# Geometric frustration in power transmission networks

My question is, why must the cyclic sum of the phase differences add up to a multiple of $2 \pi$ for the phases to be well defined?
• I appriciate the answer, but I fail to see why it has to be a multiple of $2 \pi$ in a cyclic network? Why would a generator short itself out if not? In the article they write the statement as if it was well-known that phases does in fact add up to a $2 \pi$ multiple. – Fabric Jun 22 '16 at 7:21
• Imagine you are connecting an AC current source to one end of a transmission line. That transmission line delays the signal by a certain phase. Now you also connect the other end of the transmission line to the same source. Unless the phase was exactly $2\pi$, the two signals will have different voltages because $\cos(x)= \cos(x+\varphi)$ only if $\varphi = n\times 2\pi$. In practice a generator will short itself out, if such a loop differs in its electrical phase shift by anything that's not close to $2\pi$ (and multiples). – CuriousOne Jun 22 '16 at 18:21