How correct are these concepts about circuits?

[mohamed el teir]

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 ?

 

[phinds]

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.

 

[phinds]

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.

 

[mohamed el teir]

@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

 

[mohamed el teir]

let me rephrase my point in a way acceptable by all, for this simplest circuit :

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

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.

 

[Dale]

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.

 

[Dale]

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.

 

[Perq]

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.

 

[Steve Brown]

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.

 

[lim23472]

* 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.

 

[swampwiz]

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.

 

[Steve Brown]

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.

 

[mohamed el teir]

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

 

[Steve Brown]

...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.