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Inductance of composite-conductor single-phase line

Short answer

Compute self GMD of each side using r' = 0.7788r and mutual GMD from distances. L_X = 6.212×10^-7 H/m, L_Y = 8.505×10^-7 H/m, total loop inductance L = 1.472×10^-6 H/m.

One circuit of a single-phase transmission line is composed of three solid 0.25-cm-radius wires. The return circuit is composed of two 0.5-cm-radius wires. The arrangement of the conductors is shown below. Find the inductance due to the current in each side of the line and the inductance of the complete line in henrys per metre. [Figure: side X has conductors a, b, c in a vertical line, 6 m apart (a–b = 6 m, b–c = 6 m); side Y has conductors d and e, 9 m to the right of the side X column, d level with a and e level with b.]

Each side of the line is a composite conductor. The inductance of side X is LX=2×10−7ln⁡(Dm/DsX)L_X = 2\times10^{-7}\ln(D_m/D_{sX}), and similarly for side Y; the loop inductance is L=LX+LYL = L_X + L_Y.

  side X           side Y
   a o------9 m------o d
     |6 m
   b o------9 m------o e
     |6 m
   c o

Data: side X: r=0.25r = 0.25 cm, r′=0.7788×0.0025=0.001947r' = 0.7788 \times 0.0025 = 0.001947 m. Side Y: r=0.5r = 0.5 cm, r′=0.003894r' = 0.003894 m.

Distances

PairDistance (m)
ad, be9
ae, bd, ce92+62=10.817\sqrt{9^2+6^2} = 10.817
cd92+122=15\sqrt{9^2+12^2} = 15
ab, bc (X)6
ac (X)12
de (Y)6

Mutual GMD (6 distances)

Dm=(DadDaeDbdDbeDcdDce)1/6=(9×10.817×10.817×9×15×10.817)1/6=10.743 m\begin{aligned} D_m &= (D_{ad}D_{ae}D_{bd}D_{be}D_{cd}D_{ce})^{1/6} \\ &= (9 \times 10.817 \times 10.817 \times 9 \times 15 \times 10.817)^{1/6} = 10.743\ \text{m} \end{aligned}

Self GMD of side X (32=93^2 = 9 terms)

DsX=[(r′)3×64×122]1/9=[0.0019473×64×122]1/9=0.4810 m\begin{aligned} D_{sX} &= \left[(r')^3 \times 6^4 \times 12^2\right]^{1/9} \\ &= \left[0.001947^3 \times 6^4 \times 12^2\right]^{1/9} = 0.4810\ \text{m} \end{aligned}

Self GMD of side Y (22=42^2 = 4 terms)

DsY=[(0.003894)2×62]1/4=0.003894×6=0.15285 mD_{sY} = \left[(0.003894)^2 \times 6^2\right]^{1/4} = \sqrt{0.003894 \times 6} = 0.15285\ \text{m}

Inductances

LX=2×10−7ln⁡10.7430.4810=6.212×10−7 H/mLY=2×10−7ln⁡10.7430.15285=8.505×10−7 H/mL=LX+LY=14.718×10−7 H/m\begin{aligned} L_X &= 2\times10^{-7}\ln\frac{10.743}{0.4810} = 6.212\times10^{-7}\ \text{H/m} \\ L_Y &= 2\times10^{-7}\ln\frac{10.743}{0.15285} = 8.505\times10^{-7}\ \text{H/m} \\ L &= L_X + L_Y = 14.718\times10^{-7}\ \text{H/m} \end{aligned}

Answer: LX=6.212×10−7L_X = 6.212\times10^{-7} H/m, LY=8.505×10−7L_Y = 8.505\times10^{-7} H/m, complete line L=1.472×10−6L = 1.472\times10^{-6} H/m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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