How correct are these concepts about circuits?

  • #26
phinds
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we took norton equivalent in the lectures in a different way, i don't invent examples, i am telling you the style of questions they give us and the steps of solution they told us, first of all in the question it tells you what is the external load to work on (2 ohm here) then to solve these are the steps explained to us in the lectures :
1- remove this load and replace batteries by short circuits and current sources by open circuits and get R between a,b (here it will be 0)
2- replace the place of the load with SC and calculate the current in that SC (here the SC will be parallel to the battery and that is what i was asking about)
you took the load as the battery but as i am saying the examiner enforces you to choose a specific load (2 ohm here), they forbid us in the lectures from using voltage sources as external loads in norton just resistances
In that case, the question makes no sense. A Norton equivalent has a current source in series with the load and a resistor in parallel with the load and an undefined load voltage. You can't do that in this circuit if the 2 ohm resistor is the load and the voltage source stays in the circuit.
 
  • #27
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4
I have problem understanding what OP's problem is.
Wire is not source of voltage, no.

Also, there is no "voltage" in point. Voltage of a point is 0, because voltage is difference of electric potential.

On your 2), voltage between "a" and "b" will be 0 (assuming perfect wire, with no resistance).

In the end, such question makes no sense - in most simple words, loss of voltage of the wire (its resistance multiplied by current flowing through it) must be equal to voltage between the ends of voltage source. You cannot say that "wire is source of voltage", as if you were to disconnect the circuit (as you did on "2)") you'd get 0 voltage.
 
  • #28
I have problem understanding what OP's problem is.
Wire is not source of voltage, no.

Also, there is no "voltage" in point. Voltage of a point is 0, because voltage is difference of electric potential.

On your 2), voltage between "a" and "b" will be 0 (assuming perfect wire, with no resistance).

In the end, such question makes no sense - in most simple words, loss of voltage of the wire (its resistance multiplied by current flowing through it) must be equal to voltage between the ends of voltage source. You cannot say that "wire is source of voltage", as if you were to disconnect the circuit (as you did on "2)") you'd get 0 voltage.
i didn't say wire is source of voltage, what i meant from the very beginning that in numbers 3 and 4 current and volt will be 0 which means there are no voltage or current sources just wires, didn't say source = wire, and i said also the purpose of the question and the relation with norton theorem so if you want to understand just read all posts i will not repeat what i said
 
  • #29
The problem with 3 and 4 is not about whether you want to allow infinity as a valid value. The problem is that both circuit 3 and circuit 4 are not self-consistent.
i have just remembered something, i understand the consistency principle and i am totally convinced that number 3 is wrong because the wire must have the current 0 and I at the same time, but regarding number 4 i think the concept of infinity has a rule, now we have short circuit saying voltage is zero, but this saying is based on the resistance being zero so multiplying it by current gives zero, but this in case current has a real value, if current is infinite we can't say voltage is zero, like in open circuit, we have zero current in open circuit, but we can't say voltage drop is zero, because resistance is infinite, so what do you see ?
 
  • #30
jbriggs444
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i have just remembered something, i understand the consistency principle and i am totally convinced that number 3 is wrong because the wire must have the current 0 and I at the same time, but regarding number 4 i think the concept of infinity has a rule, now we have short circuit saying voltage is zero, but this saying is based on the resistance being zero so multiplying it by current gives zero, but this in case current has a real value, if current is infinite we can't say voltage is zero, like in open circuit, we have zero current in open circuit, but we can't say voltage drop is zero, because resistance is infinite, so what do you see ?
What do we see? One sentence, with no proper capitalization. A string of run on thoughts ending in a mistake. If resistance is zero then resistance is not infinite.

You could decide to extend the rules of circuit analysis this way: A finite voltage applied across a zero resistance results in an infinite current whose sign is the same as the voltage. But such an extension would not be predictive in the other direction. What voltage results from an infinite current applied across a zero resistance?

This problem is essentially the reason why the real numbers are not normally augmented by plus and minus infinity -- the resulting number system is not a field. It does not obey all of the normally expected rules of arithmetic.
 
  • #31
If resistance is zero then resistance is not infinite.
do you mean : If resistance is zero then "current" is not infinite. ??
 
  • #32
jbriggs444
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do you mean : If resistance is zero then "current" is not infinite. ??
No. I mean that if resistance is zero then resistance is not infinite.

You wrote:

i have just remembered something, i understand the consistency principle and i am totally convinced that number 3 is wrong because the wire must have the current 0 and I at the same time, but regarding number 4 i think the concept of infinity has a rule, now we have short circuit saying voltage is zero, but this saying is based on the resistance being zero so multiplying it by current gives zero, but this in case current has a real value, if current is infinite we can't say voltage is zero, like in open circuit, we have zero current in open circuit, but we can't say voltage drop is zero, because resistance is infinite, so what do you see ?
 
  • #33
we have short circuit saying voltage is zero, but this saying is based on the resistance being zero so multiplying it by current gives zero, but this in case current has a real value, if current is infinite we can't say voltage is zero
"we have short circuit saying....."
like in open circuit, we have zero current in open circuit, but we can't say voltage drop is zero, because resistance is infinite
"like in open circuit......"
 
  • #34
jbriggs444
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this in short circuit

this in open circuit
Thus demonstrating the need for proper punctuation.
 
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  • #35
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i have just remembered something, i understand the consistency principle and i am totally convinced that number 3 is wrong because the wire must have the current 0 and I at the same time, but regarding number 4 i think the concept of infinity has a rule, now we have short circuit saying voltage is zero, but this saying is based on the resistance being zero so multiplying it by current gives zero, but this in case current has a real value, if current is infinite we can't say voltage is zero, like in open circuit, we have zero current in open circuit, but we can't say voltage drop is zero, because resistance is infinite, so what do you see ?
Take one step back. Before you even get to the point of figuring out how much current goes through the voltage source you have to set up a system of equations which describes the circuit. This is done before solving. The equation for the wire says that the difference in voltage is 0. The equation for the source says that the same difference in voltage is V. So the equations describing the components are inconsistent, even before you ever go to solve for current.
 
  • #36
Take one step back. Before you even get to the point of figuring out how much current goes through the voltage source you have to set up a system of equations which describes the circuit. This is done before solving. The equation for the wire says that the difference in voltage is 0. The equation for the source says that the same difference in voltage is V. So the equations describing the components are inconsistent, even before you ever go to solve for current.
okay i am with you, let's write the equations before figuring out the current, the equation for source says voltage difference is V, the equation of wire says the same difference of voltage is RI = 0(I), so I is V/0, we return to the same point ?
 
  • #37
phinds
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okay i am with you, let's write the equations before figuring out the current, the equation for source says voltage difference is V, the equation of wire says the same difference of voltage is RI = 0(I), so I is V/0, we return to the same point ?
Yes we do. That point being that it is invalid to try to solve contradictory equations, which is what everyone has been telling you all along.
 
  • #38
phinds
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Yes we do. That point being that it is invalid to try to solve contradictory equations, which is what everyone has been telling you all along.
You can continue to simply repeat your erroneous point of view over and over, but that will never make it right. You would be better off listening to what everyone is telling you.
 
  • #39
@phinds : firstly we are not fighting, if my posts annoy you can ignore them or block me
secondly i am not repeating anything, i am replying to his point and didn't come up with anything new
thirdly regarding erroneous equations : so why in open circuit connected to voltage source we have alike set of equations but considered valid, in open circuit : equation for the source tells that voltage difference is V, and equation for the open circuit tells that same voltage difference is I*R = 0*R or I*(infinite resistance), so R is V/0 = infinity or I is V/infinity = 0
 
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  • #40
let me rephrase my point in a way acceptable by all, for this simplest circuit :
classic_circuit_1000.jpg


I = lim(R→0) V/R = infinity, meaning that as R approaches zero I approaches infinity, this is represented by this curve :
4b0274cc573ea72178e3394830841c7b.png


the division by zero is actually a limit, the R approaches zero but never reaches it, my mistake was getting a circuit with 0 resistance and then evaluating a limit of R approaching zero, the corresponding resistance to this limit should approach zero not equal zero.
 
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  • #41
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the equation of wire says the same difference of voltage is RI = 0(I)
There is no R in the equation for the wire. If ##V_a## is the voltage on one terminal of the voltage source and ##V_b## is the voltage on the other side then the equation for the voltage source is ##V_a-V_b=V## and the equation for the wire is ##V_a-V_b=0##. There is no R involved and the equations are inconsistent and no value of I can be determined.
 
  • #42
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let me rephrase my point in a way acceptable by all, for this simplest circuit :
View attachment 91082

I = lim(R→0) V/R = infinity, meaning that as R approaches zero I approaches infinity, this is represented by this curve :
View attachment 91094

the division by zero is actually a limit, the R approaches zero but never reaches it, my mistake was getting a circuit with 0 resistance and then evaluating a limit of R approaching zero, the corresponding resistance to this limit should approach zero not equal zero.
This is a good approach. When an ideal wire seems to make something nonsensical, then replace it with a resistor and take a limit as the resistor goes to zero. Often you will find that the new circuit has a consistent set of equations and sensible limiting behavior.
 
  • #43
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4
I still have no idea what this thread is about.
Not want to be rude, or something, but there is importance of formatting and proper punctuation.
 
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  • #44
In drawing #4, unless you specify resistance of wire and internal resistance of voltage source, both are assumed to be zero. That gives infinite current for any voltage greater than zero.
 
  • #45
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* number 1 : the current will flow normally: Correct
* number 2 : the voltage source will not give current but Vb-Va = V of the voltage source : Correct
* number 3 : no current will flow: Incorrect, there will be current and the voltage will be infinite across the current source
* number 4 : the voltage source will be considered as a wire Va-Vb=0: Incorrect, voltage is V and the current flowing is infinite

I like these types of questions, it test your theoretical understanding of physical problems. With infinite solutions, we know that these are not practically realizable, but I think it is still important to understand the mathematical theory all the way up to limiting cases.
 
  • #46
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Case #4 as *ideal* is akin to division by zero. Case #4 as *semi-real* (i.e., an ideal wire from a & b to the battery, but a real wire for the rest of the circuit) is akin to the ideal in the form of there being a resistor of very, very low resistance, and hence Va & Vb would have a voltage difference equal to that of the battery.
 
  • #47
In drawing #4, unless you specify resistance of wire and internal resistance of voltage source, both are assumed to be zero. That gives infinite current for any voltage greater than zero.
...in the theoretical sense. In the practical sense, no wire has zero resistance, and if the wire is the limiting resistance in the circuit, it will likely get hot enough to melt and open the circuit, at which point current falls to zero. That is how a fuse works.
 
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  • #48
I still have no idea what this thread is about.
Not want to be rude, or something, but there is importance of formatting and proper punctuation.
this is ambiguous, there were lots of messages posted here,so i don't know exactly where your problem is
 
  • #49
...in the theoretical sense. In the practical sense, no wire has zero resistance, and if the wire is the limiting resistance in the circuit, it will likely get hot enough to melt and open the circuit, at which point current falls to zero. That is how a fuse works.
In the special case where the wire is a superconductor having zero resistance, the voltage across the source is clamped to zero, and it becomes a current source. Internal resistance of the source will cause its temperature to rise, which could cause it to self destruct, burn out, or explode.
 

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