BS. Put the test leads of a high impedance meter into a receptacle in your house, with no other loads on the circuit. It's an open circuit, excluding the impedance of the meter, which is in the mega ohms and totally insignificant. That small effect is less than the accuracy of the meter. And if you want to use more exotic instruments, we could measure it without even that negligible meter load.
My God, why do we need to go down even more silly rat holes?
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.
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.
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?
Thanks for recognizing the fact that you're in way over your head here, brainiac.
trader_4 laid this down on his screen :
Then what does that make you?
No, it isn't.
Undefined, not zero.
Ah, so now I can see your math skill level. You finally got one right.
How many oranges does that make, brainiac?
LOL
Yes, but it is not a 'voltage drop' because there is no current across the open fuse and no energy dissipation taking place.
Which is still wrong no matter how many times you say it.
All of them. Just because you take them away doesn't mean they don't exist anymore. They still exist at the place you took them away to.
Ah, more math, and correct this time - you're learning after all, albeit s-l-o-w-l-y.
It sure is, LOL.
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're the only one that doesn't understand basic algebra. No need to go into semiconductor physics.
Which it is.
Yes it is, the voltage across an open circuit like across a blown fuse is not determined by the dissipation of energy through it, but 'voltage drop'is (only if there is current). You or gfretwell or Clare brought the term loss into the discussion, not me. If I had said loss it would have been 'copper loss' not voltage loss or power loss. Somebody (Clare?) said they were the same as 'voltage drop' - which they aren't.
It doesn't matter what resistance is dissipating the energy, it is all 'voltage drop'.
No they are not.
No, it is "undefined".
Maybe they just expect everyone to know about it already? That doesn't make *them* the village idiots now does it?
If you're going to give definitions, give the right ones please.
See above comment regarding correct definitions. IMO the word "typically" does not belong in a proper definition, but maybe that's just me.
See the comment above the comment above for further enlightenment.
That's something I really like about Diesel, and you're right - it is a rarity.
You're way in over your head here. Obviously you've never passed algebra or physics 101. From Newton's Laws, we can calculate the velocity of a ball that is thrown straight up in the air at an initial velocity of 20m/sec. We can calculate it's velocity at every point in time. At the peak, Newton's Law gives a velocity of zero. Are you going to tell us that value of zero has no meaning. Of course it does, it means the ball isn't moving. Just like a voltage drop of zero means there is no voltage loss. Capiche?
And a voltage drop of ZERO from Ohm;s LAw with a current of Zero, says that. Good grief.
FromTheRafters wrote in news:nii8ff$38m$1 @news.albasani.net:
Complete nonsense. If R = 0, then D = 0T = zero. If T = 0, then D = 0R = zero.
There's no division in D= RT. That's multiplication. [...]
More utter nonsense. If I or R is zero, then so is E.
Starting with E = IR and deriving from it I = E/R is valid *if and only if* R is unequal to zero. Likewise, R = E/I can be derived from E = IR *if and only if* I is nonzero.
FromTheRafters wrote in news:niij47$m7j$1 @news.albasani.net:
No, it is not the same relationship, because there are different sets of permissible values for the three variables in the three equations. D = RT is valid for all real values of D, R, and T; T = D/R is valid for all real values of D and T, and R 0; R = D/T is valid for all real values of D and R, and T 0.
FromTheRafters wrote in news:nik91o$q95$1 @news.albasani.net:
[...] *Any* two? Really?Voltage = 120, current = zero. Calculate resistance, please.
In the discussion of the 100 ft of wire going to the heater load, the terms voltage drop and voltage loss relative to the wire are one and the same. Everyone here gets that, except you.
You don't need current to have a voltage drop of zero. In physics, do you need energy, acceleration, to have a velocity of zero? Ever take physics 101?
Simple algebra test for Johnny:
Distance Traveled = Rate x Time
1 - At a rate of 2m/sec, how far does the object move in one minute?Answer: 2m/sec x 60 secs = 120 meters
2 - At a rate of 0 m/sec, how far does the object move in one minute?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.
Explain that to us. Gfre, CL and I are on the same page. V = IR. If there is zero current, there is zero voltage There is zero voltage drop across that resistor and there is no voltage drop. It all means exactly the same thing, but then we're in the real world.
Idiot.
Again, you failed even basic math. Trivial test question:
V = IR. R is .16, I is zero.
What is V?
All the rest of us say it's ZERO. Are you sitting here, saying that the correct answer on a 7th grade math exam to that question is not zero, but "it's undefined"? Good grief.
I wouldn't be laughing, you're demonstrating to everyone that's reading this that you can't handle simple 7th grade math.
Only to the village idiot who can't pass a trivial 7th grade math question.
V = IR. R=.16, I= 0
What is V?
Our answer, the correct answer even a grade school student knows, is ZERO.
Tell us your answer?
More silly duplicity.
BTW, did you plot that graph for Ohm's Law, of Voltage versus Current? It's a straight line, that goes right through the origin. We all see a solid line. You claim there is some unique singularity at the origin, where there is hole in the line. There isn't.
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.
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.
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