Sooner or later a network arrives that will not reduce. Look at a bridge: five resistors, and no two of them are cleanly in series or cleanly in parallel. Series and parallel have run out, and this transform is what you reach for next.

Three resistors can be wired between three terminals in exactly two ways. A delta (a triangle, also called a pi) puts one resistor between each PAIR of nodes: $R_{ab}$ between A and B, $R_{bc}$ between B and C, $R_{ca}$ between C and A, all in ohms — the double subscript names the two nodes the leg spans. A wye (a star, also called a tee) runs one resistor from each node to a common centre: $R_A$, $R_B$, $R_C$, also in ohms, and here the single subscript names the one node that arm touches. Learn those two conventions and the formulas stop looking like alphabet soup.

Delta to wye: $R_A = \dfrac{R_{ab} R_{ca}}{R_{ab} + R_{bc} + R_{ca}}$ — R-A equals R-ab R-ca, over the sum of all three legs. In words, and this is the version to remember: the arm at a node is the product of the two legs that MEET at that node, divided by the sum of all three. Going back the other way, $R_{ab} = \dfrac{R_A R_B + R_B R_C + R_C R_A}{R_C}$: the leg between two nodes is the sum of all three pairwise products, divided by the arm at the node the leg does NOT touch. Both transforms preserve exactly what the network looks like from outside the three terminals, and nothing else.

Two rails keep you honest. Going delta to wye the values SHRINK — every arm comes out below both of the legs that meet at it. Going wye to delta they GROW — every leg comes out above the two arms it spans, added together. And for the balanced case the whole thing collapses to $R_Y = R_\Delta/3$, which is worth knowing and worth distrusting: it is true only when all three are equal, and the exam knows you know it.