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On 5/31/2016 4:04 PM, Mark Lloyd wrote:

For whatever it's worth...

http://www.wolframalpha.com/input/?i 0%2F0

Apparently nothing, since it's incorrect.

No, it is not the same relationship, because there are different sets of permissible values for the three variables in the three equations. D = RT is valid for all real values of D, R, and T; T = D/R is valid for all real values of D and T, and R <> 0; R = D/T is valid for all real values of D and R, and T <> 0.

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

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

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

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

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

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

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

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

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

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

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

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

#### Site Timeline

- posted on May 31, 2016, 11:26 pm

For whatever it's worth...

http://www.wolframalpha.com/input/?i 0%2F0

- posted on June 1, 2016, 12:06 am

Apparently nothing, since it's incorrect.

- posted on June 1, 2016, 6:34 am

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

I agree.

http://www.wolfram.com/programming-lab/explorations/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]

I agree.

http://www.wolfram.com/programming-lab/explorations/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: <nb7u27$crn$ snipped-for-privacy@boaterdave.dont-email.me>

Hmmm. I most certainly don't understand how I can access a copy of a

MID: <nb7u27$crn$ snipped-for-privacy@boaterdave.dont-email.me>

Hmmm. I most certainly don't understand how I can access a copy of a

Click to see the full signature.

- posted on June 1, 2016, 10:43 am

Diesel formulated on Wednesday :

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. :)

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. :)

- posted on May 31, 2016, 5:16 pm

No, it is not the same relationship, because there are different sets of permissible values for the three variables in the three equations. D = RT is valid for all real values of D, R, and T; T = D/R is valid for all real values of D and T, and R <> 0; R = D/T is valid for all real values of D and R, and T <> 0.

- posted on May 31, 2016, 2:53 am

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

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.

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.

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

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.

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.

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

- posted on May 31, 2016, 10:03 am

snipped-for-privacy@snyder.on.ca formulated the question :

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

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/questions/62664/how-can-ohms-law-be-correct-if-superconductors-have-0-resistivity

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

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

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.

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

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/questions/62664/how-can-ohms-law-be-correct-if-superconductors-have-0-resistivity

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

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

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.

- posted on May 31, 2016, 3:03 pm

On Tuesday, May 31, 2016 at 6:03:55 AM UTC-4, FromTheRafters wrote:

Only the village idiot thinks V = IR is defeated with I = 0. Even after pointing out his failed example, where he diverted to Distance traveled = rate X time, he has no answer for that yet persists. With a rate = 0, we get a distance traveled of 0. That formula, like Ohm's Law is not defeated, it works. A distance traveled of zero means the object is still in the same place.

Now the village idiot is telling us about reactance and super conductors.

I think I see the problem here. You either had no course in algebra or you never made it to word problems. We solve problems all the time where an equation gives a value of ZERO and that means there is no voltage drop, no current flowing, no movement of a train, no apples left in the box. Capiche?

Only the village idiot thinks V = IR is defeated with I = 0. Even after pointing out his failed example, where he diverted to Distance traveled = rate X time, he has no answer for that yet persists. With a rate = 0, we get a distance traveled of 0. That formula, like Ohm's Law is not defeated, it works. A distance traveled of zero means the object is still in the same place.

Now the village idiot is telling us about reactance and super conductors.

I think I see the problem here. You either had no course in algebra or you never made it to word problems. We solve problems all the time where an equation gives a value of ZERO and that means there is no voltage drop, no current flowing, no movement of a train, no apples left in the box. Capiche?

- posted on May 31, 2016, 8:38 pm

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

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.

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.

- posted on May 31, 2016, 9:37 pm

snipped-for-privacy@snyder.on.ca has brought this to us :

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

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.

- posted on June 1, 2016, 4:46 am

On Tuesday, May 31, 2016 at 5:37:27 PM UTC-4, FromTheRafters wrote:

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.

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.

- posted on June 1, 2016, 10:57 am

trader_4 formulated on Wednesday :

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.

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.

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

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. :)

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.

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.

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

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. :)

- posted on June 1, 2016, 1:44 pm

On Wednesday, June 1, 2016 at 6:57:42 AM UTC-4, FromTheRafters wrote:

Which implies that completion of the circuit is necessary to be able to measure the voltage source, hence my reply. If you're not implying that, what's the point of bringing it up at all? More FUD like bringing up calculus when you can't do simple math?

See, I was right, you do want to go off into more nits, when the real problem here is that you think V = IR, with I=0 somehow involves division. And that you can't recognize that when an equation like that gives a value of zero, the zero has meaning. Often in science and engineering the zero results are among the most interesting, that solve the problem, etc.

Voltage can exist without

And Ohm's Law works at zero too. I've asked 6 times now, have you drawn a graph of Ohm's Law, plotted, I vs V? You get a straight line, it goes right through the origin.

That's like saying you can't measure the speed of your car, without the car moving. The rest of us recognize that as a speed of ZERO.

Assignment for a grade school student. Keep a tabulation of the speed of a car while dad is driving, record the observations every minute. What speed entry do you make when the time for an observation occurs when the car is stopped at a light? The rest of us would put in zero. You? Apparently you'd argue that the speed is "undefined" because to make that entry would require division by zero.

Next assignment, calculate the average speed using those recorded measurements. We'd take all the tabulated entries, add them up and divide by the number of entries. Let's say there were ten. We have

35, 42, 47, 0, 40, 50, 30, 0

We get (35+42+47+0+40+50+30+0)/8 as the answer

WTF do you get? Speed is undefined because the car is not moving, right?

(35+ 42+47+undefined+40+50+30+undefined)/8 So, answer is undefined?

(35+42+47+40+50+30)/6 ?

That's the screwy world you're living in.

If anyone is a troll, it's you. It's been a long time since we've seen a total moron here like you. BTW, we're all still waiting for you to explain how 1 squared can be 2 sometimes, another one of your claims.

Which implies that completion of the circuit is necessary to be able to measure the voltage source, hence my reply. If you're not implying that, what's the point of bringing it up at all? More FUD like bringing up calculus when you can't do simple math?

See, I was right, you do want to go off into more nits, when the real problem here is that you think V = IR, with I=0 somehow involves division. And that you can't recognize that when an equation like that gives a value of zero, the zero has meaning. Often in science and engineering the zero results are among the most interesting, that solve the problem, etc.

Voltage can exist without

And Ohm's Law works at zero too. I've asked 6 times now, have you drawn a graph of Ohm's Law, plotted, I vs V? You get a straight line, it goes right through the origin.

That's like saying you can't measure the speed of your car, without the car moving. The rest of us recognize that as a speed of ZERO.

Assignment for a grade school student. Keep a tabulation of the speed of a car while dad is driving, record the observations every minute. What speed entry do you make when the time for an observation occurs when the car is stopped at a light? The rest of us would put in zero. You? Apparently you'd argue that the speed is "undefined" because to make that entry would require division by zero.

Next assignment, calculate the average speed using those recorded measurements. We'd take all the tabulated entries, add them up and divide by the number of entries. Let's say there were ten. We have

35, 42, 47, 0, 40, 50, 30, 0

We get (35+42+47+0+40+50+30+0)/8 as the answer

WTF do you get? Speed is undefined because the car is not moving, right?

(35+ 42+47+undefined+40+50+30+undefined)/8 So, answer is undefined?

(35+42+47+40+50+30)/6 ?

That's the screwy world you're living in.

If anyone is a troll, it's you. It's been a long time since we've seen a total moron here like you. BTW, we're all still waiting for you to explain how 1 squared can be 2 sometimes, another one of your claims.

- posted on June 1, 2016, 3:42 pm

On Wednesday, June 1, 2016 at 9:44:18 AM UTC-4, trader_4 wrote:

Correction. In my last post I gave an example of samples taken from a moving car where I said:

" Let's say there were ten. "

That should be eight.

Correction. In my last post I gave an example of samples taken from a moving car where I said:

" Let's say there were ten. "

That should be eight.

- posted on June 1, 2016, 3:53 pm

On 06/01/2016 08:44 AM, trader_4 wrote:

1. make a square 1 unit on each side

2. measure the diagonal. If this is 2, stop taking drugs and go to step 5

3. hold by the other 2 corners (other than the ones you're measuring between)

4. squeeze until measurement becomes 2

5. say you're doing advanced math, and only you understand it

1. make a square 1 unit on each side

2. measure the diagonal. If this is 2, stop taking drugs and go to step 5

3. hold by the other 2 corners (other than the ones you're measuring between)

4. squeeze until measurement becomes 2

5. say you're doing advanced math, and only you understand it

- posted on May 31, 2016, 5:12 pm

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

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

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

Starting with E = IR and deriving from it I = E/R is valid

- posted on May 31, 2016, 6:34 pm

Doug Miller explained :

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. Since time is T=D/R the time is undefined by division by zero - or - as time is infinite in the multiplication D=RT it is undefined by multiplying by infinity.

You can't get around this without using calculus, and when you use calculus you have admitted to having non-zero values where in algebra you had stipulated zero values. Ohm's Law, as it relates to 'voltage drop' will require some current 'approaching zero' rather than a stipulated zero value.

As I or R approach zero, so does E.

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

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. Since time is T=D/R the time is undefined by division by zero - or - as time is infinite in the multiplication D=RT it is undefined by multiplying by infinity.

You can't get around this without using calculus, and when you use calculus you have admitted to having non-zero values where in algebra you had stipulated zero values. Ohm's Law, as it relates to 'voltage drop' will require some current 'approaching zero' rather than a stipulated zero value.

As I or R approach zero, so does E.

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

- posted on May 31, 2016, 6:54 pm

On Tuesday, May 31, 2016 at 2:34:35 PM UTC-4, FromTheRafters wrote:

Doug said it best:

"Complete nonsense."

He and I have explained it to you:

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

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

BS. No calculus required.

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

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

Doug said it best:

"Complete nonsense."

He and I have explained it to you:

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

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

BS. No calculus required.

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

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

- posted on May 31, 2016, 9:34 pm

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

Correct, but not relevant to the equation D = RT which is

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

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

You can't get around it

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

And when either I or R

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

I'm not getting into

- posted on May 31, 2016, 10:07 pm

Doug Miller explained :

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.

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]

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.

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]

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