Convert the star circuit into its equivalent delta circuit. [Figure: star network with centre N; arm N–A = 2 Ω (A at the top), arm N–B = 4 Ω (B at lower left), arm N–C = 6 Ω (C at lower right).]

R_A = 2 Ω, R_B = 4 Ω, R_C = 6 Ω.

Sum of products: \( R_AR_B + R_BR_C + R_CR_A = 2(4) + 4(6) + 6(2) = 8 + 24 + 12 = 44\ \Omega^2 \).

Star → delta: \( R_{AB} = \dfrac{44}{R_C} = \dfrac{44}{6} = \) 7.33 Ω; \( R_{BC} = \dfrac{44}{R_A} = \dfrac{44}{2} = \) 22 Ω; \( R_{CA} = \dfrac{44}{R_B} = \dfrac{44}{4} = \) 11 Ω.

          A
         / \
  7.33 Ω/   \ 11 Ω
       /     \
      B-------C
        22 Ω

Check (delta → star): R_A = (7.33 × 11)/(7.33 + 22 + 11) = 80.67/40.33 = 2 Ω ✓.