do the calculations
using a current of 0.000000(100,000,000 zeros)01 amps and the
resistance being 0.0000000100,000,000 zeros)01 ohms for an open
circuit and everything works.

Sane as with a dead short=0.99999o(1000,000,000 nines)99 ohms because
we don't have superconductors.
In reality you don't need to go nearly as far ac the impedence of any
meter is sigmificantly lower than that - with sensitivity being in the
megohms per volt range.on digitals and kilohms per volt on analogs.

On Mon, 30 May 2016 19:55:22 -0400, FromTheRafters
wrote:

notX has brought this to us :
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.

You might be correct in that particular context.

https://en.wikipedia.org/wiki/Divisi...ter_arithmetic

If you are happy with the 'voltage drop' being somewhere between zero
and a very large integer (or infinity) then that's a plus for you.

Still, it can't really be a 'voltage drop' without energy being
dissipated by the device under consideration.
The voltage drop across the wire is ZERO. The total voltage impressed
acroos the circuit is the supply voltage. Since the voltage drop
across the conductors is zero, the voltage drop across the "open" is
supply voltage. (the voltage drop MUST equal the impressed voltage)
This is true with zero amps current flow. Because there is zero amps
current you can not solve for the resistance of the conductor because
(supply voltage assumed to be 12) 12/0 is indefineableand the
reistance of the open circuit is infinite (in a perfect world)

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

After serious thinking trader_4 wrote :
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.

Of course there isn't, because there is no current. You can't have a
voltage drop when there is no current. Thanks for finally agreeing with
me.
No voltage drop across the wire does not mean no voltage drop across
the circuit thogh.

You've ( it appears) been arguing both sides of the equation.

To avoid confusion - WHOEVER said there would be a voltage loss in an
unloaded wire is WRONG.
Also whoever said there is no voltage drop in an open circuit is ALSO
WRONG.
And to top it all off, in order to measure the voltage across an open
circuit, you MUST close the circuit - meaning it is no longer an
"open" circuit, AND
In the real world there is no such thing as zero ohms. You can get
REAL close - but "in the wild" it does not exist. - so you are never
REALLY deviding by or multiplying by ZERO when solving ohm's law

On Mon, 30 May 2016 20:40:56 -0400, FromTheRafters
wrote:

Mark Lloyd formulated on Monday :
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.

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.


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.

On Mon, 30 May 2016 21:08:12 -0400, FromTheRafters
wrote:

Sam E brought next idea :
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.

I have zero problems with dividing by no.

But with 'voltage drop' you can't have zero current, because 'voltage
drop' is all about the energy delivered to and dissipated by the
device, not the capability of the source to deliver voltage.

A stepper contactor relay's burnt contacts do not have a voltage drop
unless there is current flowing through them no matter how much voltage
the source can deliver to those closed contacts. There is no voltage
drop between the poles of a car battery, unless there is current
through it and some internal resistance to dissipate some of the
energy. Voltage drop is not a static thing like voltage is.
"voltage drop" testing of a circuit indicates an open circuit by
reading source voltage across the open connection.
A full source voltage drop across a "load" indicates zero current
flow if the resistance is infinite and infinite current if resistance
is zero - with the non- zero and non-infinite values between being
calculatable using ohm's law.

On Mon, 30 May 2016 21:28:19 -0400, FromTheRafters
wrote:

pretended :
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.

That's what *I* said, and you said I was wrong. Sheeesh, make up your
mind.
You appear to have said, I believe "You have voltage drop due to the
length and size of the wire. Because the wire isn't a super
conductor. "

You also appear to have said "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. "

And the context was an "open circuit."

Under a load, you are correct. On an open circuit, you are just plain
wrong.

If you did not say what you appear to have said, accept my apologies.

formulated the question :
On Mon, 30 May 2016 16:40:12 -0400, FromTheRafters
wrote:


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)

An open circuit does not have 0 ohms.

True, but it has zero amps. With zero amps the formula is defeated.

and the only time you device by
current is to determine resistance knowing only voltage and amperage -
which only works if you have current flow - which means I does not
equal 0. - and an "ideal source" feeding 0 ohms gives you an answer of
infinite current.

Yes, if there is no inductive reactance, so you are agreeing with me.
Check out the superconductor (zero ohms) and Ohm's Law discussions.

http://physics.stackexchange.com/que...-0-resistivity

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'.

Undefined or infinite - the current drawn from an "ideal supply" into
0 ohms would be undefined or infinite for a split second untill the
resistance would change due to the heating effect of the current and a
split second later the resistance would become infinite and the
current zero as the "fuse" opened.

Where would that heat be coming from with no resistance there?

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.

Technically trhe voltage drop across an open fuse is considered to be
supply voltage.

Exactly, because 'voltage drop' is about current traveling through the
device under consideration and dissipating energy.

If you have a series circuit with zero current on all
the defined resistances and therefore no voltage drop across any of
them the total supply voltage is dropped across the "infinite" or
"undefined" resistance element (talking DC) In an AC circuit there
will be a capacitance between the 2 terminals that can be measured,
and the capacitive reactance will cause a miniscule but measurable
current flow

Agreed except for the 'voltage is dropped' part. There is no voltage
drop when there is no current. You even said so yourself many times.

on 5/30/2016, supposed :
On Mon, 30 May 2016 19:55:22 -0400, FromTheRafters
wrote:

notX has brought this to us :
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.

You might be correct in that particular context.

https://en.wikipedia.org/wiki/Divisi...ter_arithmetic

If you are happy with the 'voltage drop' being somewhere between zero
and a very large integer (or infinity) then that's a plus for you.

Still, it can't really be a 'voltage drop' without energy being
dissipated by the device under consideration.
The voltage drop across the wire is ZERO. The total voltage impressed
acroos the circuit is the supply voltage. Since the voltage drop
across the conductors is zero, the voltage drop across the "open" is
supply voltage. (the voltage drop MUST equal the impressed voltage)
This is true with zero amps current flow. Because there is zero amps
current you can not solve for the resistance of the conductor because
(supply voltage assumed to be 12) 12/0 is indefineableand the
reistance of the open circuit is infinite (in a perfect world)

Why call it a voltage drop when you have previously stated several
times(correctly) that it requires a current to make a 'voltage drop'?

It happens that formulated :
On Mon, 30 May 2016 17:28:23 -0400, FromTheRafters
wrote:

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?
Deviding by zero is not involved because, as stated before, by
measuring the voltage you are introducing a non-zero value to both
resistance and current.

Who said anything about measuring? It is well known that the act of
detecting a thing affects the thing being detected. We were talking
about the existence of 'voltage drop' when there is zero (by
definition) current.

[snipped the what if almost zero scenarios]

explained :
On Mon, 30 May 2016 17:26:52 -0400, FromTheRafters
wrote:

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.
The voltage drop across a perfect semiconductor is not affected by
current - other than requiring the current to be NON zero.

There you go then. Besides, semiconductors are sometimes incompatible
with Ohm's Law.

https://www.physicsforums.com/thread...ms-law.310101/

- and it is
vityually impossible to read voltage anywhere in a circuit without
completing the circuit - in a DC circuit requiring a resistive load -
an an AC circuit either a resistive, inductive or capacitive load. -
which means you can NOT measure voltage in a totally unloaded or
"open" circuit.

True, but that is not the point. Does voltage drop even exist
theoretically when there is theoretically no current?

By definition, you need current to have energy dissipation in the
device and thus a 'voltage drop'. If a device is 'heating up' you can
be sure there is current through it without attaching a multimeter and
completing a previously open circuit (which isn't actually a circuit -
since it is open).

After serious thinking wrote :
On Mon, 30 May 2016 20:13:21 -0400, FromTheRafters
wrote:

After serious thinking trader_4 wrote :
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.

Of course there isn't, because there is no current. You can't have a
voltage drop when there is no current. Thanks for finally agreeing with
me.
No voltage drop across the wire does not mean no voltage drop across
the circuit thogh.

You've ( it appears) been arguing both sides of the equation.

To avoid confusion - WHOEVER said there would be a voltage loss in an
unloaded wire is WRONG.

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

Also whoever said there is no voltage drop in an open circuit is ALSO
WRONG.

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

And to top it all off, in order to measure the voltage across an open
circuit, you MUST close the circuit - meaning it is no longer an
"open" circuit, AND

I'm not talking about measurements at all.

In the real world there is no such thing as zero ohms. You can get
REAL close - but "in the wild" it does not exist. - so you are never
REALLY deviding by or multiplying by ZERO when solving ohm's law

You're wrong about that too.

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

has brought this to us :
On Mon, 30 May 2016 20:40:56 -0400, FromTheRafters
wrote:

Mark Lloyd formulated on Monday :
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.

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.


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.

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.

Simple physics.

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.

submitted this idea :
On Mon, 30 May 2016 21:08:12 -0400, FromTheRafters
wrote:

Sam E brought next idea :
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.

I have zero problems with dividing by no.

But with 'voltage drop' you can't have zero current, because 'voltage
drop' is all about the energy delivered to and dissipated by the
device, not the capability of the source to deliver voltage.

A stepper contactor relay's burnt contacts do not have a voltage drop
unless there is current flowing through them no matter how much voltage
the source can deliver to those closed contacts. There is no voltage
drop between the poles of a car battery, unless there is current
through it and some internal resistance to dissipate some of the
energy. Voltage drop is not a static thing like voltage is.
"voltage drop" testing of a circuit indicates an open circuit by
reading source voltage across the open connection.

There's no doubt that you can measure the source voltage across an
opening, that is not the issue here. However, a "voltage drop" is due
to dissipation of energy which doesn't happen in an open circuit (which
isn't even actually a circuit at all).

A full source voltage drop across a "load" indicates zero current
flow if the resistance is infinite and infinite current if resistance
is zero - with the non- zero and non-infinite values between being
calculatable using ohm's law.

Right, for the non-zero and non-infinite values - which we aren't
discussing here. However, for the zero current condition, there is no
"voltage drop" at all. The cumulative "voltage drops" in a *closed*
circuit must equal the source voltage - but they are *not* the same
thing. Source voltage does not require current to be flowing, but
'voltage drop' does.

wrote :
On Mon, 30 May 2016 16:46:39 -0400, 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.

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.


Correct - the voltage drop across the open circuit is the supply
voltage..

No. they are two separate things. The supply voltage can exist without
any current flow, and the 'voltage drop' cannot. Voltage drop is
because of energy dissipation in a device with current flowing through
it.

There is no voltage drop across the conductors, but the sum
of all voltage drops in a circuit MUST equal the supply voltage.

Equal, yes, but the same thing, no.

[snipped]

submitted this idea :
On Mon, 30 May 2016 21:28:19 -0400, FromTheRafters
wrote:

pretended :
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.

That's what *I* said, and you said I was wrong. Sheeesh, make up your
mind.
You appear to have said, I believe "You have voltage drop due to the
length and size of the wire. Because the wire isn't a super
conductor. "

No, that most assuredly wasn't me.

You also appear to have said "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. "

Nope, again not me.

And the context was an "open circuit."

Under a load, you are correct. On an open circuit, you are just plain
wrong.

Nope, not me. I responded to Al Gebra's:

"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."

With:

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

I say it is wrong because "voltage drop" depends on the dissipation of
energy by a device with current flowing through it. No current, then
you are not even talking about "voltage drop".

If Al Gebra had said "voltage" instead of "voltage drop", or if nobody
had argued about my opinion (stated clearly as my opinion) then much of
this conversation would not have taken place.

If you did not say what you appear to have said, accept my apologies.

Accepted.

Aside, do you think that Ohm's Law works for zero current or zero
resistance values? I fully accept that it works for the case of zero
voltage, but not for current or resistance because division by zero is
undefined.

Thanks in advance.

On Monday, May 30, 2016 at 7:42:03 PM UTC-4, FromTheRafters wrote:
trader_4 brought next idea :
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.

D=RT is a relationship and can be written as T=D/R or R=D/T and it is
still the same relationship.

http://mathforum.org/dr.math/faq/faq.distance.html

See this:

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

Okay, so if I is zero, what is R? Can you show that the relationship
still holds?


R is the finite resistance of the wire, in this case, .16 ohms as
has been stated many times now. Idiot.



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

Well duh! The thing is that it is *not* "voltage drop" because an open
fuse does not dissipate energy.

"Ohm's law states that the *current through a conductor* . . ."

Where's the current through an open fuse, brainiac?

It's zero of course.

The problem is you stated:

"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."

If a fuse blows, the voltage across it after it blows is the full
open circuit voltage. Try using a meter and see. It's one of the
ways to find a blown fuse.




"Voltage drop describes how the supplied energy of a voltage source is
reduced *as electric current moves through the passive elements* . . ."

Show me how Ohm's law holds when the current is zero, brainiac.

We all have shown you, sady you don't understand basic algebra.

V = IR. When I =0, V = 0.

Like Clare said, you're really way in over your head here.

You brought up distance traveled = rate x time, claimed that doesn't work
with a rate or time of zero either. It obviously does, if the rate is
zero, the equation gives a distance traveled of zero. That zero has
meaning, it means the distance traveled is zero. Capiche?

On Monday, May 30, 2016 at 8:13:30 PM UTC-4, FromTheRafters wrote:


V = IR is the formula for voltage drop. Put in zero for I, any finite
resistance for R, you get V = 0. That tells you there is no voltage drop.

Of course there isn't, because there is no current. You can't have a
voltage drop when there is no current. Thanks for finally agreeing with
me.

--
Getting there was like pulling teeth though

You really are the village idiot. Having no voltage drop and a voltage
drop of zero are the same thing. The equation can't produce words, it
produces a value of ZERO. YOU claimed that value had no meaning. THAT
is what's BS. It has meaning, it means the voltage drop is zero.
As I said many posts ago, it's like having 3 apples and taking 3 away.
Subtraction yields a number of zero. There are zero apples. It has
meaning. I could also say there are no apples. Or in your example of
distance traveled a result of zero means the distance traveled is
zero. I could state that as the object has not moved. Capiche?
No, of course not.

On Monday, May 30, 2016 at 8:19:32 PM UTC-4, notX wrote:
On 05/30/2016 06:11 PM, trader_4 wrote:

[snip]

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.

Did you consider that if the heater is 100 feet from the power source,
there's 200 feet of wire?

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.

On Monday, May 30, 2016 at 8:41:04 PM UTC-4, FromTheRafters wrote:
Mark Lloyd formulated on Monday :
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.

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.

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.

On Monday, May 30, 2016 at 8:23:53 PM UTC-4, FromTheRafters wrote:
trader_4 was thinking very hard :
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.

Sorry to have to tell you this, but there can be no power without any
current either.

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.

On Monday, May 30, 2016 at 9:08:20 PM UTC-4, FromTheRafters wrote:
Sam E brought next idea :
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.

I have zero problems with dividing by no.

But with 'voltage drop' you can't have zero current, because 'voltage
drop' is all about the energy delivered to and dissipated by the
device, not the capability of the source to deliver voltage.

A stepper contactor relay's burnt contacts do not have a voltage drop
unless there is current flowing through them no matter how much voltage
the source can deliver to those closed contacts. There is no voltage
drop between the poles of a car battery, unless there is current
through it and some internal resistance to dissipate some of the
energy. Voltage drop is not a static thing like voltage is.

Neither Sam nor I ever said it was. We said that V = IR. With a current
of zero, V, the voltage drop is zero. YOU are the one that claims that
has "no meaning". I have 3 apples, I take 3 away, how many are left?

Aplles to start - apples taken away = apples left

3 - 3 = 0

That zero, in your world, apparently has no meaning either. It's really
sad, the state of education in America. Even worse when someone so
ignorant has the nerve to try to explain math and science to those of
us that understand it.