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I have come across a strange derivation of Kirchhoff's voltage law. You assume that Tellegen's theorem and Kirchhoff's current law both hold. (Bold letters signify a vector.)
The proof:
Let L be an arbitrary loop in the graph G. Consider the i obtained by assigning zero current to all branches of G except for those of loop L; depending on whether the reference direction of branch j in loop L agrees with that of loop L, we assign each branch current to be either 1A or -1A. The resulting i satisfies KCL at all nodes of G. Tellegen's theorem gives: $$ \sum_{j=1}^{b}v_{j}i_{j}= \sum_{overBranchesInLoopL}\pm v_{j}=0 $$ thus the algebraic sum of the branch voltages around loop L is zero, i.e., KVL holds for loop L. Since L is arbitrary, we have shown that KVL holds for all loops of G.*

The Problem I have with this proof is that he just chooses that the current is zero in all branches, but the branches of L. Can anyone elaborate on why one is allowed to do this?

This proof was obtained from the book Linear and Nonlinear Circuits by Leon O. Chua, Charles A. Desoer, Ernest S. Kuh on page 30.

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At the beginning of the referenced section, the authors state the assumption on which the derivation is based (emphasis mine):

  1. If, for all $\mathbf{i}$ satisfying KCL, $\mathbf{v}^T\mathbf{i} = 0$, then $\mathbf{v}$ satisfies KVL.

The choice of currents that is used in the derivation is just a case of $\mathbf{i}$ satisfying KCL: the currents in the branches external to the loop don't matter because the assumption is that $\mathbf{v}^T\mathbf{i} = 0$ holds true for any choice of currents satisfying KCL.

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  • $\begingroup$ But let us say we have a vector i satisfying Tellegen's theorem. Why can we then assign all the currents to be the same? i.e. 1 or -1. In general a circuit may have different currents flowing in different branches. $\endgroup$ Commented Jul 20, 2017 at 19:34
  • $\begingroup$ Correct me if I am wrong, but I think we can do this as Tellegen's theorem is valid for all possible i and v (vectors). Here, i belongs to the subspace satisfying KCL. Now take a special vector as described. You get that any v would belong to a subspace satisfying exactly the condition of KVL. $\endgroup$ Commented Sep 30, 2018 at 8:12
  • $\begingroup$ @AlphaRomeo The Tellegen's theorem is not valid for all possible $\mathbb{i}$ and $\mathbb{v}$, but it is valid only if the vectors are associated to the same network digraph and satisfy KCL and KVL. The assumption given assumes that the vectors $\mathbb{i}$ and $\mathbb{v}$ are associated to a certain network graph. $\endgroup$ Commented Sep 30, 2018 at 19:47

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