Star-delta transformation

Why it matters

The unbalanced Wheatstone bridge is the core circuit of strain-gauge, RTD and thermistor measurement. It cannot be reduced by series–parallel rules alone, because its bridge arm creates a delta that hides inside the network. Star–delta (Y–Δ, also called T–Π) conversion removes that obstacle in one step. The same conversion is used for three-phase loads and for T and π attenuator and filter sections.

Key ideas

The two forms.

• A star (Y, T) has three resistors meeting at a common internal node N, with outer ends at terminals 1, 2 and 3. Call them R1, R2 and R3, named after their terminal.
• A delta (Δ, π) has three resistors connected directly between the terminal pairs: R12, R23 and R31.

Equivalence. The two forms are equivalent if the resistance measured between every pair of terminals, with the third terminal floating, is the same in both. Equating the three pairs gives the conversion formulas. The equivalence holds only at the three terminals: the internal star node N does not exist in the delta.

Delta to star. Each star arm equals the product of the two delta resistors touching that terminal, divided by the sum of all three: R1 = R12·R31/(R12 + R23 + R31), and similarly for R2 and R3.

Star to delta. Each delta resistor equals the sum of the pairwise products of the star arms, divided by the star arm at the opposite terminal: R12 = (R1R2 + R2R3 + R3R1)/R3, and similarly for R23 and R31.

In conductance form the star-to-delta rule mirrors the delta-to-star rule: G12 = G1·G2/(G1 + G2 + G3).

Balanced case. If all three resistors are equal, RY = RΔ/3 and RΔ = 3·RY.

Impedances. The same formulas hold for complex impedances in AC steady state. The arms may be mixed R, L and C, and the result may contain frequency-dependent and even negative-resistance-looking terms. That is mathematically fine at a single frequency.

When to use it.

• Bridge networks where no element is purely in series or parallel.
• Converting a Δ-connected three-phase load to an equivalent Y so that per-phase analysis can be used.
• Converting T sections to π sections in attenuator and filter design.

Balanced bridge shortcut. If the bridge arms satisfy R_AC·R_DB = R_AD·R_CB (the Wheatstone balance condition), C and D are at the same potential. The bridge arm then carries no current and can be removed or shorted, so no conversion is needed.

Formulas

• Delta to star: R1 = R12·R31/ΣRΔ, R2 = R12·R23/ΣRΔ, R3 = R23·R31/ΣRΔ, where ΣRΔ = R12 + R23 + R31 (all in Ω).
• Star to delta: R12 = S/R3, R23 = S/R1, R31 = S/R2, where S = R1R2 + R2R3 + R3R1 (Ω²).
• Balanced: RY = RΔ/3.
• Bridge balance: R_AC·R_DB = R_AD·R_CB.
• All formulas apply to impedances Z in place of R in sinusoidal steady state.

Worked examples

Example 1 (standard): equivalent resistance of an unbalanced bridge. Given: between terminals A and B there is a bridge. A–C is 10 Ω, A–D is 20 Ω, C–D (the bridge arm) is 30 Ω, C–B is 15 Ω and D–B is 5 Ω. Find R_AB.

1. Check balance: 10 × 5 = 50 and 20 × 15 = 300. These are unequal, so the bridge is unbalanced.
2. Convert delta A–C–D to a star (sum = 10 + 20 + 30 = 60 Ω):
   • R_A = 10·20/60 = 3.333 Ω
   • R_C = 10·30/60 = 5 Ω
   • R_D = 20·30/60 = 10 Ω
3. From the star node N to B there are two parallel paths: 5 + 15 = 20 Ω and 10 + 5 = 15 Ω. In parallel they give 20·15/35 = 8.571 Ω.
4. R_AB = 3.333 + 8.571 = 11.905 Ω.

Answer: R_AB ≈ 11.9 Ω.

Example 2 (GATE level): current in the bridge arm. Given: the bridge of Example 1 with 12 V applied, A positive and B at 0 V. Find the current in the 30 Ω arm.

1. Total current: I = 12/11.905 = 1.008 A.
2. Voltage across the N–B parallel section: V_NB = 1.008 × 8.571 = 8.64 V.
3. Current in the path through C: 8.64/20 = 0.432 A. So V_C = 0.432 × 15 = 6.48 V.
4. Current in the path through D: 8.64/15 = 0.576 A. So V_D = 0.576 × 5 = 2.88 V.
5. The node voltages C and D are the same in the original circuit, because they are terminals of the equivalence. So the original 30 Ω arm carries I_CD = (6.48 − 2.88)/30 = 0.12 A, from C to D.
6. Check by nodal analysis on the original circuit. At C: (VC − 12)/10 + (VC − VD)/30 + VC/15 = 0. At D: (VD − 12)/20 + (VD − VC)/30 + VD/5 = 0. These give VC = 6.48 V and VD = 2.88 V. ✓

Answer: 0.12 A from C to D.

Common mistakes

• In star-to-delta, dividing by the adjacent arm instead of the arm at the opposite terminal.
• In delta-to-star, multiplying the wrong pair. Use the two delta resistors that touch the terminal in question.
• Trying to find currents in the original delta resistors directly from the star currents. Go back to terminal voltages first.
• Converting when the bridge is balanced. Check R_AC·R_DB = R_AD·R_CB first, because the bridge arm may simply drop out.
• Applying RY = RΔ/3 to an unbalanced set.

For GATE IN

Frequent questions: equivalent resistance of a bridge, a lattice or a "cube/tetrahedron" network; the current in a bridge arm; the balance condition of a Wheatstone or AC bridge; converting a Δ-connected three-phase load to Y. Look for symmetry and balance before converting. Practise doing one conversion and then reducing by series–parallel in under two minutes.

Quick check

1. A balanced delta has 30 Ω per arm. What is the equivalent star arm?
2. A star has arms R1 = 1 Ω, R2 = 2 Ω and R3 = 3 Ω. What is R12 in the equivalent delta?
3. When does the bridge arm of a Wheatstone bridge carry zero current?
4. A delta has R12 = 6 Ω, R23 = 12 Ω and R31 = 9 Ω. What is star arm R1?

Answers: 1. 10 Ω. 2. 11/3 = 3.67 Ω. 3. When R_AC·R_DB = R_AD·R_CB. 4. 6·9/27 = 2 Ω.