1
\$\begingroup\$

enter image description here

The derivation of the delta-star transformation rules is based on the fact that the resistance between terminals 1,3 is R1+R3 I can't see how we can consider R1,R3 just two resistances connected in series between terminals 1 and 3 with R2 having no significance in this arrangement. Can someone enlighten on this?

\$\endgroup\$
2
\$\begingroup\$

There are different ways to derive the delta-star transformation equations. In any case, you must ensure that the two circuits are equivalent, i.e. they don't alter the behavior of "the rest of the world" (whatever is connected to terminals 1,2,3).

In the derivation you refer to, you start by getting some equations that must be true if the two circuits are to be equivalent. Once you get enough independent equations, you can isolate the parameter of interests (either \$[R_1,R_2,R_3]\$ or \$[R_{12},R_{23},R_{31}]\$) using usual algebraic methods.

If you understand what I just told you, you should understand that, if you keep terminal 2 disconnected in both circuits, the resistance that you see between terminal 1 and 3 must be the same (otherwise the delta and the star circuit cannot be equivalent).

Hence \$ R_1 + R_3 = R_{31} \parallel (R_{12} + R_{23})\$. This gets you an equation. Repeating the same reasoning with terminals 1 and 3 open, gets you another two similar (but independent) equations. With a total of three independent equations you can isolate any 3 variables, hence you can express the parameters of the delta circuit using those of the star circuit or vice-versa.

If you don't understand why the presence of \$R_2\$ is irrelevant when terminal 2 is disconnected, keep in mind that in this case there is no current flowing in \$R_2\$. Hence the KCL equation at the junction between \$R_1\$, \$R_2\$, and \$R_3\$ is the same as if \$R_2\$ weren't there! Therefore, even if \$R_1\$ and \$R_3\$ are not topologically in series, they behave as if they were in series, because the equations you would write with or without \$R_2\$ would be equivalent.

\$\endgroup\$

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.