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GMR of 5-strand conductor, 0.5 cm strands

Short answer

GMR of the five-strand conductor is 0.956 cm.

Determine the GMR of the stranded conductor for calculation of inductance shown in figure below if the radius of each sub conductor is 0.5 cm. [Figure: five identical touching strands, two in the top row resting on three in the bottom row.]

The GMR (self-GMD for inductance) of an nn-strand conductor is the n2n^2-th root of the product of all inter-strand distances, with self-distance r′=0.7788rr' = 0.7788r.

        (4)   (5)
     (1)   (2)   (3)
  touching strands, centre spacing 2r

Data: r=0.5r = 0.5 cm, 2r=12r = 1 cm, r′=0.7788×0.5=0.3894r' = 0.7788 \times 0.5 = 0.3894 cm.

Distances (cm)

PairDistance
1-2, 2-3, 4-5, 1-4, 2-4, 2-5, 3-52r=12r = 1
1-34r=24r = 2
1-5, 3-423r=1.7322\sqrt3 r = 1.732

Check: 7 + 1 + 2 = 10 pairs.

GMR

Ds=[(r′)5×(17×2×1.7322)2]1/25=[0.38945×62]1/25=0.9557 cm\begin{aligned} D_s &= \left[(r')^5 \times \left(1^7 \times 2 \times 1.732^2\right)^2\right]^{1/25} \\ &= \left[0.3894^5 \times 6^2\right]^{1/25} \\ &= 0.9557\ \text{cm} \end{aligned}

In terms of rr: Ds=1.9114rD_s = 1.9114r.

Answer: GMR =0.956= 0.956 cm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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