Kirchoff's Law

1. Jun 6, 2004

brentd49

I'm having a problem with Kirchoff's law because It seems it can be interpreted 2 ways. Perhaps they both arrive at the same answer, but if you could help me out as to which way to do it or if it matters.
1. the Integral[E*dl]=0, in a closed loop
2. the algebraic sum of the voltages in a closed loop equals zero.

The difference arrises in the sign. For instance, for the Integral[E*dl] going across a voltage source is -V because you are going against the electric field. But for the algebraic sum of voltages you are going to a higher potential so it is +V. Similarly for resitors the electric field (Integral[E*dl]) gives a +V, but for algebraic sum of voltages it is -V because there is a loss in potential. I don't understand I have never heard a explanation for this in my EE circuits class or physics class. What do you think?

2. Jun 6, 2004

baffledMatt

It doesn't matter because in both case the answer is zero. If you want you could write instead:
$$-\oint \underline E . \underline dl = 0$$
which is exactly the same thing and corresponds to the second form you have there.

Matt

3. Jun 6, 2004

turin

I concur with baffledMatt.

I do want to mention, though, that the first form that you have mentioned will lead to a non-trivial violation for high speed circuits. There are various generalizations that will allow it to show up in the second form, but it is better explained by a generalization of the first (called Faraday's Law). If you are an EE undergrad, you may never run into it. I think they give you the criterion for being able to use the circuit theory that you are learning as applying to circuits smaller than about a tenth of a wavelength. This takes care of the low speed approximation.