How will current flow in super conducting wire in this circuit ?

Question

Imagine this if we connect a super conducting wire (with zero resistance) across two terminals of an ideal battery(again with no resistance) and consider Potential Diffrence across points A & B .(see diagram below)
there will be no potential drop from A to B. Then, Why will the current flow from A to B points considering the fact that current flows from higher to lower potential point.

(+) ------>------A-----------B------->----- (-)

Diagram

Thanks Larry for ur reply but u didnt get my point I am asking why current flows from point A to B as potential at point A is same as point B.

Answer 1

You're confusing resistance with pressure. If indeed there is a difference in voltage between A and B then there will be flow.
Think of a tank filled with water held a few feet above another tank that has a pipe joining the two. Water will flow to the lower container thru the pipe. In this case the pipe has no resistance, which means water flows at maximum possible speed.
If both tanks are at the same level, then, sure, there will be no water flow. That's not how the problem was stated, however.

Answer 2

My electronics teacher described the ideal voltage source as, "when you short it out, the Sun dims". In other words the current would be infinite. Basically an ideal battery is a perfect voltage source. A zero resistance wire is an ideal short. In reality you cannot have either thing, there is always some small resistance, and the smaller it is, the higher the current is. In theory, if you say resistance is actually zero, you end up dividing by zero to try and calculate the current. V=IR rearranges to I=V/R. V is the voltage of the battery, and R the total series resistance of the wire and battery. In your theoretical example, R is 0, so you are dividing by 0. To try and bring some reality to the example, just imagine that the battery would discharge immediately and it's terminal voltage would go to zero. If the resistance really was 0, it would discharge in 0 seconds. The result would be that as soon as the perfect wire shorted the perfect battery, it would instantly change to a perfectly dead battery, which of course has no voltage and makes no current flow. This is all kind of like asking how many angels could dance on the head of a pin, don't you think?

Answer 3

The water theory is correct. The current will flow as long as one potential is different from the other. When the current flows it actually is reducing the potential at the terminal it came from, so given sufficient time (from nanoseconds to hours) eventually the two terminals will be at the same potential. So as long as there is current flow, there will be a potential difference. When you make the statement that they are at the same potential (because they are connected together through a conductor) you are describing its steady state condition.

Answer 4

In super conductors we dont use battery to induce current.

This web site may be useful to understand the conduction in semi conductor.

When R = 0, E=0. It is not necessary I=0.

Extract from the web site.
An interesting aspect of the phenomenon is the continued flow of current in a superconducting circuit after the source of current has been shut off: for example, if a lead ring is immersed in liquid helium, an electric current that is induced magnetically will continue to flow after the removal of the magnetic field. Powerful electromagnets, which, once energized, retain magnetism virtually indefinitely, have been developed using several superconductors. The 1972 Nobel Prize in Physics was awarded to J. Bardeen, L. Cooper, and S. Schrieffer for their theory (known as the BCS theory) of classical superconductors. This quantum-mechanical theory proposes that at very low temperatures electrons in an electric current move in pairs. Such pairing enables them to move through a crystal lattice without having their motion disrupted by collisions with the lattice. Several theories of high-temperature superconductors have been proposed, but none has been experimentally confirmed.

Answer 5

right here's a suitable information: assume that the voltage of the cellular is U. Then the present contained in the finished device is, by Ohm's regulation, i = U/(R+r). The voltage U is sent over both resistances contained in the ratio of the resistances; for this reason the voltage over R is u = U*R/(R+r). the potential presented to R for this reason is P = u i = U²R/(R+r)² to locate the optimal, we derive this and set the end result equivalent to 0; to make it a lot less difficult, we rewrite it as ... = U²(R+r-r)/(R+r)² = U² [ a million/(R+r) - r/(R+r)² ] P' = U² * [ -a million/(R+r)² + 2r/(R+r)³ ] = 0 this would in straight forward words be 0 if -a million/(R+r)² + 2r/(R+r)³ = 0 ... multiply with -(R+r)³ [ok as R, r > 0] -(R+r) + 2r = 0 -R-r+2r = 0 -R+r = 0 R = r That it is a optimal would properly be shown by doing the 2d spinoff; or by consumer-friendly experience: If R = 0 or R = countless, the potential presented to R is for sure 0. So the "in-between fee" might want to be a optimal. right here's yet another information: assume that R isn't equivalent to r, yet fairly smaller or more desirable. enable the version be d, so as that R = r + d. Then the present is i = U/(2r+d); and the voltage over R is u = U*(r+d)/(2r+d). the potential, hence, is U²*(r+d)/(2r+d)². We subtract this now from a similar fee the position d = 0: U²*(r)/(2r)² - U²*(r+d)/(2r+d)² = U² * (r(2r+d)² - (r+d)*(2r)²)/ (2r)²(2r+d)² = U² * (r(4r²+4rd+d²) - (r+d)*4r²)/ (2r)²(2r+d)² = U² * (4r³+4r²d+rd² - 4r³-4dr²)/ (2r)²(2r+d)² = U² * (rd²)/ (2r)²(2r+d)² = U² * d²/ 4r(2r+d)² Now, it really is continually an excellent type: hence, the potential for d=0 is continually more desirable than the single the position d > 0. So the potential optimal is at R = r.

Answer 6

Your connected to a battery so current will flow from negative to positive (differenc of potential), A and B are points along your conductor and thus current will flow.