Finding voltage and current in a somewhat unusual circuit

[gerbi]

If there is a wire with 1 micro-ohm of resistance, would you be able to solve the problem?

Suppose you then allow the resistance of the wire to become less and less until it becomes zero ohms as if it were a superconducting wire. Wouldn't there still be a current in that wire?

Go back and look at what I said in post #26,

There is no resistance, no voltage drop.. means no current. Unless it could be a superconductor.. but again, it's just a theory.

Ok, let's end this. Everyone has own rights. The essence here is interpretation.

 

[The Electrician]

There is no resistance, no voltage drop.. means no current.

You have a serious misconception if you think that when there is no voltage drop there can't be a current. If you are still a student, you should discuss this with your instructor.

 

[gerbi]

You have a serious misconception if you think that when there is no voltage drop there can't be a current. If you are still a student, you should discuss this with your instructor.

read my post again friend... I guess I'm not a student for a long time

 

[Bassalisk]

[PLAIN]http://pokit.org/get/3cf3c9bf3d978f78e44f8bbcf9d435b3.jpg

Ok, this is almost the same case. I short circuited a resistor. So the current is 0 here too? You have those 2 nodes like in the problem before. That means that no current will flow in the circuit?

We all know that it will.

 

[gerbi]

[PLAIN]http://pokit.org/get/3cf3c9bf3d978f78e44f8bbcf9d435b3.jpg

Ok, this is almost the same case. I short circuited a resistor. So the current is 0 here too? That means that no current will flow in the circuit?

We all know that it will.

I'm sorry.. the problem stated at the beginnign of this thread is static or transient ? Don't mix it please because You're right. It's alomst the same.

 

[Bassalisk]

Well I have to drift of the problem, because you are going into the theoretical stuff of the circuits, where not everything is perfect. Guy asked for a current through the wire, a problem that students do at the beginning of their journey of learning.

You went in with that no current will flow stuff and probably confused a guy. Current will flow through that wire. That is an experimental fact. This is not a question of theory, this is a question of exercise

 

[Studiot]

Bassalisk I am proud of you!

Excellent analytical presentations.

I particularly like the observation

That is not the same, problem ASKS to calculate the current through WIRE.

 

[gerbi]

Well I have to drift of the problem, because you are going into the theoretical stuff of the circuits, where not everything is perfect. Guy asked for a current through the wire, a problem that students do at the beginning of their journey of learning.

You went in with that no current will flow stuff and probably confused a guy. Current will flow through that wire. That is an experimental fact. This is not a question of theory, this is a question of exercise

Know what ? I won't answer that entirely because It would be a really long discussion. Stay at your opinion and I will stay at mine.

Anyway.. studies are studies. So students.. listen to Your professors and do not try to be smarter than tey are because that often ends in pain ;)

 

[Bassalisk]

Bassalisk I am proud of you!

Excellent analytical presentations.

I particularly like the observation

:) Thank you.

 

[Bassalisk]

Know what ? I won't answer that entirely because It would be a really long discussion. Stay at your opinion and I will stay at mine.

Anyway.. studies are studies. So students.. listen to Your professors and do not try to be smarter than tey are because that often ends in pain ;)

I am not saying you are completely wrong. Just, the problem you are stating is for a different thread.

 

[OkuuEE]

(Meh typo in picture R₅ should be R₄)

Question was how current flows when there is no voltage drop?

I try to answer by using this picture. Because of KVL U₁=U₂ and U₃=U₄ and because R₁ != R₂ current is forced divide unbalancly(0.6A and 1.4A in this case). I think this part is very clear.

If there wouldn't be current in middle node, same 0.6A and 1.4A currents would flow trought R₃ and R₄. But that would broke ohm's law because then R₃*0.6A != R₄*1.4A. Current is forced to flow trought node because it is needed to make balance between voltages.

I made another picture to demonstrate idea:

Should I₂ be zero because there is no voltage drop?

 

[gerbi]

(Meh typo in picture R₅ should be R₄)

Question was how current flows when there is no voltage drop?

I try to answer by using this picture. Because of KVL U₁=U₂ and U₃=U₄ and because R₁ != R₂ current is forced divide unbalancly(0.6A and 1.4A in this case). I think this part is very clear.

If there wouldn't be current in middle node, same 0.6A and 1.4A currents would flow trought R₃ and R₄. But that would broke ohm's law because then R₃*0.6A != R₄*1.4A. Current is forced to flow trought node because it is needed to make balance between voltages.

I made another picture to demonstrate idea:

Should I₂ be zero because there is no voltage drop?

Dear friend..
Like I said, I treat this discussion as closed. I have some opinions about what You're presenting above but I'll keep it for myself.

I'm sorry if my opinions mislead the thread starter. Keep solving circuts like guys here showed. It will keep You safe from troubles.

This all thread is about some asumptions in circuts. You take other than I was taught to. That's all. Can we leave it now ?

 

[OkuuEE]

Dear friend..
Like I said, I treat this discussion as closed. I have some opinions about what You're presenting above but I'll keep it for myself.

I'm sorry if my opinions mislead the thread starter. Keep solving circuts like guys here showed. It will keep You safe from troubles.

This all thread is about some asumptions in circuts. You take other than I was taught to. That's all. Can we leave it now ?

No hard feelings bro. I just wanted to clear things out, not to insult you or anything. Everyone makes mistakes and has brainfarts.

 

[sophiecentaur]

If I can be a bit sloppy with the Maths, the answer to that question is
I = V/R
for the piece of zero resistance wire with zero volts across it.
0/0 can be anything you like if you have no other info but, if you take the limit as R approaches 0, in that circuit, you get a perfectly finite value for the current which is the one that I_am_learning came up with and in a very good, short hand way. You can just treat the four resistor ends as coming together in a node and then apply Kirchoff 1 to show how the current is re-distributed on its way through.
If you had been asked "what is the difference between the currents flowing through one of the series pairs" the answer might have been more obvious (?).

Actually - the argument that no volts and no resistance gives no current could be applied to every piece of 'ideal' wire in every circuit and would imply that no circuit would ever let current pass through it.!

 

[The Electrician]

The circuit is essentially an unbalanced Wheatstone bridge. A nifty technique for solving it is to break the two vertical resistor combinations apart as shown in the attachment.

The top half of the image shows how to break the two halves of the bridge and apply identical 50 volt sources.

The bottom half of the image shows the Thevenin equivalent of each half. Since we have two voltage sources of 100/9 volts and 50/7 volts respectively, with corresponding Thevenin resistances of 140/9 ohms and 30/7 ohms, we can find the current in Rwire by applying a voltage of (100/9-50/7) volts to a resistance of (140/9+30/7+Rwire) ohms and solving for the current.

The current is given by (100/9-50/7)/(140/9+30/7+Rwire) = (250/63)/(1250/63 + R) = 250/(63*R+1250)

We can even get a value for current if Rwire is 1 micro-ohm or even zero. In case Rwire is zero, the current is 1/5 = .2 amp.

 

[Bassalisk]

The circuit is essentially an unbalanced Wheatstone bridge. A nifty technique for solving it is to break the two vertical resistor combinations apart as shown in the attachment.

The top half of the image shows how to break the two halves of the bridge and apply identical 50 volt sources.

The bottom half of the image shows the Thevenin equivalent of each half. Since we have two voltage sources of 100/9 volts and 50/7 volts respectively, with corresponding Thevenin resistances of 140/9 ohms and 30/7 ohms, we can find the current in Rwire by applying a voltage of (100/9-50/7) volts to a resistance of (140/9+30/7+Rwire) ohms and solving for the current.

The current is given by (100/9-50/7)/(140/9+30/7+Rwire) = (250/63)/(1250/63 + R) = 250/(63*R+1250)

We can even get a value for current if Rwire is 1 micro-ohm or even zero. In case Rwire is zero, the current is 1/5 = .2 amp.

Yes I agree, I was going to go after Thevenin but since mate told us that he is beginner i dropped it. But well done!

 

[MATLABdude]

There's an even simpler and more germane illustration of the "current within node" problem. [URL]http://upload.wikimedia.org/wikipedia/commons/4/4f/Current_division_example.svg[/URL]
From the Wikipedia article on current divider:
http://en.wikipedia.org/wiki/Current_divider

When you change the current source for a voltage source, you can see that all the resistors and the source share the same two common nodes. However, it's plain to see that current must move within the two common nodes to supply current to each of the resistors, even when you assume the wire has zero resistance.

As others have said, an ideal node just implies that everything connected to it is at the same voltage / potential, it says nothing about the currents within. Don't be too non-plussed though, 0A was my answer when I first glanced at it, too!