Will any charge flow through the circuit?

[donaldparida]

https://drive.google.com/open?id=0B3IenBvKWb4PZW9rRWg4cXptT2c
I had to find the resistance of this circuit between A and B. My teacher said that since the blank wires have zero resistance, the potential difference across the ends of the blank wire is zero(according to v = ir) and thus 1,4 are equipotent and 2,5 are equipotent and the circuit is thus simplified.The equivalent resistance comes out to be 2r/5.

MY QUESTION:Will charge flow only through the circuit having resistance or will it flow through the circuit having no resistance or will it flow though both the branches.

MY REASONING:Actually i am confused.
One way in which i can reason is:the potential difference across each branch of a parallel circuit is same and in this case it is 0 and hence no current can flow thorough 1-4 and 2-5.
Second way i can reason is that current always chooses the least resistant path and hence all the current will flow through 1-4 and 2-5.

 

[vanhees71]

Somehow the figure got lost, and this kind of questions should be posted to the homework forum!

 

[donaldparida]

@vanhees71 should i shift this to homework forum?

 

[cnh1995]

@vanhees71 should i shift this to homework forum?

It would be better if you did. Also, please upload the image in the thread itself instead of providing its link.

Will charge flow only through the circuit having resistance or will it flow through the circuit having no resistance or will it flow though both the branches.

Look up the terms 'open circuit' and 'short circuit'. In an open circuit, voltage is (or can be) nonzero but current is zero while in a short circuit, voltage is zero but current is (or can be) non zero. Ideal wires are kind of a short circuit.

 

[donaldparida]

Homework Statement

https://drive.google.com/open?id=0B3IenBvKWb4PZW9rRWg4cXptT2c

I had to find the resistance of this circuit between A and B. My teacher said that since the blank wires have zero resistance, the potential difference across the ends of the blank wire is zero(according to v = ir) and thus 1,4 are equipotent and 2,5 are equipotent and the circuit is thus simplified.The equivalent resistance comes out to be 2r/5.
Will charge flow only through the circuit having resistance or will it flow through the circuit having no resistance or will it flow though both the branches.

2. Homework Equations
V=IR

The Attempt at a Solution

Actually i am confused.
One way in which i can reason is:the potential difference across each branch of a parallel circuit is same and in this case it is 0 and hence no current can flow thorough 1-4 and 2-5.
Second way i can reason is that current always chooses the least resistant path and hence all the current will flow through 1-4 and 2-5.

 

[BvU]

The drawing is intentionally misleading. Manipulate it a little bit to make it clearer:

1) Put 2-4 upright by pulling 2 down and 4 up. You get a diamond shape with 1-2 on the lower left and 4-5 on the upper right; 2-3-4 is the vertical axis.

2) shorten 1-4 and 2-5 until grokking is (*)

(*) insight breaks through -- borrowed from Heinlein

 

[cnh1995]

Second way i can reason is that current always chooses the least resistant path and hence all the current will flow through 1-4 and 2-5.

Instead of doing such reasoning, I'd siggest you redraw the circuit in such a way that you could easily simplify it. Replace the equipotentials with a single node.

Edit: I see BvU suggested the same while I was typing...

 

[Merlin3189]

:Will charge flow only through the circuit having resistance or will it flow through the circuit having no resistance or will it flow though both the branches.

I don't know the answer to that, though I understand that very large currents can flow in superconductors, which people seem to say have zero resistance.
I think the "effort" with superconductors goes into changing the current. If there is no current, you'll need something, like a PD, to start current flowing. Once you have a current, no PD is required to maintain it. But another emf is required to reduce the current again.

In fact, for your circuit the wire links 1-4 and 2-5 will have some resistance (supposedly very small), so you can assume current flows through them in any case.
If you assumed a very small resistance, say r/10^6, you could work out the more complex circuit and get the right answer to 5sf.
I think we just assume connecting wires are zero resistance to simplify the calculation for all practical purposes. And then we allow it to conduct any current without any PD (even though, as with a superconductor, it would need a transient PD to initiate current flow.).

One way in which i can reason is:the potential difference across each branch of a parallel circuit is same and in this case it is 0 and hence no current can flow thorough 1-4 and 2-5.

I don't understand this. I agree about equal PD across parallel circuits, but see no reason why you then assume it is zero here nor why you assume no current flows.

Second way i can reason is that current always chooses the least resistant path and hence all the current will flow through 1-4 and 2-5.

Well your premise is simply wrong. Current does not choose the path of least resistance nor any path. Current flows along any and all paths available to it.

I think the notion of current flow along the path of least resistance may come from currents in insulators? In an ideal insulator no current can flow anywhere. But if the field strength is great enough, you get dielectric breakdown and the insulator can conduct.
Since field strength is PD/distance (volts/metre) then a homogeneous insulator breaks down along the shortest path between the sources of the PD. If there are variations in dielectric strength, then some parts will breakdown sooner (at lower field strength) and, by becoming conductive allow the field to increase elsewhere causing breakdown in higher dielectric strength areas. This can lead to a very irregular shaped current path - the path of least aggregate dielectric strength?

 

[rude man]

One way in which i can reason is:the potential difference across each branch of a parallel circuit is same and in this case it is 0 and hence no current can flow thorough 1-4 and 2-5.
Second way i can reason is that current always chooses the least resistant path and hence all the current will flow through 1-4 and 2-5.

The potential across an equipotential (like an ideal wire) is zero but so is its resistance, giving i = V/R = 0/0 which is indeterminate so you can't associate zero potential with zero current.

 

[gneill]

Second way i can reason is that current always chooses the least resistant path and hence all the current will flow through 1-4 and 2-5.

This is actually a bit of folklore that is not true. It's based upon lay observations of how water flows through terrain, or how lightning seems to find a single shortest route to ground once it strikes. Neither is strictly true, and the false generalization is based on observing the end result of a "carved path" left in the wake of the largest current.

In a circuit, current will take every path available to it provided there is a potential difference to drive it. Parallel paths are no problem, and the current will divide amongst the available paths according to the relative conductivities of the paths. That is to say, while a lower resistance path will take more current than a higher resistance one, every path still conducts some current. The exception to this is when one of the parallel paths happens to be a perfect conductor (such as an ideal wire in a circuit diagram). Then the zero resistance path will clamp the potential difference across all the paths to zero. There being no potential difference across the other paths, there will be no potential to drive current through those path's components, while the zero resistance path allows all the current to flow through itself unimpeded.

 

[donaldparida]

@BvU could you please provide a diagram. I am finding it hard to imagine the changes.

 

[phinds]

@BvU could you please provide a diagram. I am finding it hard to imagine the changes.

No, this is EXACTLY what the problem is supposed to teach you --- how to change circuit diagrams so as to show clearly what is serial and what is parallel. If we redraw the diagram we are spoon feeding you the answer to this question and that is against forum rules.

And by the way, this specific example is a classic for teaching circuit redrawing. We've seen it here before several times. You need to figure this one out on your own now.

 

[cnh1995]

@BvU could you please provide a diagram. I am finding it hard to imagine the changes.

Well, go step-by-step.
Name the resistor terminals as a,b,c and so on and connect the equipotential terminals together.
Edit: I see the terminals are already numbered. Nvm, you can use the numbers directly.

 

[vanhees71]

Just to let you know, I can't do anything of this, and I also don't see your figures or links.

 

[davenn]

@vanhees71 should i shift this to homework forum?

start a new thread in homework, use the homework template and upload any images to the thread (as one other person said)

I will ask a mentor to delete this thread once you have that sorted out Dave

 

[berkeman]

Two threads merged after the Mentors struggled with a server issue all day. Yay!

 

[donaldparida]

@BvU @phinds , I did it. After struggling for about half an hour i was able to imagine pulling and reshaping the circuit into the diamond shaped structure. I also discovered another approach in which the nodes are taken at convenient positions and then the resistors between the nodes are drawn by tallying it with the original circuit. What do you think of this?

 

[cnh1995]

Did you get the answer as 2r/5? Or do you need some more help?

What do you think of this?

Is there any circuit diagram you wanted to show us with this post? Because I can't see any image attached.

 

[donaldparida]

@cnh1995 , No. I am not referring to any circuit. I want to know what others think of the second approach which i used.

I also discovered another approach in which the nodes are taken at convenient positions and then the resistors between the nodes are drawn by tallying it with the original circuit. What do you think of this?

By the way i think that this approach is more convenient than pulling the nodes and reshaping the circuit.

 

[ehild]

Homework Statement

https://drive.google.com/open?id=0B3IenBvKWb4PZW9rRWg4cXptT2c
View attachment 110473

I had to find the resistance of this circuit between A and B. My teacher said that since the blank wires have zero resistance, the potential difference across the ends of the blank wire is zero(according to v = ir) and thus 1,4 are equipotent and 2,5 are equipotent and the circuit is thus simplified.The equivalent resistance comes out to be 2r/5.
Will charge flow only through the circuit having resistance or will it flow through the circuit having no resistance or will it flow though both the branches.

As the currents have to reach the resistors, current should flow through the wires. The ends of the wires are at the same potential, but their resistance is zero, so any current can flow resulting zero potential across them. Simplify the circuit, determine the current (magnitude and direction) through each resistor, and determine the current flowing through the wires in the original set-up using Kirchhoff's nodal law.

 

[phinds]

@BvU @phinds , I did it. After struggling for about half an hour i was able to imagine pulling and reshaping the circuit into the diamond shaped structure. I also discovered another approach in which the nodes are taken at convenient positions and then the resistors between the nodes are drawn by tallying it with the original circuit. What do you think of this?

I think you are likely over-thinking it and making it harder than it is. Perhaps this will help you see where it is best to go with such problems. Once you get used to following the lines from A to B, this sort of thing becomes trivial.

 

[donaldparida]

@gneill , i am facing one problem. Due to the branch with zero resistance, the net resistance between the two branches becomes zero [(r1r2)/(r1+r2)]. Then how can we calculate the equivalent resistance between A and B since it should also be 0 then. Shouldn't it be?

 

[phinds]

@gneill , i am facing one problem. Due to the branch with zero resistance, the net resistance between the two branches becomes zero [(r1r2)/(r1+r2)]. Then how can we calculate the equivalent resistance between A and B since it should also be 0 then. Shouldn't it be?

Did you not believe the validity of the diagram I posted? If not, please show where you think it is wrong.

 

[gneill]

@gneill , i am facing one problem. Due to the branch with zero resistance, the net resistance between the two branches becomes zero [(r1r2)/(r1+r2)]. Then how can we calculate the equivalent resistance between A and B since it should also be 0 then. Shouldn't it be?

Can you draw a path from A to B that does not pass through at east one resistor?

 

[donaldparida]

@phinds The zero resistance branch has has same potential across its ends( zero potential difference across it) and is a short circuit. When you are representing the equi-potent points by a single point you are ignoring the short circuit, which has zero resistance (through which the current should flow) and which should make the equivalent resistance of the parallel branches zero and that's where i have a problem.

 

[phinds]

@phinds The zero resistance branch has has same potential across its ends( zero potential difference across it) and is a short circuit. When you are representing the equi-potent points by a single point you are ignoring the short circuit, which has zero resistance (through which the current should flow) and which should make the equivalent resistance of the parallel branches zero and that's where i have a problem.

ONE MORE TIME ... show me what is wrong, EXACTLY with the circuit I drew and do with with a drawing. There IS no short circuit from A to B. If you think there is, show it.

Seriously, you are making this a whole lot harder than it is. Just follow the paths in my diagram and mark them with a red pen or something on the original and you'll see where your mistake is.

 

[cnh1995]

When you are representing the equi-potent points by a single point you are ignoring the short circuit, which has zero resistance (through which the current should flow) and which should make the equivalent resistance of the parallel branches zero

When the equipotential points are represented by a single point, it means those points are "connected" together. A conducting wire in a circuit can be represented by a single point or 'node'. So the long wire joining 2 and B is just a node in the circuit.
Also, it is not shorting out the intermediate resistors. Yes, V_2B=0V, but that doesn't short out the resistors.
V_2B=V₂₃+V₃₄+V₄₅.
Now V_2B=0 doesn't mean the RHS voltages are zero. There will be current through those resistors.

 

[donaldparida]

@phinds
[NOPARSE]https://drive.google.com/open?id=0B3IenBvKWb4PZEI1VFVmMkxVbHE2UFRXeEJLQ3lydW5oSXRV [/NOPARSE]

 

[phinds]

@phinds
[PLAIN]https://drive.google.com/open?id=0B3IenBvKWb4PZEI1VFVmMkxVbHE2UFRXeEJLQ3lydW5oSXRV[/QUOTE]
Sorry, my browser is not showing your image, just the "broken image" symbol

 

[donaldparida]

@phinds https://drive.google.com/open?id=0B3IenBvKWb4PZEI1VFVmMkxVbHE2UFRXeEJLQ3lydW5oSXRV

 

[gneill]

Your "short circuits" are simply wires connecting components. There's nothing special about these wire connections. Neither of them are in parallel with any components in the strict definition of "parallel connection". Geometric layout on a drawing does not make things parallel; It's the connection topology that counts.

Contiguous wire pathways on a circuit diagram comprise nodes. It's the connections of components between nodes that define the circuit topology. You can stretch or bend or otherwise contort the drawing of these nodes in any fashion you like so long as you don't change what connects to them (components). You can move any connection to a given node to any location on the same node without changing the circuit topology.

You need to practice "morphing" circuit diagrams by bending or stretching node wires, or "sliding" connections along a node to make the circuit more easy to comprehend. With a bit of experience this will become much easier, and you'll wonder what all the fuss was about.

You should be able to follow three distinct paths from A to B on your original diagram. By distinct I mean that they each pass uniquely through a subset of the components, never passing through the same component twice or passing though a component that is covered by another path. Wires do not count as components: you are allowed to use the same wire for multiple paths. Those three distinct paths will be three parallel branches (in the strict definition of parallel connection) that connect node A to node B.

 

[phinds]

You don't seem to get that 1 and 3 are the same thing and then 4 and 2 are the same thing. You did not do what I suggested and follow ALL the lines and figure out which of the original are which in the modified circuit.

 

[vanhees71]

Yes, I also don't see a mistake in @phinds 's redrawing of the circuit. It's just the topology that matters here!

 

[BvU]

Was my post #6 totally in vain ?

 

[phinds]

Was my post #6 totally in vain ?

Both of our efforts were long in vain. Maybe he gets it now.