Skip to main content

Self GMD of 7-strand conductor, 4 mm strands

Short answer

Self-GMD (for inductance) = 2.1767 × 4 mm = 8.71 mm (0.871 cm). Computed from 49 distance terms with strand r'=0.7788×4 mm.

Compute the self-GMD for the following conductor. Radius of each strand is 4 mm. [Figure: seven identical strands: one central strand surrounded by six touching strands.]

The self-GMD for inductance of a 7-strand conductor (1 central + 6 around) is found from all 72=497^2 = 49 distances, with self-distance r′=0.7788rr' = 0.7788r.

        (2)   (3)
     (7)   (1)   (4)
        (6)   (5)
  1 = centre, outer ring radius 2r

Data: r=4r = 4 mm, r′=0.7788×4=3.115r' = 0.7788 \times 4 = 3.115 mm.

Distances

Pair typeDistanceOrdered pairs
Selfr′r'7
Centre to outer2r2r12
Adjacent outers2r2r12
Outer to next-but-one23r2\sqrt3 r12
Outer to opposite4r4r6

Self-GMD

Ds=[(0.7788r)7(2r)24(23r)12(4r)6]1/49=2.1767 r=2.1767×4=8.707 mm\begin{aligned} D_s &= \left[(0.7788r)^7 (2r)^{24} (2\sqrt3 r)^{12} (4r)^6\right]^{1/49} \\ &= 2.1767\,r = 2.1767 \times 4 = 8.707\ \text{mm} \end{aligned}

For capacitance (self-distance rr): Ds=2.2558r=9.023D_s = 2.2558r = 9.023 mm.

Answer: self-GMD =8.71= 8.71 mm =0.871= 0.871 cm (for inductance).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Power System Analysis I chapter 4 questions →