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

No argument on that point.

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

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

D=RT, so when the rate (velocity) is zero then D=zero

Consider this:

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

trader_4 pretended :

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

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

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

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

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

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

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

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

Not if you use the D=RT formula, it is 'undefined' when either R or T is zero (I did

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

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

Simple math, you are

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'

I, of course, agree about 'approaching' these limits as being worthy of consideration. However, that was

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

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

Idiot.

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

See this:

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

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

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

Is there zero resistance in that heater circuit wire?

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

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

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

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

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

Distance Traveled = Rate x Time.

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

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

Idiot.

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

See this:

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

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

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

Is there zero resistance in that heater circuit wire?

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

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

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

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

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

Distance Traveled = Rate x Time.

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

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

trader_4 brought next idea :

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

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

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

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

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

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

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

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

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

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

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

Well duh! The thing is that it is

"Ohm's law states that the

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

"Voltage drop describes how the supplied energy of a voltage source is reduced

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

On Monday, May 30, 2016 at 7:42:03 PM UTC-4, FromTheRafters wrote:

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

It's zero of course.

The problem is you stated:

*">You can have *

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

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

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

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

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

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

It's zero of course.

The problem is you stated:

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

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

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

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

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

trader_4 presented the following explanation :

Not that specific case, brainiac. I'm talking about the general case of Ohm's Law where you can take any two known quantities and calculate the third.

Duh, again. But it is***not*** a 'voltage drop', brainiac.

But I do. I don't know who sady is though.

Clare apparently thought she was talking to someone else.

D=RT, R=D/T, T=D/R are all the same "algebraically" because you can divide both sides of D=RT by R (to get T=D/R) or by T (to get R=D/T) as long as T or R are***NON-ZERO***.

It's the same relationship where any two given's should allow you to find the third 'not given' value. The values that are in the denominators must be non-zero or it doesn't work.

http://www.sosmath.com/algebra/fraction/frac3/frac31/frac311/frac311.html

http://www.onemathematicalcat.org/algebra_book/online_problems/frac_zero.htm

Saying that the rate is zero is the same as saying that the distance over time is zero because (R=D/T) which is saying that the time is non-zero because it must be or you have broken the rules. So this non-zero time must be the distance over the rate (T=D/R) and the rate must be non-zero or you have broken the rules. So there is a contradiction because you already stated that the rate was zero. QED

Thanks for playing, brainiac.

Not that specific case, brainiac. I'm talking about the general case of Ohm's Law where you can take any two known quantities and calculate the third.

Duh, again. But it is

But I do. I don't know who sady is though.

Clare apparently thought she was talking to someone else.

D=RT, R=D/T, T=D/R are all the same "algebraically" because you can divide both sides of D=RT by R (to get T=D/R) or by T (to get R=D/T) as long as T or R are

It's the same relationship where any two given's should allow you to find the third 'not given' value. The values that are in the denominators must be non-zero or it doesn't work.

http://www.sosmath.com/algebra/fraction/frac3/frac31/frac311/frac311.html

http://www.onemathematicalcat.org/algebra_book/online_problems/frac_zero.htm

Saying that the rate is zero is the same as saying that the distance over time is zero because (R=D/T) which is saying that the time is non-zero because it must be or you have broken the rules. So this non-zero time must be the distance over the rate (T=D/R) and the rate must be non-zero or you have broken the rules. So there is a contradiction because you already stated that the rate was zero. QED

Thanks for playing, brainiac.

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On Tuesday, May 31, 2016 at 11:02:21 AM UTC-4, FromTheRafters wrote:

You can't even understand the simple circuit with a resistance of .16 ohms, apply Ohm's Law, do basic algebra. No need to go anywhere else.

V = IR

I = 0, V = 0. End of story. And the V = 0 has meaning, the voltage drop is zero.

Did you try what I told you to do. Make a graph of Ohm's Law, plot voltage versus current. You get a straight line through the origin. Yet you claim the formula doesn't work with I=0. How is that, when we can plot it and we all see that with I=0, V=0?

Keep jumbling them all around, any way you like. What you're implying is that any equation where one of the variables, R in the example above, can be zero in some case, means that equation has no meaning because through algebra it can be manipulated so that to find some other value, you wind up with R being in the denominator. Did you even take algebra? In the examples there IS NO DIVISION BY ZERO required. YOU are the only one that is desperately trying to divide by zero, to discredit Ohm's Law.

V = IR. I = 0, then V=0 There is no zero in any denominator. D = RT R =0, then D =0, there is no zero in any denominator.

Any time, village idiot.

You can't even understand the simple circuit with a resistance of .16 ohms, apply Ohm's Law, do basic algebra. No need to go anywhere else.

V = IR

I = 0, V = 0. End of story. And the V = 0 has meaning, the voltage drop is zero.

Did you try what I told you to do. Make a graph of Ohm's Law, plot voltage versus current. You get a straight line through the origin. Yet you claim the formula doesn't work with I=0. How is that, when we can plot it and we all see that with I=0, V=0?

Keep jumbling them all around, any way you like. What you're implying is that any equation where one of the variables, R in the example above, can be zero in some case, means that equation has no meaning because through algebra it can be manipulated so that to find some other value, you wind up with R being in the denominator. Did you even take algebra? In the examples there IS NO DIVISION BY ZERO required. YOU are the only one that is desperately trying to divide by zero, to discredit Ohm's Law.

V = IR. I = 0, then V=0 There is no zero in any denominator. D = RT R =0, then D =0, there is no zero in any denominator.

Any time, village idiot.

[...]

***Any*** two? Really?

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

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

Doug Miller laid this down on his screen :

It can't be done, and that's my point. There must be current for there to be a 'voltage drop'.

That is the case where I don't even need to use division by zero to show that Ohm's Law is 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. The result is undefined because zero times infinity is undefined.

http://electronics.stackexchange.com/questions/98779/is-ohms-law-violating-itself

It can't be done, and that's my point. There must be current for there to be a 'voltage drop'.

That is the case where I don't even need to use division by zero to show that Ohm's Law is 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. The result is undefined because zero times infinity is undefined.

http://electronics.stackexchange.com/questions/98779/is-ohms-law-violating-itself

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

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.

It's

False. This is an impossible situation. If current is zero, then voltage

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.

Doug Miller pretended :

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.

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.

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

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

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.

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.

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

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

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.

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

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

On Tuesday, May 31, 2016 at 5:45:39 PM UTC-4, Doug Miller wrote:

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

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

+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.

+1

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

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

+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.

+1

On 05/31/2016 12:21 PM, Doug Miller wrote:

Resistance is infinite.

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

Resistance is infinite.

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

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That is incorrect. Resistance is any finite number: e.g. 5 ohms

No it did not.

And yours is one of them.

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

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.

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

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

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.

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

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.

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No, my calculator throws an error because division by zero

root

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

Not necessarily three

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

[snip]

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).

[snip]

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).

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