This seems more like a proof about induction than a proof by induction.

But what is this proof really about?

That is, what circuit with an inductance of 4 henries between A and B does that blob with a ? question mark represent?   What do the lemma and steps 1 through 4 mean?

In a sense this is a riddle. The answer is not just any old 4-henry circuit.


I think the configuration we are looking at is

a Sierpinski-gasket-like thing. Call it $T$. Then $T$ consists of three half-size copies of $T$ (I mean actually literally shrunk, so in particular all the inductors are half the physical size -- so if they are coils of wire, they will have half the inductance because inductance scales like area/length) connected together via three 1-henry inductors in the "obvious" places. And then the question is: what is the inductance between two corners of $T$?

Here, courtesy of the OP, is a diagram showing how this works:


being a three-terminal circuit built out of impedances, $T$ is equivalent to some other circuit where each of the three terminals is connected via a two-terminal blob $U$ to a central node. $U$ is the same for all three of them, by symmetry. If the corner-to-corner inductance of $T$ is $L$ then the inductance of $U$ must be $L/2$. (This is essentially the "Lemma".) So the inductance of $U$ in a scaled-down copy of $T$ is, because of the scaling, half of this, or $L/4$.

So diagram 1 is obtained by

taking $T$ (with corner-to-corner inductance $L$), exposing its three half-size $T$s (each with corner-to-corner inductance $L/2$), and turning each of them into three half-size $U$s (each with inductance $L/4$).

Then diagram 2

just adds up the inductances along the resulting "inner" paths. Diagram 3 does the usual star-delta thing to give us an equivalent circuit that's easier to work with. Diagram 4 then works out what the corner-to-corner inductance of the resulting thing has to be. Equating this to $L$ gives us the equations at bottom right, and hence enables us to determine the value of $L$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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