The voltage drop across a perfect semiconductor is not affected by current - other than requiring the current to be NON zero. - 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.
Deviding by zero is not involved because, as stated before, by measuring the voltage you are introducing a non-zero value to both resistance and current. - Resistance WAY off from zero - approaching (but never COMPLETELY reaching infinite (by the very definition of infinity) meaning current- for all practical reasons being ZERO - but in reality just being infinitesimally small, Do the calculations using a current of 0.000000(100,000,000 zeros)01 amps and the resistance being 0.0000000100,000,000 zeros)01 ohms for an open circuit and everything works.
Sane as with a dead short=0.99999o(1000,000,000 nines)99 ohms because we don't have superconductors. In reality you don't need to go nearly as far ac the impedence of any meter is sigmificantly lower than that - with sensitivity being in the megohms per volt range.on digitals and kilohms per volt on analogs.
No voltage drop across the wire does not mean no voltage drop across the circuit thogh.
You've ( it appears) been arguing both sides of the equation.
To avoid confusion - WHOEVER said there would be a voltage loss in an unloaded wire is WRONG. Also whoever said there is no voltage drop in an open circuit is ALSO WRONG. And to top it all off, in order to measure the voltage across an open circuit, you MUST close the circuit - meaning it is no longer an "open" circuit, AND In the real world there is no such thing as zero ohms. You can get REAL close - but "in the wild" it does not exist. - so you are never REALLY deviding by or multiplying by ZERO when solving ohm's law
It HAS tyo have, because with zero(or as close to zero as the real world can produce) current flow in the conductors and an ipressed voltage of 120, or whatever volts across the cirduit, the voltage HAS to drop across something - in this case the "infinite" or "undrfined" resistance across the blown fuse.
Simple physics.
"voltage drop" testing of a circuit indicates an open circuit by reading source voltage across the open connection. A full source voltage drop across a "load" indicates zero current flow if the resistance is infinite and infinite current if resistance is zero - with the non- zero and non-infinite values between being calculatable using ohm's law.
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.
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.
It happens that snipped-for-privacy@snyder.on.ca formulated :
Who said anything about measuring? It is well known that the act of detecting a thing affects the thing being detected. We were talking about the existence of 'voltage drop' when there is zero (by definition) current.
[snipped the what if almost zero scenarios]snipped-for-privacy@snyder.on.ca explained :
There you go then. Besides, semiconductors are sometimes incompatible with Ohm's Law.
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).
I'm not talking about a voltage loss, but a 'voltage drop'.
Then you are calling yourself wrong. There is no current in an open circuit, and you have agreed several times (correctly) that you need current to have a 'voltage drop'.
I'm not talking about measurements at all.
You're wrong about that too.
snipped-for-privacy@snyder.on.ca has brought this to us :
It's not a 'voltage drop' when there is no current. Voltage drop exists because of the dissipation of energy across the device under consideration.
There's not much simple about physics.
From Wikipedia:
"Zero electrical DC resistance
[...]The simplest method to measure the electrical resistance of a sample of some material is to place it in an electrical circuit in series with a current source I and measure the resulting voltage V across the sample. The resistance of the sample is given by Ohm's law as R = V / I. If the voltage is zero, this means that the resistance is zero."
If there was no voltage (a voltage drop actually) measured, you would be fooled into believing that there is zero resistance if you leave out the part about the need to have current flowing through the device.
With no current, you can't trust Ohm's Law to give a meaningful result. See above where R = V / I. If I is zero, R is *undefined* not "zero" or "infinity".
The above example uses a "circuit" (not an open circuit which isn't actually a circuit at all) and a "current source" with current flowing through the device.
snipped-for-privacy@snyder.on.ca submitted this idea :
There's no doubt that you can measure the source voltage across an opening, that is not the issue here. However, a "voltage drop" is due to dissipation of energy which doesn't happen in an open circuit (which isn't even actually a circuit at all).
Right, for the non-zero and non-infinite values - which we aren't discussing here. However, for the zero current condition, there is no "voltage drop" at all. The cumulative "voltage drops" in a *closed* circuit must equal the source voltage - but they are *not* the same thing. Source voltage does not require current to be flowing, but 'voltage drop' does.
No. they are two separate things. The supply voltage can exist without any current flow, and the 'voltage drop' cannot. Voltage drop is because of energy dissipation in a device with current flowing through it.
Equal, yes, but the same thing, no.
[snipped]
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?
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.
A blown fuse that's in a circuit has the full open circuit voltage across it. Use a meter and see. It's one of the ways we find blown fuses in a circuit.
Yes, and that's why P = R*I^2 gives a value of ZERO, when I = 0. That zero is the value of the power and it too has meaning.
Neither Sam nor I ever said it was. We said that V = IR. With a current of zero, V, the voltage drop is zero. YOU are the one that claims that has "no meaning". I have 3 apples, I take 3 away, how many are left?
Aplles to start - apples taken away = apples left
3 - 3 = 0That zero, in your world, apparently has no meaning either. It's really sad, the state of education in America. Even worse when someone so ignorant has the nerve to try to explain math and science to those of us that understand it.
OK, let's look at them one at a time. The first guy is just a poor writer who doesn't even define the terms he's using. What he's saying is "voltage drop" is the theoretical voltage drop given by Ohm's Law for a circuit. He then uses the term "voltage loss" for the actual MEASURED voltage difference in that circuit. But there is no such industry standard definition and it has nothing to do with what we are talking about here. That measured "voltage loss" is just the theoretical voltage drops from Ohm's Law, plus other voltage drops, from other resistance in the circuit other than the theoretical resistance of the wire. 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, 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.
The second guy is a true village idiot. Note that again, no definition of voltage drop or voltage loss is ever given. 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. Voltage loss is typically used to describe unwanted voltage drop in a circuit due to the resistance of the wire, connections, other components, etc.
In the simple case we're talking about the voltage drop across the wire is also the voltage loss. Capiche?
Thank you for admitting your mistake. It's something rarely done around here.
Sigh. Sadly that is incorrect too. The ideal diode equation for the perfect semiconductor shows the relationship between the current through the diode and the voltage across it. It's exponential. The current is a function of the voltage, it's just that the current goes up a lot for a small change in voltage.
- and it is
Of course you can measure voltage in an open circuit. Stick the test leads of a high impedance meter into a receptacle with no loads on that circuit. For all practical purposes, you're measuring the open circuit voltage, you'll see ~120V. The only load is the insignificant impedance of the meter, in the millions of ohms. That load is so small, it's less than the accuracy of the meter itself. And if you want to use more exotic test equipment, clearly we could measure it without even that insignificant load.
We have a guy here who doesn't understand basic algebra and Ohm's Law. IDK why you're dragging in semiconductor physics, AC circuit dynamics and leading the discussion into rat holes.
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