Star-Delta Conversion: the trick for networks that won't simplify
When nothing is in series and nothing is in parallel, you need this.
Sooner or later you meet a network — the bridge circuit is the classic — where no two resistors are cleanly in series or parallel. Everything is tangled. Rather than falling back on simultaneous equations, there's a transformation that reshapes three resistors into an equivalent three, turning the tangle back into something you can collapse the normal way.
Delta → Star
Each star arm = (product of the two delta resistors touching that node)
÷ (sum of all three delta resistors)
R₁ (at A) = (R_AB × R_CA) / (R_AB + R_BC + R_CA)
R₂ (at B) = (R_AB × R_BC) / (R_AB + R_BC + R_CA)
R₃ (at C) = (R_BC × R_CA) / (R_AB + R_BC + R_CA)
The memory hook: for the star arm at node A, multiply the two delta resistors that touch A (that's R_AB and R_CA), and divide by the sum of all three. Same pattern at every node — you don't need three separate formulas, just one rule.
Star → Delta
Each delta resistor = (sum of pairwise products of star resistors)
÷ (the star resistor at the OPPOSITE node)
Let S = R₁R₂ + R₂R₃ + R₃R₁
R_AB = S / R₃ (R₃ is at C — opposite the AB side)
R_BC = S / R₁ (R₁ is at A — opposite the BC side)
R_CA = S / R₂ (R₂ is at B — opposite the CA side)
The balanced case — check this first
If all three are equal:
R_delta = 3 × R_star R_star = R_delta / 3
Example: a 30 Ω delta becomes a 10 Ω star.
Many exam problems use equal resistors precisely so this shortcut applies. Look for it before grinding through the general formula.
Worked numerical 1 — delta to star
Convert a delta of R_AB = 10 Ω, R_BC = 20 Ω, R_CA = 30 Ω into its equivalent star.
Sum = 10 + 20 + 30 = 60 Ω
R₁ (at A) = (R_AB × R_CA)/Sum = (10 × 30)/60 = 5 Ω
R₂ (at B) = (R_AB × R_BC)/Sum = (10 × 20)/60 = 3.33 Ω
R₃ (at C) = (R_BC × R_CA)/Sum = (20 × 30)/60 = 10 Ω
Sanity check: each star arm should be smaller than either
delta resistor that formed it. 5 < 10 ✔ 3.33 < 10 ✔ 10 < 20 ✔
Worked numerical 2 — star to delta
Convert a star of R₁ = 5 Ω, R₂ = 3.33 Ω, R₃ = 10 Ω back to delta (it should return the values above).
A bridge has 10 Ω between A–B, 20 Ω B–C, 30 Ω C–A (a delta), with a 40 Ω from B and 50 Ω from C going to a common output node. Find resistance seen from A.
Step 1 — replace the delta ABC with its star (from numerical 1):
R₁ = 5 Ω at A, R₂ = 3.33 Ω at B, R₃ = 10 Ω at C
Step 2 — now B and C branches are plain series:
B path: 3.33 + 40 = 43.33 Ω
C path: 10 + 50 = 60 Ω
Step 3 — those two are in parallel:
R_p = (43.33 × 60)/(43.33 + 60) = 2600/103.33 = 25.16 Ω
Step 4 — add the star arm at A, which is in series:
R_total = 5 + 25.16 = 30.16 Ω
The transformation did the real work: after step 1 the network was ordinary series-parallel.
💡 Exam angle: the delta→star direction is asked far more often, because it's what unlocks bridge networks. Memorise the "product of the two touching, over the sum of all three" rule and you have both directions. Always check whether the resistors are equal first — if so, the ÷3 or ×3 shortcut saves several minutes.
Syllabus points
Y↔Δ transformation formulas (numerical)
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