# Defining a Unique Voltage

Discussion in 'Physics' started by RdAdr, Jul 28, 2015.

May 19, 2013
214
1
The book "Foundations of Analog and Digital Electronic Circuits" by Anant Agarwal says the following things regarding the defining of an unique voltage between two points (a voltage that doesnt depend on the path taken).

First, it defines the voltage as being the line integral going from x to y of the electric field.
Then it presents the Faraday's law of induction: the line integral over a closed path of the electric field is minus the rate of change of the magnetic flux through a surface delimited by the closed path.

Until now all is clear.

Then the author says a thing which I put in the picture attached. This I dont understand.
(ps: he says that it doesnt have useful meaning because in circuit theory the voltage must be uniquely defined between the two terminals of an electric element).
What does he try to say by choosing x and y to be the same? Like x=y? And what would be the surface in this case?

Then the author tries to define a unique voltage and I understand that to have a unique voltage between two points (so that the voltage doesnt depend on the path taken from x to y), the magnetic field must be constant so that the closed loop line integral is zero and not equal to minus the rate of change of the magnetic flux. I understand this. Because if it's zero then I can say that the line integral going from x to y + the line integral going from y to x is 0.

But I really dont understand what he tries to say in the picture attached.

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2. ### WBahn Moderator

Mar 31, 2012
20,057
5,644
All it is saying is that the path integral of the electric field around any closed path must be zero in order for there to be a unique potential difference between two points in the field. This means that the electric field under consideration is "conservative". If you have a changing magnetic field then the electric field is "non-conservative" and this condition doesn't hold and you can't define a unique potential difference between two points in such a field, since the potential difference is now path dependent.

May 19, 2013
214
1
I understood that. But if we choose the two points to be the same, will not that integral evaluate to 0?

For example:

from 1 to 1 of xdx = 0

Isn't the same for the line integral? If I evaluate the line integral from one point to the same point, would I not obtain a zero value? And not a nonzero value?

LE:
Ok, I think I got it. First he defines the voltage as a line integral over an open path. Then he makes the two points the same and by doing so he obtains a closed path.
From the net: "A path C is called closed if its initial and final points are the same point. For example a circle is a closed path."

So, when he says "if we choose the two points x and y to be the same" he is not saying x = y in the sense that the length of the path is 0, but in the sense of a closed path and of length different than 0.

My confusion arose from the fact that I considered this path having length 0.

Last edited: Jul 29, 2015
4. ### WBahn Moderator

Mar 31, 2012
20,057
5,644
Not if the field is non-conservative. That is the practical definition of a non-conservative field -- namely that the integral of the field is path dependent. As soon as that is the case, then the path integral along a closed path need no longer be either zero or unique.