On Tuesday, May 31, 2016 at 3:58:39 PM UTC-4, FromTheRafters wrote:
trader_4 was thinking very hard :
On Tuesday, May 31, 2016 at 11:44:46 AM UTC-4, FromTheRafters wrote:


Each connection, switch, etc will have some resistance, which gets
added to the total. So, you wind up with the measured voltage being
less than the theoretical from the wire alone, ie the measured voltage
drop is larger than the theoretical. That is all he's saying,
but he's not very clear about it and it has nothing to do with the
example here. Voltage drop and voltage loss here are one and the
same,

No they are not.


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



because we are only considering the ideal circuit, without those
additional resistances. We could put an additional resistor in to
model that, but it would make no difference in the fact that with
zero current, the voltage drop or voltage loss is zero.

No, it is "undefined".


Here Johnny, 7th grade math test:

V = IR r=.16 ohms, I = 0

What is the value of V?

All of us say it's ZERO.




The second guy is a true village idiot. Note that again, no definition
of voltage drop or voltage loss is ever given.

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


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


He just goes off to
diagrams and it's not clear from looking at some of those diagrams,
what his point is.

Let me do for you what neither of these two did, nor can you, which is
give definitions. Voltage drop is the voltage across a component, which
in the case of a resistance load is given by Ohm's Law, V = IR.

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


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


Voltage loss is typically used to describe unwanted voltage drop
in a circuit due to the resistance of the wire, connections,
other components, etc.

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


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



In the simple case we're talking about the voltage drop across the wire
is also the voltage loss. Capiche?

See the comment above the comment above for further enlightenment.

See the fact that the neither the village idiot nor his cites can
define the terms he's trying to argue about.

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

HAND

Yeah, I do tend to get upset when we have village idiots here that
are clueless trying to explain to those of us who are degreed electrical
engineers, electrical inspectors, etc how ohm's law works and basic algebra.

BTW, did you make that graph yet of V = IR, plotting V vs I? It goes
right through the origin, does it not? 0,0 is that "undefined"?

Still waiting for you to explain your new claim too, where you claim
that 1 squared can equal 2. I have bets it's only in your special little
universe.

Sam E writes:
On 05/31/2016 08:49 AM, trader_4 wrote:

[snip]

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.


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


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

Perhaps you meant mass?

On Tue, 31 May 2016 13:32:05 -0700 (PDT)
trader_4 wrote:

who are degreed electrical
engineers, electrical inspectors, etc

Who apparently never work as they post here day and night.

On Tue, 31 May 2016 06:03:47 -0400, FromTheRafters
wrote:

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

On Tue, 31 May 2016 06:07:03 -0400, FromTheRafters
wrote:

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'?
Infinite resistance and zero current gives supply voltage as the
drop. Doesn't matter what the theoretical math is.

On Tuesday, May 31, 2016 at 4:30:29 PM UTC-4, Scott Lurndal wrote:
trader_4 writes:
For example, we can
apply Newton's Law to give the velocity of a ball shot up in the air.
At the peak, the equation give a value of Zero. To you and I, that's a
valid answer, it means the velocity is zero, the ball isn't moving, etc.

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

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

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

On Tue, 31 May 2016 06:24:56 -0400, FromTheRafters
wrote:

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

On Tue, 31 May 2016 06:33:36 -0400, FromTheRafters
wrote:

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

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

On Tue, 31 May 2016 07:05:35 -0400, FromTheRafters
wrote:

wrote :
I'll leave you to your "mental masturbation"

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

Doug Miller explained :
FromTheRafters wrote in news:nii8ff$38m$1
@news.albasani.net:

trader_4 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).

Complete nonsense. If R = 0, then D = 0T = zero. If T = 0, then D = 0R =
zero.

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

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

If the rate is zero by stipulating that it is so, and the distance is
zero by multiplication using Ohm's Law, the time must be infinite since
we aren't moving.

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

Since time is T=D/R the time is undefined by division
by zero

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

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

- or - as time is infinite in the multiplication D=RT it is
undefined by multiplying by infinity.

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

You can't get around this without using calculus,

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

and when you use
calculus you have admitted to having non-zero values where in algebra
you had stipulated zero values.

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

Ohm's Law, as it relates to 'voltage
drop' will require some current 'approaching zero' rather than a
stipulated zero value.

[...]

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'.

More utter nonsense. If I or R is zero, then so is E.

As I or R approach zero, so does E.

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

Starting with E = IR and deriving from it I = E/R is valid *if and only if* R
is unequal to zero. Likewise, R = E/I can be derived from E = IR *if and
only if* I is nonzero.

Correct, so Ohm's Law doesn't work for these zero values

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

and calculus
must be resorted to. Then you have actual non-zero values approaching
limits.

So, we're back to my original statement that 'voltage drop' as defined
by Ohm's Law requires that there be a current flowing through the
device and energy being dissipated by that device.

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

has brought this to us :
On Tue, 31 May 2016 06:03:47 -0400, FromTheRafters
wrote:

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

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

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

Doug Miller laid this down on his screen :
FromTheRafters wrote in news:nik91o$q95$1
@news.albasani.net:

[...]
I'm talking about the general case of
Ohm's Law where you can take any two known quantities and calculate the
third.

*Any* two? Really?

Voltage = 120, current = zero. Calculate resistance, please.

It can't be done, and that's my point.

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

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

There must be current for there
to be a 'voltage drop'.

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

That is the case where I don't even need to use division by zero to
show that Ohm's Law is broken.

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

When there is a non-zero voltage stipulated, and the current is
stipulated as zero, the resistance must be infinite by Ohm's law.

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

The
result is undefined because zero times infinity is undefined.

http://electronics.stackexchange.com...olating-itself

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

Mark Lloyd wrote in :

On 05/31/2016 12:21 PM, Doug Miller wrote:
FromTheRafters wrote in news:nik91o$q95$1
@news.albasani.net:

[...]
I'm talking about the general case of
Ohm's Law where you can take any two known quantities and calculate the
third.

*Any* two? Really?

Voltage = 120, current = zero. Calculate resistance, please.


Resistance is infinite.

That is incorrect. Resistance is any finite number: e.g. 5 ohms * 0 amps = 0 volts. The one
thing R *cannot* be is infinite.

My computer did the calculation right.

No it did not.

Some don't.

And yours is one of them.

I pressed '1', '2', '0', '/', '0', '=' and the display showed 'inf'.

That's because:
(a) you did a mathematically undefined operation (division by zero), and
(b) your calculator is constructed incorrectly, and giving you the wrong answer. The display
*should* have showed "undefined".

*My* calculator does it right. Pressing the same sequence of keys gives this result:
ERRIVIDE BY 0

My wife's calculator also does it right, with the result -E-, same as it gives for the square root
of a negative number, logarithm of a nonpositive number, inverse sine of 2, etc.

FromTheRafters wrote in news:nikm9f$mae$1
@news.albasani.net:

trader_4 formulated on Tuesday :
2 - At a rate of 0 m/sec, how far does the object move in one minute?

D=RT

Undefined,


"Undefined"?? You have to be kidding. Do you really mean that you don't know, and can't
figure out, how far it moved? that in real life if an object is not moving then no matter how
much time elapses the distance it moves is ZERO?

If something is not moving, the distance it moves is NOT "undefined". The distance it
moves is zero.

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

Absolute nonsense. If the rate R = D / T = 0, that does *not* mean that T is infinite. It means
that D = 0.

Our answer: 0 m/sec X 60 secs = 0 meters.

Your answer: That is undefined! It's dividing by zero! You can't
have a distance of zero!


Sorry Johnny, you fail.

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

How much farther do you need it "broken down"? What part of 0 * 60 = 0 do you find
confusing?

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

It's not necessary to use calculus. 0 * 60 = 0 is 4th-grade arithmetic.

FromTheRafters wrote in news:nikq90$udm$1
@news.albasani.net:

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

The set of natural numbers is the infinite set {1, 2, 3, 4, ...} and its sum is likewise infinite.

And under no circumstances at all is one to the second power equal to two.

Perhaps you should wait to post again until after you've sobered up. ;-)

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

Doug Miller explained :
FromTheRafters wrote in news:nii8ff$38m$1
@news.albasani.net:

trader_4 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).

Complete nonsense. If R = 0, then D = 0T = zero. If T = 0, then D = 0R =
zero.

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

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

If the rate is zero by stipulating that it is so, and the distance is
zero by multiplication using Ohm's Law, the time must be infinite since
we aren't moving.

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

Right, so we're left with an answer of undefined.

So, if the T is as you state 'any finite number' there must be a rate
of D/T and if that finite number T is zero we're back at division by
zero and if that finite number T is not zero we're back at some
positive or negative rate which is a contradiction of your original
premise of R being zero.

Since time is T=D/R the time is undefined by division
by zero

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

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

Okay, you got me there. I was treating D=RT as a formula not just as an
equation. My mistake. E=MC^2 so if C=0 E=0 and M doesn't matter. Got it
now, sorry for wasting your time.

[snip]

On 05/31/2016 03:45 PM, wrote:
On Tue, 31 May 2016 06:33:36 -0400, FromTheRafters
wrote:

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

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


If a pointless pin still a pin? How about an infinite number of monkeys
sharing 0 bananas? All right - it is a silly post. Some silliness is
essential. 42. HAL is "once removed" from IBM.

--
"Multitasking Attempted. System confused."

FromTheRafters wrote in news:nil1vt$d8d$1
@news.albasani.net:

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

Doug Miller explained :
FromTheRafters wrote in news:nii8ff$38m$1
@news.albasani.net:

trader_4 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).

Complete nonsense. If R = 0, then D = 0T = zero. If T = 0, then D = 0R =
zero.

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

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

If the rate is zero by stipulating that it is so, and the distance is
zero by multiplication using Ohm's Law, the time must be infinite since
we aren't moving.

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

Right, so we're left with an answer of undefined.

No, we're not, because T is *not* infinite. T can be *any* *finite* value. The one thing it
*cannot* be is infinite precisely because 0 * infinite is undefined -- but the left side of this
equation, D, is NOT undefined. It is very much defined, and it is EQUAL TO ZERO.

So, if the T is as you state 'any finite number' there must be a rate
of D/T and if that finite number T is zero we're back at division by
zero

NO WE ARE NOT! Where in the equation D = RT do you see any division anywhere? This
is *multiplication*. And as I have already noted, transforming D = RT into D / T = R is valid
only for NON-ZERO values of T.

and if that finite number T is not zero we're back at some
positive or negative rate which is a contradiction of your original
premise of R being zero.

Complete nonsense. D = RT. D = 0, R = 0, substitute any finite value for T and the equation
is true.

Since time is T=D/R the time is undefined by division
by zero

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

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

Okay, you got me there. I was treating D=RT as a formula not just as an
equation. My mistake. E=MC^2 so if C=0 E=0 and M doesn't matter. Got it
now, sorry for wasting your time.

No, you still don't have it. C isn't a variable in that equation. C is a constant. If M = 0, then E =
0 and C doesn't matter.

The only thing you got right here is that you're certainly wasting a lot of time.

Doug Miller wrote :
FromTheRafters wrote in news:nikq90$udm$1
@news.albasani.net:

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

The set of natural numbers is the infinite set {1, 2, 3, 4, ...} and its sum
is likewise infinite.

Sure, doing it *that* way. Of course you would never get there to *all*
natural numbers since it is an infinite process. It doesn't make sense
in arithmetic to us seventh gradeers now just like the previous
division by zero being undefined wouldn't make sense to a seventh
grader. That was my point. however with some redefining of things into
a different system of which seventh graders are not aware you can get
different results. Results that work in that system.

And under no circumstances at all is one to the second power equal to two.

Perhaps you should wait to post again until after you've sobered up. ;-)

In an additive group in the integers (which the naturals are a subset
of) exponentiation is the repeated application of the group operator
just as it is in multiplicative groups. It just so happens that
repeated addition is equivalent to multiplication and we aren't used to
thinking of it as exponentiation. I said exponent and right
circumstances. I didn't say second power, because that would have been
misleading.

Scott Lurndal expressed precisely :
Sam E writes:
On 05/31/2016 08:49 AM, trader_4 wrote:

[snip]

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.


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


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

Perhaps you meant mass?

Nice! Watch out though, E=MC^2 can take on all sorts of funny stuff
when it is just an equation where C can be set to zero and solved by
seventh graders.

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

Doug Miller laid this down on his screen :
FromTheRafters wrote in news:nik91o$q95$1
@news.albasani.net:

[...]
I'm talking about the general case of
Ohm's Law where you can take any two known quantities and calculate the
third.

*Any* two? Really?

Voltage = 120, current = zero. Calculate resistance, please.

It can't be done, and that's my point.

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

You got me again, what I meant was Ohm's Law is not the right tool if
you are trying to use it outside of its limitations. Within its
limitations you can do as I said. Using zero like you did in the
example is outside of its limitations.

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

Within Ohm's Law you are right about that. There is no arrangemnt of
the terms where V is problematic as far as I'm aware. It always seems
to be in the numerator or standing alone.

There must be current for there
to be a 'voltage drop'.

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

That is the case where I don't even need to use division by zero to
show that Ohm's Law is broken.

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

When there is a non-zero voltage stipulated, and the current is
stipulated as zero, the resistance must be infinite by Ohm's law.

False. This is an impossible situation.

Outside the limitations of Ohm's Law because the current can't be zero.

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

The
result is undefined because zero times infinity is undefined.

http://electronics.stackexchange.com...olating-itself

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

I wasn't posting it for the question, but for the answers.

Doug Miller explained :
FromTheRafters wrote in news:nil1vt$d8d$1
@news.albasani.net:

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

Doug Miller explained :
FromTheRafters wrote in news:nii8ff$38m$1
@news.albasani.net:

trader_4 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).

Complete nonsense. If R = 0, then D = 0T = zero. If T = 0, then D = 0R =
zero.

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

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

If the rate is zero by stipulating that it is so, and the distance is
zero by multiplication using Ohm's Law, the time must be infinite since
we aren't moving.

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

Right, so we're left with an answer of undefined.

No, we're not, because T is *not* infinite. T can be *any* *finite* value.
The one thing it *cannot* be is infinite precisely because 0 * infinite is
undefined -- but the left side of this equation, D, is NOT undefined. It is
very much defined, and it is EQUAL TO ZERO.

So, if the T is as you state 'any finite number' there must be a rate
of D/T and if that finite number T is zero we're back at division by
zero

NO WE ARE NOT! Where in the equation D = RT do you see any division anywhere?
This is *multiplication*. And as I have already noted, transforming D = RT
into D / T = R is valid only for NON-ZERO values of T.

Which is why Ohm's Law is a formula not just an equation.

and if that finite number T is not zero we're back at some
positive or negative rate which is a contradiction of your original
premise of R being zero.

Complete nonsense. D = RT. D = 0, R = 0, substitute any finite value for T
and the equation is true.

Since time is T=D/R the time is undefined by division
by zero

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

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

Okay, you got me there. I was treating D=RT as a formula not just as an
equation. My mistake. E=MC^2 so if C=0 E=0 and M doesn't matter. Got it
now, sorry for wasting your time.

No, you still don't have it. C isn't a variable in that equation. C is a
constant. If M = 0, then E = 0 and C doesn't matter.

It looks just like a letter in an equation not a constant in a formula.
As a formula the relationships between the terms become important just
like they do in Ohm's Law.

The only thing you got right here is that you're certainly wasting a lot of
time.

Yeah, I guess it's time to stop now.

It's been fun.

hah brought next idea :
On 05/31/2016 03:45 PM, wrote:
On Tue, 31 May 2016 06:33:36 -0400, FromTheRafters
wrote:

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

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


If a pointless pin still a pin? How about an infinite number of monkeys
sharing 0 bananas? All right - it is a silly post. Some silliness is
essential. 42. HAL is "once removed" from IBM.

That's the spirit! Sorry if I upset anyone.

On 5/31/2016 4:04 PM, Mark Lloyd wrote:
On 05/31/2016 12:21 PM, Doug Miller wrote:
FromTheRafters wrote in news:nik91o$q95$1
@news.albasani.net:

[...]
I'm talking about the general case of
Ohm's Law where you can take any two known quantities and calculate the
third.

*Any* two? Really?

Voltage = 120, current = zero. Calculate resistance, please.


Resistance is infinite.

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


For whatever it's worth...

http://www.wolframalpha.com/input/?i=120%2F0

FromTheRafters wrote in news:nil5s9$jon$1
@news.albasani.net:

Doug Miller explained :
FromTheRafters wrote in news:nil1vt$d8d$1
@news.albasani.net:

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

Doug Miller explained :
FromTheRafters wrote in news:nii8ff$38m$1
@news.albasani.net:

trader_4 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).

Complete nonsense. If R = 0, then D = 0T = zero. If T = 0, then D = 0R =
zero.

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

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

If the rate is zero by stipulating that it is so, and the distance is
zero by multiplication using Ohm's Law, the time must be infinite since
we aren't moving.

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

Right, so we're left with an answer of undefined.

No, we're not, because T is *not* infinite. T can be *any* *finite* value.
The one thing it *cannot* be is infinite precisely because 0 * infinite is
undefined -- but the left side of this equation, D, is NOT undefined. It is
very much defined, and it is EQUAL TO ZERO.

So, if the T is as you state 'any finite number' there must be a rate
of D/T and if that finite number T is zero we're back at division by
zero

NO WE ARE NOT! Where in the equation D = RT do you see any division anywhere?
This is *multiplication*. And as I have already noted, transforming D = RT
into D / T = R is valid only for NON-ZERO values of T.

Which is why Ohm's Law is a formula not just an equation.

And the difference is --?

and if that finite number T is not zero we're back at some
positive or negative rate which is a contradiction of your original
premise of R being zero.

Complete nonsense. D = RT. D = 0, R = 0, substitute any finite value for T
and the equation is true.

Since time is T=D/R the time is undefined by division
by zero

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

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

Okay, you got me there. I was treating D=RT as a formula not just as an
equation. My mistake. E=MC^2 so if C=0 E=0 and M doesn't matter. Got it
now, sorry for wasting your time.

No, you still don't have it. C isn't a variable in that equation. C is a
constant. If M = 0, then E = 0 and C doesn't matter.

It looks just like a letter in an equation not a constant in a formula.

I suppose it does -- to you. And to everyone else who doesn't understand it.

As a formula the relationships between the terms become important just
like they do in Ohm's Law.

The only thing you got right here is that you're certainly wasting a lot of
time.

Yeah, I guess it's time to stop now.

It's been fun.

No, it hasn't.

FromTheRafters wrote in news:nil3lr$g3j$1
@news.albasani.net:

Doug Miller wrote :
FromTheRafters wrote in news:nikq90$udm$1
@news.albasani.net:

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

The set of natural numbers is the infinite set {1, 2, 3, 4, ...} and its sum
is likewise infinite.

Sure, doing it *that* way. Of course you would never get there to *all*
natural numbers since it is an infinite process. It doesn't make sense
in arithmetic to us seventh gradeers now just like the previous
division by zero being undefined wouldn't make sense to a seventh
grader. That was my point. however with some redefining of things into
a different system of which seventh graders are not aware you can get
different results. Results that work in that system.

And under no circumstances at all is one to the second power equal to two.

Perhaps you should wait to post again until after you've sobered up. ;-)

In an additive group in the integers (which the naturals are a subset
of) exponentiation is the repeated application of the group operator
just as it is in multiplicative groups. It just so happens that
repeated addition is equivalent to multiplication and we aren't used to
thinking of it as exponentiation. I said exponent and right
circumstances. I didn't say second power, because that would have been
misleading.

You really ought to confine your discussions to subjects you know something about.

If there are any.

FromTheRafters wrote in news:nil57k$ims$1
@news.albasani.net:

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

Doug Miller laid this down on his screen :
FromTheRafters wrote in news:nik91o$q95$1
@news.albasani.net:

[...]
I'm talking about the general case of
Ohm's Law where you can take any two known quantities and calculate the
third.

*Any* two? Really?

Voltage = 120, current = zero. Calculate resistance, please.

It can't be done, and that's my point.

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

You got me again, what I meant was Ohm's Law is not the right tool if
you are trying to use it outside of its limitations. Within its
limitations you can do as I said. Using zero like you did in the
example is outside of its limitations.

More nonsense. Nothing invalid in mathematics, or in Ohm's Law, about multiplying by zero.

You don't seem to understand the difference between multiplication by zero and division by
zero.

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

Within Ohm's Law you are right about that. There is no arrangemnt of
the terms where V is problematic as far as I'm aware. It always seems
to be in the numerator or standing alone.

There must be current for there
to be a 'voltage drop'.

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

That is the case where I don't even need to use division by zero to
show that Ohm's Law is broken.

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

When there is a non-zero voltage stipulated, and the current is
stipulated as zero, the resistance must be infinite by Ohm's law.

False. This is an impossible situation.

Outside the limitations of Ohm's Law because the current can't be zero.

OF COURSE current can be zero. If no current flows, I is not undefined. I is equal to zero.

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

The
result is undefined because zero times infinity is undefined.

http://electronics.stackexchange.com...olating-itself

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

I wasn't posting it for the question, but for the answers.

And you were given correct answers repeatedly, and continued to argue with them.

1+1 wrote in news:tbWdna-Ij5YHgNPKnZ2dnUU7-
:

On 5/31/2016 4:04 PM, Mark Lloyd wrote:
On 05/31/2016 12:21 PM, Doug Miller wrote:
FromTheRafters wrote in news:nik91o$q95$1
@news.albasani.net:

[...]
I'm talking about the general case of
Ohm's Law where you can take any two known quantities and calculate the
third.

*Any* two? Really?

Voltage = 120, current = zero. Calculate resistance, please.


Resistance is infinite.

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


For whatever it's worth...

Apparently nothing, since it's incorrect.

http://www.wolframalpha.com/input/?i=120%2F0

The result of division by zero is undefined. It's not "equal to infininity".

On 05/31/2016 04:52 PM, Doug Miller wrote:
[snip]

*My* calculator does it right. Pressing the same sequence of keys gives this result:
ERRIVIDE BY 0

Your calculator throws an error because it doesn't know how to do it. I
didn't need a calculator for something so simple.

My wife's calculator also does it right, with the result -E-, same as it gives for the square root
of a negative number, logarithm of a nonpositive number, inverse sine of 2, etc.

I know how to find the square root of a nonpositive number (actually,
both of them). Every number had TWO square roots. A little hint: when is
360 equal to 0?

BTW, every number has three cube roots too, but that may be a bit more
complicated.

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

"The activities engaged in by the Christian Coalition...were a vital
part of why we had a revolution at the polls on November 8, 1994." [Newt
Gingrich]

Mark Lloyd wrote in :

On 05/31/2016 04:52 PM, Doug Miller wrote:
[snip]

*My* calculator does it right. Pressing the same sequence of keys gives this result:
ERRIVIDE BY 0

Your calculator throws an error because it doesn't know how to do it. I
didn't need a calculator for something so simple.

No, my calculator throws an error because division by zero *is* an error.

My wife's calculator also does it right, with the result -E-, same as it gives for the square
root
of a negative number, logarithm of a nonpositive number, inverse sine of 2, etc.

I know how to find the square root of a nonpositive number (actually,
both of them).

Yes, I do too -- but most calculators operate only on real numbers.

Every number had TWO square roots. A little hint: when is
360 equal to 0?

BTW, every number has three cube roots too, but that may be a bit more
complicated.

Not necessarily three *distinct* cube roots...

On 05/31/2016 08:00 PM, Doug Miller wrote:

[snip]

BTW, every number has three cube roots too, but that may be a bit more
complicated.

Not necessarily three *distinct* cube roots...


The same number, but different numerals.

The 3 cube roots of a number have angles of 0 (positive), 120, 240
degrees. In this case, 0 at any angle is still 0.

BTW, when I mentioned 360 = 0, somehow I thought of the Xbox 360. What I
actually meant there was about angles (in degrees).

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

"The activities engaged in by the Christian Coalition...were a vital
part of why we had a revolution at the polls on November 8, 1994." [Newt
Gingrich]

On Tuesday, May 31, 2016 at 5:37:27 PM UTC-4, FromTheRafters wrote:
has brought this to us :
On Tue, 31 May 2016 06:03:47 -0400, FromTheRafters
wrote:

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

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

No completion of that circuit is necessary to measure the voltage
potential. The meter has an impedance in the meg ohms and for all
practical purposes can be ignored for the purposes of this discussion.
And there are non-contact measuring techniques and instruments as well.
You can measure electric potential without a circuit. You really
are in way over your head here.

On Tuesday, May 31, 2016 at 5:45:39 PM UTC-4, Doug Miller wrote:
FromTheRafters wrote in news:niklg0$kke$1
@news.albasani.net:

Doug Miller laid this down on his screen :
FromTheRafters wrote in news:nik91o$q95$1
@news.albasani.net:

[...]
I'm talking about the general case of
Ohm's Law where you can take any two known quantities and calculate the
third.

*Any* two? Really?

Voltage = 120, current = zero. Calculate resistance, please.

It can't be done, and that's my point.

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


Exactly. So then he moves the goal post and talks about something else.


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

There must be current for there
to be a 'voltage drop'.

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

Agree, he doesn't realize it. Note his insistence that one can't
measure the potential of a voltage source without completing the
circuit.



That is the case where I don't even need to use division by zero to
show that Ohm's Law is broken.

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

+1

I even suggested that he plot V = IR, V vs R on a graph. It's a straight
line that goes through the origin.




When there is a non-zero voltage stipulated, and the current is
stipulated as zero, the resistance must be infinite by Ohm's law.

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

The
result is undefined because zero times infinity is undefined.

http://electronics.stackexchange.com...olating-itself

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


+1

On Tuesday, May 31, 2016 at 6:00:17 PM UTC-4, Doug Miller wrote:
FromTheRafters wrote in news:nikm9f$mae$1
@news.albasani.net:

trader_4 formulated on Tuesday :
2 - At a rate of 0 m/sec, how far does the object move in one minute?

D=RT

Undefined,


"Undefined"?? You have to be kidding. Do you really mean that you don't know, and can't
figure out, how far it moved? that in real life if an object is not moving then no matter how
much time elapses the distance it moves is ZERO?


+1

If something is not moving, the distance it moves is NOT "undefined". The distance it
moves is zero.


+1



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

Absolute nonsense. If the rate R = D / T = 0, that does *not* mean that T is infinite. It means
that D = 0.

Our answer: 0 m/sec X 60 secs = 0 meters.

Your answer: That is undefined! It's dividing by zero! You can't
have a distance of zero!


Sorry Johnny, you fail.

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

How much farther do you need it "broken down"? What part of 0 * 60 = 0 do you find
confusing?

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

It's not necessary to use calculus. 0 * 60 = 0 is 4th-grade arithmetic.

Rafters doing calculus that would be something. He also said that
1 squared is sometimes equal to 2. I'm sure you'd like to see that
explained too.

On Tuesday, May 31, 2016 at 6:08:02 PM UTC-4, FromTheRafters wrote:
Doug Miller explained :
FromTheRafters wrote in news:niklfm$kju$1
@news.albasani.net:

Doug Miller explained :
FromTheRafters wrote in news:nii8ff$38m$1
@news.albasani.net:

trader_4 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).

Complete nonsense. If R = 0, then D = 0T = zero. If T = 0, then D = 0R =
zero.

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

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

If the rate is zero by stipulating that it is so, and the distance is
zero by multiplication using Ohm's Law, the time must be infinite since
we aren't moving.

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

Right, so we're left with an answer of undefined.

So, if the T is as you state 'any finite number' there must be a rate
of D/T and if that finite number T is zero we're back at division by
zero and if that finite number T is not zero we're back at some
positive or negative rate which is a contradiction of your original
premise of R being zero.

Since time is T=D/R the time is undefined by division
by zero

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

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

Okay, you got me there. I was treating D=RT as a formula not just as an
equation. My mistake. E=MC^2 so if C=0 E=0 and M doesn't matter. Got it
now, sorry for wasting your time.

[snip]

Call it formula or equation, you're still wrong, period. You must have
taken quibbling 101 instead of algebra 101.

On Tuesday, May 31, 2016 at 6:36:49 PM UTC-4, FromTheRafters wrote:
Doug Miller wrote :
FromTheRafters wrote in news:nikq90$udm$1
@news.albasani.net:

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

The set of natural numbers is the infinite set {1, 2, 3, 4, ...} and its sum
is likewise infinite.

Sure, doing it *that* way. Of course you would never get there to *all*
natural numbers since it is an infinite process. It doesn't make sense
in arithmetic to us seventh gradeers now just like the previous
division by zero being undefined wouldn't make sense to a seventh
grader. That was my point. however with some redefining of things into
a different system of which seventh graders are not aware you can get
different results. Results that work in that system.

And under no circumstances at all is one to the second power equal to two.

Perhaps you should wait to post again until after you've sobered up. ;-)

In an additive group in the integers (which the naturals are a subset
of) exponentiation is the repeated application of the group operator
just as it is in multiplicative groups. It just so happens that
repeated addition is equivalent to multiplication and we aren't used to
thinking of it as exponentiation. I said exponent and right
circumstances. I didn't say second power, because that would have been
misleading.

OMG, more stupidity.

You posted:

"The natural number one with an exponent of two can equal two under the
right circumstances."

An exponent of two is a number raised to the second power. It's the
number times itself. Now explain to us how 1 x 1 = 2.

Doug Miller
Wed, 01 Jun 2016
00:06:18 GMT in alt.home.repair, wrote:


The result of division by zero is undefined. It's not "equal to
infininity".

I agree.

http://www.wolfram.com/programming-l.../circlescapes/

I did find that nostalgic. A trip down memory lane. I remember typing
in the demo source from the owners manual to my color computer 3 when I
was a kid to do the same thing. [g]


--
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

FromTheRafters
Mon, 30 May 2016 18:18:28 GMT in
alt.home.repair, wrote:

55484A76596D466962486B67626D3867633256756332556762 32596761485674623
34967 5A576C30614756794943413D


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--
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

Diesel formulated on Wednesday :
Doug Miller
Wed, 01 Jun 2016
00:06:18 GMT in alt.home.repair, wrote:


The result of division by zero is undefined. It's not "equal to
infininity".

I agree.

http://www.wolfram.com/programming-l.../circlescapes/

I did find that nostalgic. A trip down memory lane. I remember typing
in the demo source from the owners manual to my color computer 3 when I
was a kid to do the same thing. [g]

Complex infinity (the result from the Wolfram calculator) is not the
same as infinity (or infininity whatever that is). It actually means
undefined or unknown. So it is correct, not incorrect as claimed by the
poster you are responding to.

That is to say it was actually in agreement with Doug Miller and
myself, and he just didn't realize it. For what it's worth.

trader_4 formulated on Wednesday :
On Tuesday, May 31, 2016 at 5:37:27 PM UTC-4, FromTheRafters wrote:
has brought this to us :
On Tue, 31 May 2016 06:03:47 -0400, FromTheRafters
wrote:

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

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

No completion of that circuit is necessary to measure the voltage
potential.

I didn't say that there was. What I said was that the meter completed
the circuit and the measurement was taken from the 'voltage drop'
across the internal resistance of the meter.

The meter has an impedance in the meg ohms and for all
practical purposes can be ignored for the purposes of this discussion.

Maybe for your discussion, but not for mine. Voltage can exist without
current and current can exist without voltage, but 'voltage drop'
requires current and resistance and Ohm's Law works under those
conditions no matter how close to zero they get without actually being
zero.

And there are non-contact measuring techniques and instruments as well.
You can measure electric potential without a circuit.

I agree with that statement. But you can't measure 'voltage drop'
without current going through the device which is responsible for it.

You really
are in way over your head here.

Sure, if you say so. But I don't stick to the shallow end of the pool
like you do. You should stop trying to put words in my mouth so I won't
feel as if I should respond, it makes you look like a troll.