Why, for a circuit with B branches and N nodes, the number of the linearly independent KVL equations for a network is: B-N+1 ?


1 Answer 1


The linearly independent KVLs apply to each "unique loop" (as opposed just to "loop.") I think the following paper will help a lot here:

Network Topology

It's actually not too hard to follow. When looking around for the above paper to post, I was searching for graph theory terms of 'vertices' and 'edges,' which led me a bit far afield at first. But this paper looks good to me.

If you just think of the branches alone, which connect two nodes, then there are a maximum of \$2\cdot B\$ unique nodes. Of course, none of them are connected to each other in this case. But since it is required that every node in a circuit be shared by at least two branches, there must be no more than half that, or \$N\le \left(\frac{2\cdot B}{2}=B\right)\$ nodes. So \$N \le B\$. If N is exactly equal to B, then each node connects exactly two branches and there is only one loop.

If you instead think of this single series loop as instead being a chain where the start and end aren't actually connected, then you can imagine disconnecting one of the branches from the end without changing the number of loops: still 1. Now take this removed branch and select any node on the chain to attach one end and any other node on the chain for the other end. You have created one more unique loop in doing so. So the number of loops went from 1 to 2, but to do that you removed a branch and then re-added it no longer in series at the end but somewhere else.

If \$N= B\$, the number of loops is \$B-N+1=1\$. But if you now remove one branch and re-add it elsewhere as mentioned above, you haven't changed \$B\$ but you did reduce the node count \$N\$ by 1 (the tail node of that tail end branch was uniquely numbered before, but will now be attached to a previously numbered node, reducing the count of nodes by 1.) So now you have 2 unique loops and again this is \$B-N+1\$, or now 2. And so on.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.