A Δ network of 6 Ω between nodes A-B, 4 Ω B-C, and 12 Ω C-A is converted to a Y network. Which is Ra?

A Δ network of 6 Ω between nodes A-B, 4 Ω B-C, and 12 Ω C-A is converted to a Y network. Which is Ra?

• A. Rb ≈ 1.091 Ω
• B. Rc ≈ 2.182 Ω
• C. Ra ≈ 3.273 Ω
• D. Ra ≈ 4.000 Ω

Answer: (C)

Delta-to-wye conversion turns a triangle of resistors into a three-leg star centered at a new node. The resistance from node A to that center is the product of the two delta resistances connected to A, divided by the sum of all three delta resistances. Here, the two resistances touching node A are 6 Ω (between A and B) and 12 Ω (between C and A). The sum of all delta resistances is 6 + 4 + 12 = 22. Ra = (6 × 12) / 22 = 72 / 22 = 3.273 Ω (approximately). So the leg from node A to the center is about 3.273 Ω. For context, the other two legs would be (6 × 4)/22 ≈ 1.091 Ω and (4 × 12)/22 ≈ 2.182 Ω, which aligns with the given options.

Delta-to-wye conversion turns a triangle of resistors into a three-leg star centered at a new node. The resistance from node A to that center is the product of the two delta resistances connected to A, divided by the sum of all three delta resistances.

Here, the two resistances touching node A are 6 Ω (between A and B) and 12 Ω (between C and A). The sum of all delta resistances is 6 + 4 + 12 = 22.

Ra = (6 × 12) / 22 = 72 / 22 = 3.273 Ω (approximately).

So the leg from node A to the center is about 3.273 Ω. For context, the other two legs would be (6 × 4)/22 ≈ 1.091 Ω and (4 × 12)/22 ≈ 2.182 Ω, which aligns with the given options.