Current w 0 R is non-0?

anhnha

Joined Apr 19, 2012
905
Ok, I did review.

You can find a solution by inserting a resistance R between the two nodes.
Then work out the answer as R approaches the limit of zero ohms.

I have to run now so I can't do the calculation.

Someone please do this.
Now I think this method will not work in all cases. If we connect an ideal source (no internal resistance) with an ideal wire.

In order to calculate current flows through the ideal wire I use the limit:
\(I = \lim_{R \rightarrow 0} \frac{V}{R} = \propto \)
This means that when R very very small, almost zero, I will be infinity but it is still indeterminate at R = 0, right?
 

Attachments

I am a bit confused about Ohm's law. Which form is Ohm's law:
1. \(I = \frac{V}{R} \)
2. \( V = I.R\)
3. \(R = \frac{V}{I} \)

Does Ohm's law only apply for ohmic materials (as WBahn said) not ideal wire?

As my understanding till now is that if we only apply Ohm's law for ohmic materials (R ≠ 0) then all three forms above are equivalent and they are all called Ohm's law.

If I understand you correctly, you are saying that Ohm's law can be used for ideal wire but the form you are using is V=IR. I think that form is what you mean by Ohm's law.

I think if we consider ideal wires are not ohmic materials and don't apply Ohm's law for it then the problem will be easier.
But how do you know that the voltage drop across an ideal wire is zero?
1. By using Ohm's law: V = IR with R = 0 then V = 0
2. Maybe, based on experiment or something else.
I'm saying that Ohm's law gives the expected result for certain calculations involving ideal wires (wires with truly zero ohms resistance) as long as you don't try to make a calculation that results in zero in the denominator.
 

t_n_k

Joined Mar 6, 2009
5,455
There have already been quite acceptable solutions suggested to obtain the current in the zero resistance link - without resorting to "phantom" resistances which reduce to zero.

If one finds some help in the (albeit unnecessary) method of inserting a resistance which reduces to zero at the limit, then perhaps the attached schematic in which the link is replaced with an ideal super node voltage source whose value reduces to zero might be another option of interest. It is of interest (to me) in that it exemplifies a familiar technical challenge in dealing with a super node in mesh analysis.
 

Attachments

WBahn

Joined Mar 31, 2012
29,979
I am a bit confused about Ohm's law. Which form is Ohm's law:
1. \(I = \frac{V}{R} \)
2. \( V = I.R\)
3. \(R = \frac{V}{I} \)

Does Ohm's law only apply for ohmic materials (as WBahn said) not ideal wire?
Ohm's law states that the current through a resistance is proportional to the voltage across it. The resistance is the proportionality constant. If a material obeys that property, then it is ohmic, if it doesn't, then it isn't. If you double the voltage, you double the current. If you halve the voltage, you halve the current. That doesn't describe the behavior of an ideal wire, hence an ideal wire is not an ohmic material. In an ideal wire the voltage across it is zero indepedent of the current through it. In many cases it is convenient to think of an ideal wire as a zero ohm resistor, but that is a convenience and does not make it an ohmic material that obeys Ohm's law.

But how do you know that the voltage drop across an ideal wire is zero?
Because that is what DEFINES an ideal wire -- as soon as you say that you are treating something as an ideal wire, you are stating that the voltage across it is zero, not small, not tiny, not infinitesimal, but zero.
 

WBahn

Joined Mar 31, 2012
29,979
This is exactly what I did but I didn't replace R by 0 in set of equations before solving it.
The result in the case are same but I don't know if this method (replace parameter R = 0 before solving the set of equations) is valid in all cases.
I remember in an old thread The Electrician said that we have to solve set of equations with R then after get result, we find the limit R -> 0.

I tried to solve set of equations 3x3 with R but it took a lot of time and I gave up.
What R are you talking about? I didn't replace any R by 0. I worked with the four resistors and the one voltage source in the circuit. Period.
 

WBahn

Joined Mar 31, 2012
29,979
Now I think this method will not work in all cases. If we connect an ideal source (no internal resistance) with an ideal wire.

In order to calculate current flows through the ideal wire I use the limit:
\(I = \lim_{R \rightarrow 0} \frac{V}{R} = \propto \)
This means that when R very very small, almost zero, I will be infinity but it is still indeterminate at R = 0, right?
Indeterminate means that the value cannot be determined from the present form of the information. 0/∞ is not indeterminate -- it is 0. ∞/0 is not indeterminate, it is ∞. K/0 is not indeterminate, it is ∞. K/∞ is not indeterminate, it is 0. 0/0 is indeterminate. ∞/∞ is indeterminate. If you have an indeterminate form, then you must look elsewhere for what the answer is -- and the answer may be infinite, may be zero, may be a non-zero finite value, or it may simply remain indeterminate.
 

anhnha

Joined Apr 19, 2012
905
Thanks WBahn, you cleared up my mind. :cool:
Because that is what DEFINES an ideal wire -- as soon as you say that you are treating something as an ideal wire, you are stating that the voltage across it is zero, not small, not tiny, not infinitesimal, but zero.
In many cases it is convenient to think of an ideal wire as a zero ohm resistor, but that is a convenience and does not make it an ohmic material that obeys Ohm's law.
Exactly, these are what were confusing.
Another question in my picture above, if R = 0, then the configuration is impossible both in theory and practice, right? Thus, although the limit of V/R is indeterminate as R -> 0 but because the configuration is impossible, meaning that there is no other ways to find it.
 
Last edited:

anhnha

Joined Apr 19, 2012
905
What R are you talking about? I didn't replace any R by 0. I worked with the four resistors and the one voltage source in the circuit. Period.
Sorry, I was mistaken, I thought you inserted an R two ends of the ideal wire and then find limit as R-> 0.
This is the method that MrChips said in post #27. I tried to do it but finally ends up with a set of equations 3x3 with R parameter.

My intend is to solve all loop currents as a function of R then calculate the current through the ideal wire by using loop currents. Finally take the limit of it when R->0.

My confusion is that is it possible to replace R = 0 before solving the equations?
 

WBahn

Joined Mar 31, 2012
29,979
Thanks WBahn, you cleared up my mind. :cool:


Exactly, these are what were confusing.
Another question in my picture above, if R = 0, then the configuration is impossible both in theory and practice, right? Thus, although the limit of V/R is indeterminate as R -> 0 but because the configuration is impossible, meaning that there is no other ways to find it.
Have you ever heard of superconductors?
 

WBahn

Joined Mar 31, 2012
29,979
Sorry, I was mistaken, I thought you inserted an R two ends of the ideal wire and then find limit as R-> 0.
This is the method that MrChips said in post #27. I tried to do it but finally ends up with a set of equations 3x3 with R parameter.

My intend is to solve all loop currents as a function of R then calculate the current through the ideal wire by using loop currents. Finally take the limit of it when R->0.

My confusion is that is it possible to replace R = 0 before solving the equations?
As The Electrician already pointed out, if you want to take this approach (for some reason known only to Mr Chips, apparently), then you have to solve for the case of R being whatever it is (i.e., nonzero) and then take the limit as R approaches zero.

But the only time when doing this is needed is when there is some reason to believe that you can't just set R equal to zero in the first place and there is no reason to believe any such thing in this case. As you said, you get the the same set of equations, so what's the point?
 

MrChips

Joined Oct 2, 2009
30,720
Ok, I concede. I was wrong.

I solved the KCL and KVL equations.

The current through a shorting resistor of resistance R is 1000/(63R + 1250) = 0.8A when R = 0.
 

anhnha

Joined Apr 19, 2012
905
Have you ever heard of superconductors?
Yes, I know a bit about it through online courses.
I learned about Faraday's law. If magnetic flux penetrates the circular wire (now assume that it is superconductor) varies with time, then the total voltage in the wire is zero but current flowing through it is nonzero

I want to know if there is a case (practice or theory) that we have both a nonzero voltage and nonzero current across/through an ideal wire or superconductor.
I think this is impossible.
 
Last edited:

WBahn

Joined Mar 31, 2012
29,979
Yes, I know a bit about it through online courses.
I learned about Faraday's law. If magnetic flux penetrates the circular wire (now assume that it is superconductor) varies with time, then the total voltage in the wire is zero but current flowing through it is nonzero

I want to know if there is a case (practice or theory) that we have both a nonzero voltage and nonzero current across/through an ideal wire or superconductor.
I think this is impossible.
Well, that's gonna come as a surprise to the people that work with superconducting magnets on a day in and day out basis, like I did when I worked for NIST in the late 1980s involved in research into superconducting materials.

A superconducting magnet is a coil of wire, just like any other electromagnet. The one that I had the most experience with was a 10 tesla radial access split pair magnet with 54H of inductance and a core field intensity of right about 100mT/A. When we ramped it up we generally went at around 2V due to limitations in the magnet power supply. At 2V that means that we had a rate of change of current of

di/dt = 2V/54H ≈ 2.2A/min

This meant that to get to our first measurement point, which was typically about 6T, took the better part of half an hour.

So any time we were ramping or deramping a magnet we had a superconducting coil that had both a non-zero voltage across it and a non-zero current in it.
 

anhnha

Joined Apr 19, 2012
905
Then what is the difference between in ideal inductor (no internal resistance ) and a superconductor?

In the ideal inductor, if we hook it up with an ideal voltage source, in transition time both voltage and current across/through inductor is non-zero. However, in steady state voltage across the inductor is zero but how about its current at this time? Assuming that all contacts are ideal and components has zero internal resistances.
I am still confused with this case.
I think it is impossible rather than indeterminate.
 

WBahn

Joined Mar 31, 2012
29,979
Then what is the difference between in ideal inductor (no internal resistance ) and a superconductor?

In the ideal inductor, if we hook it up with an ideal voltage source, in transition time both voltage and current across/through inductor is non-zero. However, in steady state voltage across the inductor is zero but how about its current at this time? Assuming that all contacts are ideal and components has zero internal resistances.
I am still confused with this case.
I think it is impossible rather than indeterminate.
WHAT is impossible about it?

Superconducting magnets in MRI machines have currents flowing in them for years with no power supply attached. The persistent mode switches are not perfect because you have to get the current from one end superconducting wire through the nonsuperconducting support matrix and back into the other end of the wire. Even so, these magnets loose only about 1ppm of their energy per year due to these parasitic effects.

In steady state, the voltage across an ideal inductor is zero independent of the current through it, because the voltage across an inductor is NOT related to the current through it, but to the rate of change of the current though it.

You keep saying you "think" something is impossible but don't give any basis upon which hang those thoughts. You ask for either theoretical or real world examples of where what you think is impossible is possible. I then describe a very real world every day situation in which it is not only possible but routine, and you just continue to say that you "think" it is impossible.

So you just go ahead and keep denying reality and think what ever you want.
 

anhnha

Joined Apr 19, 2012
905
Sorry, I was not intended to deny what you said.
What I am confused is here:
Assuming that we have an ideal battery 5V and ideal wire connected as in the picture.
I know this is impossible in practice but let's assume that theoretically.
At the same time, let assume that all contacts also ideal.


Is this configuration possible? I mean we have still have 5 V across the battery as two ends of it is connected with an ideal wire.
There is a paradox here: An ideal voltage source keeps a constant voltage 5V across two ends of it. At the same time, the ideal wire keeps a zero voltage across it.
What will happen? Or this configuration is even impossible in theory?
 

Attachments

WBahn

Joined Mar 31, 2012
29,979
This has been covered over and over.

The notion of an ideal voltage source shorted by an ideal wire is, at best, a theoretical curiosity that results in a paradox because the definitions of the components are not compatible with the constraints of the circuit. So it is a MEANINGLESS construct!
 
Top