On Monday, May 30, 2016 at 11:46:09 AM UTC-4, FromTheRafters wrote:
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.
There is no such thing as a 'voltage drop' when there is no current
D=RT, so when the rate (velocity) is zero then D=zero*T and since
zero*T equals zero then D equals zero, so you are saying that the
distance is zero. This is obviously not the case.
What is the amperage (A) when you put zero ohms (R) resistance across
an ideal 12 volt (V) source?
Undefined! Because A=V/R and you can't divide by zero.
This is simple math, why are you struggling with it?
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
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
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.
That is true, unless you state that I or R is zero. See above and
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.
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.
D=RT is a relationship and can be written as T=D/R or R=D/T and it is
still the same relationship.
Okay, so if I is zero, what is R? Can you show that the relationship
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.
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?
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
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.
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.
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
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
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.
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.
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.
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 *your understanding* of Ohm's Law, and of junior high school algebra, that are broken.
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.
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.
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.
That is incorrect. Resistance is any finite number: e.g. 5 ohms * 0 amps = 0 volts. The one
thing R *cannot* be is infinite.
No it did not.
And yours is one of them.
(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:
ERR:DIVIDE 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.
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
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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