Transmission Line Inductance

Set 5 — Transmission Line Inductance

Twenty worked problems on the parameter that dominates every transmission calculation. A line's inductance is fixed entirely by geometry — how far apart the phases are, and how the current is distributed inside each conductor — and it enters through a logarithm, which is why a line ten times longer has ten times the reactance while a line with conductors ten times further apart has only about a third more. Everything here reduces to identifying two distances, \(D_m\) and \(D_s\), and taking their ratio.

Textbook Chapter 6 · 20 solved · 12 practice · 3 challenge · 12 MCQs

i Method Recap

The master relation. For one conductor of a group, \(L = 2\times10^{-7}\ln(D_m/D_s)\) H/m, where \(D_m\) is the geometric mean distance to the return path and \(D_s\) the self geometric mean distance of the conductor. Every problem in this set is the identification of those two numbers.
GMR of a solid round conductor is \(r' = re^{-1/4} = 0.7788r\). The factor exists to absorb the internal inductance \(\mu_0/8\pi = 0.5\times10^{-7}\) H/m, which is independent of radius. Once \(r'\) is used, the conductor may be treated as a hollow shell carrying all its current on the surface.
GMD is the geometric mean of every distance from each conductor of one group to each conductor of the other; GMR is the geometric mean of every distance within one group, including each conductor to itself, for which the distance is \(r'\). A group of \(n\) conductors therefore has \(n^2\) terms under an \(n^2\)-th root.
Convenient working forms. Per conductor, \(L = 0.2\ln(D_m/D_s)\) mH/km \(= 0.4605\log_{10}(D_m/D_s)\) mH/km. For a single-phase two-wire loop both conductors contribute, so the loop inductance is twice this.
Unsymmetrical spacing needs transposition. A transposed line behaves as if equilaterally spaced at \(D_{eq} = \sqrt[3]{D_{12}D_{23}D_{31}}\). Without transposition the three phases have unequal inductances and the line is a permanent source of negative-sequence current.
Bundling raises the GMR, not the conductor size. For \(n\) sub-conductors of GMR \(D_s\) at bundle spacing \(d\): two give \(\sqrt{D_sd}\), three give \(\sqrt[3]{D_sd^2}\), four give \(1.09\sqrt[4]{D_sd^3}\). The effect on reactance is large because \(D_s\) sits inside the logarithm.
Reactance is \(X = 2\pi fL\), and at 50 Hz that is \(314.16L\). A typical overhead line lands between 0.3 and 0.45 \(\Omega\)/km per phase, and a result far outside that band signals an error in \(D_s\) — most often a radius used where a GMR was meant.

VideoWalkthrough

Problem 1Warm-upGeometric Mean Radius

A solid round conductor has a diameter of 0.8 cm. Find its geometric mean radius, and state what physical quantity the factor 0.7788 accounts for.

Solution

The radius is half the diameter:

\[ r = \frac{0.8}{2} = 0.4\ \text{cm} \]

The geometric mean radius of a solid round conductor:

\[ D_s = r' = re^{-1/4} = 0.4 \times 0.7788 = 0.3115\ \text{cm} \]

What the factor accounts for. A real conductor carries current throughout its cross-section, so part of the magnetic flux links only part of the current. That internal flux contributes an inductance

\[ L_{\text{int}} = \frac{\mu_0}{8\pi} = 0.5\times10^{-7}\ \text{H/m} \]

independently of the radius — a result that surprises on first meeting and is examined in Problem 3.

Writing the total inductance and absorbing the internal term into the logarithm:

\[ L = 2\times10^{-7}\left[\frac14 + \ln\frac{D}{r}\right] = 2\times10^{-7}\ln\frac{D}{re^{-1/4}} = 2\times10^{-7}\ln\frac{D}{r'} \]

So \(r'\) is the radius of a fictitious hollow thin-walled conductor which, carrying all its current on the surface and therefore having no internal flux, would have exactly the same total inductance as the real solid one.

GMR is a modelling device, not a measurement. Nothing about the conductor is 0.7788 of anything — the number is the price of pretending that a distributed current is a surface current. Once paid, every later formula can treat conductors as filaments, which is what makes the geometric-mean machinery of the rest of this set possible. Manufacturers quote GMR directly on conductor data sheets precisely so that the user never has to think about internal flux again.

Answer\(D_s = r' = 0.3115\) cm; the factor absorbs the internal inductance \(\mu_0/8\pi\)

Problem 2Exam levelSingle-Phase Line

A single-phase, two-wire transmission line 15 km long is made of round conductors each 0.8 cm in diameter, separated by 40 cm. Calculate the equivalent diameter of a fictitious hollow, thin-walled conductor having the same inductance as the original line, and find the total inductance of the line.

Solution

From Problem 1, the fictitious conductor has radius \(r' = 0.3115\) cm, so its diameter is

\[ 2r' = 2(0.3115) = 0.623\ \text{cm} \]

Both conductors of the loop carry the current and both contribute inductance, so the loop inductance per metre is twice the single-conductor value:

\[ L = 4\times10^{-7}\,l\,\ln\frac{\text{GMD}}{\text{GMR}} \]

Here GMD is the spacing, 40 cm, and GMR is \(r' = 0.3115\) cm. Both are lengths, so any consistent unit will do in the ratio:

\[ \frac{\text{GMD}}{\text{GMR}} = \frac{40}{0.3115} = 128.4 \]

Substituting, with the length in metres:

\[ L = 4\times10^{-7}(15\times10^{3})\ln(128.4) = 6\times10^{-3}(4.855) = 0.02913\ \text{H} \]

\[ L = 29.13\ \text{mH} \]

The factor of four, not two, is what distinguishes a loop from a conductor. Each conductor of the pair has \(2\times10^{-7}\ln(D/r')\) H/m of its own, and the loop is the series combination of the two. In a three-phase line the corresponding factor is again \(2\times10^{-7}\) per phase, because the "return" is shared among the other two phases rather than being a dedicated conductor. Applying the single-phase factor of four to a three-phase line doubles the answer, and is one of the two commonest errors in this topic.

AnswerEquivalent diameter \(= 0.623\) cm; \(L = 29.13\) mH for 15 km

Problem 3Challenge-liteInternal and External

A solid round conductor of radius 1 cm forms part of a line whose return is 3 m away. Compute the internal and external components of its inductance separately, and verify that their sum equals the value obtained from the GMR.

Solution

Internal inductance, from the flux linking part of the current inside the conductor:

\[ L_{\text{int}} = \frac{\mu_0}{8\pi} = \frac{4\pi\times10^{-7}}{8\pi} = 0.5\times10^{-7}\ \text{H/m} = 0.05\ \text{mH/km} \]

Note that no radius appears. A thin wire and a thick one have identical internal inductance, because the smaller flux path of the thin wire is exactly offset by its higher current density.

External inductance, from the flux between the conductor surface and the return path:

\[ L_{\text{ext}} = 2\times10^{-7}\ln\frac{D}{r}\ \text{H/m} = 0.2\ln\frac{D}{r}\ \text{mH/km} = 0.2\ln(300) \]

\[ = 0.2(5.7038) = 1.1408\ \text{mH/km} \]

Total:

\[ L = 0.05 + 1.1408 = 1.1908\ \text{mH/km} \]

Verification through the GMR. With \(r' = 0.7788(0.01) = 0.007788\) m:

\[ L = 0.2\ln\frac{3}{0.007788} = 0.2\ln(385.2) = 0.2(5.9538) = 1.1908\ \text{mH/km}\ \checkmark \]

The internal part is \(0.05/1.1908 = 4.2\%\) of the total — small, but not negligible, and it is exactly this 4.2% that the GMR device exists to carry.

The internal inductance being radius-independent has a practical consequence. It means that making a conductor thicker reduces its inductance only through the external term, and only logarithmically — doubling the radius saves \(2\times10^{-7}\ln2 = 0.139\) mH/km whatever the starting size. Since doubling the radius quadruples the metal, thickening a conductor to reduce reactance is spectacularly poor value. Bundling, in Problems 10 to 12, achieves the same end far more cheaply by raising the geometric mean radius without adding proportionate metal.

Answer\(L_{\text{int}} = 0.05\), \(L_{\text{ext}} = 1.141\), total \(1.191\) mH/km — matching the GMR result

Problem 4Challenge-liteStranded Conductor

A conductor is made of seven identical copper strands, each of radius \(r\) — one central strand surrounded by six touching it. Find the GMR of the composite conductor.

Solution

The GMR of a group of \(n\) conductors is the \(n^2\)-th root of the product of all \(n^2\) distances, each conductor to every other and to itself, the self-distance being \(r' = 0.7788r\). With \(n = 7\) there are 49 terms.

The geometry. With six strands packed round one, the centres of adjacent outer strands are \(2r\) apart, the centre strand is \(2r\) from each outer one, strands two apart around the ring are \(2\sqrt3\,r\) apart, and diametrically opposite strands are \(4r\) apart.

Collecting the 49 distances by type:

\[ D_s = \sqrt[49]{(r')^{7}\,(2r)^{12}\,(2r)^{6}\,\big(2\sqrt3\,r\big)^{12}\,(4r)^{6}\,(2r)^{6}} \]

Seven self-terms; twelve centre-to-outer pairs counted both ways; the outer ring's adjacent, next-adjacent and opposite pairs likewise counted both ways.

Carrying out the product and the root — the algebra is tedious but entirely mechanical — gives the standard result:

\[ D_s = \frac{2r\sqrt[7]{3 \times 0.7788}}{\sqrt[49]{6}} = 2.177\,r \]

Comparing with the overall physical radius of the bundle, which is \(3r\):

\[ \frac{D_s}{3r} = \frac{2.177}{3} = 0.726 \]

Very close to the 0.7788 of a solid conductor — stranding changes the GMR by only a few per cent for the same outside radius.

Stranding is a mechanical decision that turns out to be electrically almost free. Conductors are stranded so they can be bent, spooled and terminated; the 7% difference between 0.726 and 0.7788 is an accidental by-product, not a design objective. This is why data sheets simply quote a measured GMR for each stranding pattern rather than expecting the user to compute one — and why a computed GMR that differs greatly from \(0.75 \times\) the physical radius should be suspected before it is trusted.

Answer\(D_s = 2.177\,r\) — about \(0.726\) of the overall radius \(3r\)

Problem 5Exam levelACSR Conductor

The outside diameter of the single layer of aluminium strands of an ACSR conductor is 5.04 cm, and each strand is 1.68 cm in diameter. Determine the 50 Hz reactance at 1 m spacing, neglecting the central steel strand, and give reasons for neglecting it.

Solution

The layer is six aluminium strands around one steel strand. Checking the geometry:

\[ \text{steel strand diameter} = 5.04 - 2(1.68) = 1.68\ \text{cm} \]

so all seven strands are the same size, \(d = 1.68\) cm — the same arrangement as Problem 4, but with the centre strand excluded from the current path.

With six current-carrying strands the GMR is a 36-term root. The distances round the hexagon are \(D_{12} = D_{16} = d\), \(D_{13} = D_{15} = \sqrt3\,d\) and \(D_{14} = 2d\):

\[ D_s = \sqrt[36]{\left[\left(\frac{d'}{2}\right)d^{2}\left(\sqrt3\,d\right)^{2}(2d)\right]^{6}} \]

Substituting \(d' = 0.7788d\) and simplifying gives the standard six-strand result:

\[ D_s = 1.155\,d = 1.155(1.68) = 1.93\ \text{cm} \]

Since the spacing of 1 m greatly exceeds the conductor size, \(D_m \approx D = 100\) cm. The inductance of each conductor:

\[ L = 0.4605\log_{10}\frac{100}{1.93} = 0.4605(1.714) = 0.789\ \text{mH/km} \]

For the two conductors of a single-phase line:

\[ L_{\text{loop}} = 2(0.789) = 1.578\ \text{mH/km} \]

\[ X = 2\pi fL = 314(1.578\times10^{-3}) = 0.495\ \Omega/\text{km} \]

Why the steel is neglected. Steel has roughly one seventh the conductivity of aluminium and, being ferromagnetic, a permeability that makes its internal impedance far higher at power frequency. It therefore carries a negligible share of the current. Its role is mechanical: it takes the tension so the aluminium need only conduct — which is the whole idea of ACSR.

The steel core is not merely negligible; it is deliberately arranged to be. Because the six aluminium strands are wound in a helix around it, their magnetic fields largely cancel within the core, keeping the flux in the steel — and hence its losses and its inductive contribution — small. This is why ACSR behaves electrically as an aluminium conductor of the same outside diameter, and why its data sheet quotes a GMR computed from the aluminium alone.

Answer\(D_s = 1.93\) cm, \(L = 1.578\) mH/km for the loop, \(X = 0.495\ \Omega\)/km

Problem 6Challenge-liteComposite Conductors

One side of a single-phase line consists of three solid wires of radius 0.25 cm, spaced 6 m apart in a line; the return consists of two wires of radius 0.5 cm, spaced 6 m apart, the two groups being 9 m apart. Find the inductance of each side and of the complete line.

Solution

GMD between the groups. Six distances, from each of \(a,b,c\) to each of \(d,e\):

\[ D_{ad} = D_{be} = 9\ \text{m}, \qquad D_{ae} = D_{bd} = D_{ce} = \sqrt{6^{2}+9^{2}} = \sqrt{117}\ \text{m}, \qquad D_{cd} = \sqrt{9^{2}+12^{2}} = 15\ \text{m} \]

Their geometric mean:

\[ D_m = \sqrt[6]{9^{2} \times 15 \times 117^{3/2}} = 10.743\ \text{m} \]

GMR of side X — nine terms, each of the three wires to itself and to the other two, with \(r' = 0.7788(0.25\times10^{-2}) = 1.947\times10^{-3}\) m:

\[ D_{sX} = \sqrt[9]{\left(r'\right)^{3}\times 6^{4}\times 12^{2}} = \sqrt[9]{(1.947\times10^{-3})^{3}\,6^{4}\,12^{2}} = 0.481\ \text{m} \]

The \(6^4\) comes from the four adjacent pairs counted both ways and the \(12^2\) from the outer pair.

GMR of side Y — four terms, with \(r' = 0.7788(0.5\times10^{-2}) = 3.894\times10^{-3}\) m:

\[ D_{sY} = \sqrt[4]{(3.894\times10^{-3})^{2}\times 6^{2}} = 0.153\ \text{m} \]

Inductance of each side:

\[ L_X = 2\times10^{-7}\ln\frac{10.743}{0.481} = 2\times10^{-7}(3.106) = 6.212\times10^{-7}\ \text{H/m} \]

\[ L_Y = 2\times10^{-7}\ln\frac{10.743}{0.153} = 2\times10^{-7}(4.252) = 8.503\times10^{-7}\ \text{H/m} \]

Total loop inductance, the two sides in series:

\[ L = L_X + L_Y = 14.715\times10^{-7}\ \text{H/m} = 1.472\ \text{mH/km} \]

The two sides have different inductances, and the larger one belongs to the smaller group. Side Y has only two wires and therefore a GMR of 0.153 m against side X's 0.481 m, so its logarithm is larger. Spreading current over more conductors raises the GMR and lowers the inductance — which is the whole mechanism of bundling, arrived at here by accident. Note also that both sides share the same \(D_m\): the mutual geometry is a property of the pair, not of either group.

Answer\(L_X = 6.212\times10^{-7}\), \(L_Y = 8.503\times10^{-7}\), total \(14.715\times10^{-7}\) H/m

Problem 7Warm-upEquilateral Spacing

A three-phase line has its conductors at the corners of an equilateral triangle of side 3 m. Each conductor has a GMR of 0.0125 m. Find the inductance per phase per km and the 50 Hz reactance.

Solution

With equilateral spacing all three distances are equal, so the geometric mean distance is simply that spacing:

\[ D_m = \sqrt[3]{(3)(3)(3)} = 3\ \text{m} \]

Inductance per phase — note the coefficient is \(2\times10^{-7}\), not \(4\times10^{-7}\), because the return is shared among the other two phases:

\[ L = 2\times10^{-7}\ln\frac{D_m}{D_s} = 2\times10^{-7}\ln\frac{3}{0.0125} = 2\times10^{-7}\ln(240) \]

\[ = 2\times10^{-7}(5.4806) = 1.0961\times10^{-6}\ \text{H/m} = 1.096\ \text{mH/km} \]

Reactance at 50 Hz:

\[ X = 2\pi fL = 2\pi(50)(1.096\times10^{-3}) = 314.16(1.096\times10^{-3}) = 0.344\ \Omega/\text{km} \]

The value sits squarely in the usual 0.3 to 0.45 \(\Omega\)/km band, so no error in \(D_s\) is indicated.

Equilateral spacing is the case in which no transposition is needed, and it is almost never used. Three conductors at the corners of a triangle require a tower whose crossarms extend in two dimensions, which is expensive and mechanically awkward; horizontal or vertical arrangements are far cheaper to build and are used instead, at the cost of the transposition that Problems 8 and 9 examine. The equilateral case survives as the reference against which those arrangements are measured, through the equivalent spacing \(D_{eq}\).

Answer\(L = 1.096\) mH/km per phase, \(X = 0.344\ \Omega\)/km

Problem 8Exam levelTransposed Line

Determine the inductance per km per phase of a transposed 50 Hz three-phase line whose conductors are arranged horizontally with spacings of 1.6 m, 1.6 m and 3.2 m between the pairs. The conductor diameter is 0.8 cm.

Solution

The three mutual distances are 1.6 m, 1.6 m and 3.2 m — the outer pair being twice the adjacent spacing, as a horizontal arrangement requires. The equivalent equilateral spacing is their geometric mean:

\[ D_{eq} = \sqrt[3]{D_{12}D_{23}D_{31}} = \sqrt[3]{(1.6)(3.2)(1.6)} = \sqrt[3]{8.192} = 2.015\ \text{m} \]

Self GMD of the conductor, converting the diameter in cm to a radius in metres:

\[ D_s = 0.7788 \times \frac{0.8}{2 \times 100} = \frac{0.7788 \times 0.4}{100} = 0.003115\ \text{m} \]

Inductance per phase:

\[ L = 0.2\ln\frac{2.015}{0.003115} = 0.2\ln(646.9) \]

\[ = 0.2(6.4722) = 1.294\ \text{mH/km} \]

The coefficient 0.2 is \(2\times10^{-7}\) H/m re-expressed in mH/km: multiply by \(10^3\) m per km and again by \(10^3\) mH per H.

The corresponding reactance:

\[ X = 314.16(1.294\times10^{-3}) = 0.407\ \Omega/\text{km} \]

The equivalent spacing 2.015 m is neither the smallest nor the average of the three distances. The arithmetic mean would be \((1.6+1.6+3.2)/3 = 2.13\) m — close here, but the geometric mean is the correct one because the inductance depends on \(\ln D\), and the mean of logarithms is the logarithm of the geometric mean. The two means diverge as the spacings become more unequal, and it is the geometric one that the physics selects.

Answer\(D_{eq} = 2.015\) m, \(L = 1.294\) mH/km, \(X = 0.407\ \Omega\)/km

Problem 9Challenge-liteTransposition

Explain why an unsymmetrically spaced three-phase line is transposed, what transposition physically consists of, and what the consequences are of leaving a line untransposed.

Solution

The problem. Consider the horizontal arrangement of Problem 8. The centre conductor is 1.6 m from each of its neighbours; each outer conductor is 1.6 m from the centre and 3.2 m from the far outer. The three conductors therefore occupy geometrically different positions, and their inductances differ:

\[ L_{\text{centre}} = 2\times10^{-7}\ln\frac{\sqrt{(1.6)(1.6)}}{D_s}, \qquad L_{\text{outer}} = 2\times10^{-7}\ln\frac{\sqrt{(1.6)(3.2)}}{D_s} \]

Evaluating with \(D_s = 0.003115\) m:

\[ L_{\text{centre}} = 0.2\ln\frac{1.6}{0.003115} = 1.257\ \text{mH/km} \]

\[ L_{\text{outer}} = 0.2\ln\frac{2.263}{0.003115} = 1.326\ \text{mH/km} \]

A spread of about 5.5% — modest, but it is a permanent asymmetry, not a fluctuation.

The consequence. Unequal phase impedances mean that a balanced set of applied voltages produces an unbalanced set of currents. Decomposed by the symmetrical components of Set 21, that unbalance is a standing negative-sequence current, which causes double-frequency rotor heating in every connected machine, and a zero-sequence component that can interfere with communication circuits and mis-operate protection.

What transposition is. The line is divided into three equal sections, and at each of the two transposition points the conductors are rotated so that each phase occupies each of the three physical positions for exactly one third of the length. Averaging the flux linkages over the three sections gives every phase the same inductance:

\[ L = 2\times10^{-7}\ln\frac{\sqrt[3]{D_{12}D_{23}D_{31}}}{D_s} \]

which is exactly the \(D_{eq}\) used in Problem 8 — the geometric mean appears because it is the mean of the three logarithms.

In practice. Full transposition needs special towers and is expensive, so on shorter lines the 5% asymmetry is simply tolerated, and on longer ones transposition is often carried out at substations rather than mid-span. Modern practice frequently omits it entirely and accepts the resulting unbalance, which is small compared with the unbalance the loads themselves impose.

Transposition is the reason a three-phase line can be described by a single number at all. Every later set — the ABCD parameters of Sets 11 and 12, the \(Y\)-bus of Set 16, the positive-sequence network of Set 22 — assumes one impedance per line. That assumption is manufactured by transposition, or by the decision to ignore its absence. It is worth knowing which, because the residual negative-sequence current is exactly what the machine-heating limits of Set 26 are set against.

AnswerEqualises the three phase inductances at \(D_{eq} = \sqrt[3]{D_{12}D_{23}D_{31}}\); untransposed gives a 5.5% spread and standing negative-sequence current

Problem 10Warm-upTwo-Conductor Bundle

Each phase of a line uses a two-conductor bundle, the sub-conductors having GMR 0.0125 m and being spaced 0.45 m apart. Find the GMR of the bundle, and compare it with that of a single sub-conductor.

Solution

The bundle is a group of two conductors, so its GMR is the fourth root of four distances: each to itself (\(D_s\)) and each to the other (\(d\)):

\[ D_{sb} = \sqrt[4]{D_s \cdot d \cdot d \cdot D_s} = \sqrt{D_s\,d} \]

Substituting:

\[ D_{sb} = \sqrt{(0.0125)(0.45)} = \sqrt{0.005625} = 0.075\ \text{m} \]

Comparing with a single sub-conductor:

\[ \frac{D_{sb}}{D_s} = \frac{0.075}{0.0125} = 6 \]

A sixfold increase in effective radius, achieved by adding one more conductor rather than by making the conductor six times thicker — which would have needed thirty-six times the metal.

The geometric mean of a small number and a large one is dominated by neither. That is what makes bundling so effective: \(\sqrt{D_sd}\) sits midway between the conductor's own radius and the bundle spacing on a logarithmic scale, so a modest 0.45 m spacing lifts a 12.5 mm GMR to 75 mm. Since inductance depends on \(\ln(D_m/D_s)\), that sixfold rise in \(D_s\) removes \(2\times10^{-7}\ln6 = 0.358\) mH/km from every kilometre of the line — around a quarter of the total, as Problem 12 confirms.

Answer\(D_{sb} = 0.075\) m — six times the sub-conductor GMR

Problem 11Exam levelLarger Bundles

Repeat Problem 10 for a three-conductor bundle at the corners of an equilateral triangle of side 0.45 m, and for a four-conductor bundle at the corners of a square of side 0.45 m. Take \(D_s = 0.0125\) m throughout.

Solution

Three-conductor bundle. Nine distances: three self-terms and six mutual, all equal to \(d\):

\[ D_{sb} = \sqrt[9]{D_s^{3}\,d^{6}} = \sqrt[3]{D_s\,d^{2}} \]

\[ = \sqrt[3]{(0.0125)(0.45)^{2}} = \sqrt[3]{0.0025313} = 0.136\ \text{m} \]

Four-conductor bundle. Sixteen distances: four self-terms, eight sides of length \(d\) and four diagonals of length \(\sqrt2\,d\):

\[ D_{sb} = \sqrt[16]{D_s^{4}\,d^{8}\,(\sqrt2\,d)^{4}} = 2^{1/8}\sqrt[4]{D_s\,d^{3}} = 1.09\sqrt[4]{D_s\,d^{3}} \]

\[ = 1.09\sqrt[4]{(0.0125)(0.45)^{3}} = 1.09\sqrt[4]{0.0011391} = 1.09(0.1837) = 0.200\ \text{m} \]

Collecting all four cases:

\[ \begin{array}{lcc} \text{Bundle} & D_{sb}\ (\text{m}) & \text{Ratio to } D_s \\ \hline \text{single} & 0.0125 & 1.0 \\ \text{two} & 0.075 & 6.0 \\ \text{three} & 0.136 & 10.9 \\ \text{four} & 0.200 & 16.0 \end{array} \]

The returns diminish. Going from one to two sub-conductors multiplies \(D_{sb}\) by 6; from two to three by only 1.82; from three to four by 1.47.

The 1.09 in the four-conductor formula is \(2^{1/8}\), and it is the only place a bundle's shape shows through. The three-conductor formula has no such factor because an equilateral triangle has all its mutual distances equal; the square does not, and the four diagonals contribute their \(\sqrt2\). Since inductance depends on \(\ln D_{sb}\), the diminishing returns above translate into steadily smaller reactance savings — which is why four is the practical limit and six-conductor bundles appear only at 750 kV and above, where corona rather than reactance is the driver.

AnswerThree-conductor \(0.136\) m; four-conductor \(0.200\) m — 10.9 and 16 times the sub-conductor GMR

Problem 12Exam levelBundled Line

A 400 kV line has an equivalent spacing of 12 m and uses the two-conductor bundle of Problem 10. Find its inductance and reactance per km per phase, and compare with the same line using a single conductor of GMR 0.0125 m.

Solution

Bundled line, using \(D_{sb} = 0.075\) m:

\[ L = 0.2\ln\frac{12}{0.075} = 0.2\ln(160) = 0.2(5.0752) = 1.015\ \text{mH/km} \]

\[ X = 314.16(1.015\times10^{-3}) = 0.319\ \Omega/\text{km} \]

Single conductor, same geometry, \(D_s = 0.0125\) m:

\[ L = 0.2\ln\frac{12}{0.0125} = 0.2\ln(960) = 0.2(6.8669) = 1.373\ \text{mH/km} \]

\[ X = 314.16(1.373\times10^{-3}) = 0.431\ \Omega/\text{km} \]

The saving:

\[ \frac{0.431 - 0.319}{0.431}\times100 = 26.1\% \]

The reduction can also be read directly from the logarithms, without computing either inductance:

\[ \Delta L = 0.2\ln\frac{D_{sb}}{D_s} = 0.2\ln 6 = 0.358\ \text{mH/km}\ \checkmark \]

A 26% cut in series reactance is worth far more than it sounds. The steady-state power limit of a line is \(P = EV\sin\delta/X\), so the same reduction raises the transfer capability by 35%, and it cuts the voltage drop by the same 26%. Bundling was originally adopted at EHV to control corona — the subject of Set 8 — and the reactance reduction was a by-product. It is now often the primary justification, and it is the reason that essentially every line above 220 kV is bundled.

AnswerBundled: \(1.015\) mH/km, \(0.319\ \Omega\)/km. Single: \(1.373\) mH/km, \(0.431\ \Omega\)/km — a 26.1% saving

Problem 13Exam levelDouble Circuit

A double-circuit three-phase line is arranged so that each phase's two conductors are 8 m apart, each conductor having \(D_s = 0.0125\) m, and the equivalent GMD between phases is 5.6 m. Find the inductance per phase per km and compare with a single circuit of the same conductor at 5.6 m spacing.

Solution

The two conductors of one phase form a group of two, exactly like the bundle of Problem 10 — only the spacing is far larger:

\[ D_{sb} = \sqrt{D_s \cdot D_{aa'}} = \sqrt{(0.0125)(8)} = \sqrt{0.1} = 0.316\ \text{m} \]

Inductance per phase:

\[ L = 0.2\ln\frac{5.6}{0.316} = 0.2\ln(17.72) = 0.2(2.8746) = 0.575\ \text{mH/km} \]

Single circuit at the same 5.6 m spacing:

\[ L = 0.2\ln\frac{5.6}{0.0125} = 0.2\ln(448) = 0.2(6.1048) = 1.221\ \text{mH/km} \]

The ratio:

\[ \frac{0.575}{1.221} = 0.471 \]

Slightly less than half — the second circuit does better than mere paralleling would suggest.

Two circuits in parallel give less than half the reactance, and the reason is the widened GMR. Simple paralleling of two independent lines of 1.221 mH/km would give 0.611; the actual 0.575 is lower because the two conductors of each phase, being 8 m apart, act as an enormously wide bundle. The effect works in the designer's favour and grows with the separation of the circuits — which is one reason double-circuit towers place the two circuits on opposite faces rather than side by side.

Answer\(D_{sb} = 0.316\) m, \(L = 0.575\) mH/km — 47% of the single-circuit value

Problem 14Warm-upEffect of Spacing

By how much does the inductance of the line in Problem 7 change if the spacing is doubled from 3 m to 6 m? Express the change both absolutely and as a percentage, and generalise.

Solution

At 6 m spacing:

\[ L = 0.2\ln\frac{6}{0.0125} = 0.2\ln(480) = 0.2(6.1738) = 1.235\ \text{mH/km} \]

Against the 1.096 mH/km of Problem 7:

\[ \Delta L = 1.235 - 1.096 = 0.139\ \text{mH/km}, \qquad \frac{0.139}{1.096}\times100 = 12.7\% \]

The general result. The increment depends only on the ratio of the spacings, never on their absolute size:

\[ \Delta L = 0.2\ln\frac{2D}{D} = 0.2\ln 2 = 0.1386\ \text{mH/km} \]

Doubling the spacing always adds 0.1386 mH/km, whether the line starts at 1 m or 100 m.

The logarithm is what makes overhead transmission possible at all. If inductance were proportional to spacing rather than to its logarithm, a 400 kV line with its 12 m clearances would have four times the reactance of an 11 kV line at 3 m, and long-distance transmission would be hopeless. Instead it has only about 25% more. The same insensitivity means that the designer has essentially free rein over spacing — it is set by insulation clearance, conductor swing and tower economics, and the electrical consequence is an afterthought.

Answer\(L\) rises from 1.096 to 1.235 mH/km — \(+0.139\) mH/km, or 12.7%; doubling any spacing adds \(0.2\ln2\)

Problem 15Warm-upReactance

A line has an inductance of 1.28 mH/km per phase. Find its reactance per km at 50 Hz and at 60 Hz, and the total reactance of a 200 km line at 50 Hz.

Solution

At 50 Hz:

\[ X = 2\pi fL = 2\pi(50)(1.28\times10^{-3}) = 314.16(1.28\times10^{-3}) = 0.402\ \Omega/\text{km} \]

At 60 Hz, reactance scales directly with frequency:

\[ X = 0.402 \times \frac{60}{50} = 0.483\ \Omega/\text{km} \]

Total for 200 km at 50 Hz:

\[ X_{\text{total}} = 200(0.402) = 80.4\ \Omega \]

Both per-km values sit inside the usual 0.3–0.45 \(\Omega\)/km band at 50 Hz, and the 60 Hz figure is correspondingly higher — which is why American practice quotes a slightly wider band.

Reactance scales with frequency; inductance does not. The inductance of 1.28 mH/km is a property of the geometry alone and is identical at 50 and 60 Hz. This matters when reading data sheets, which sometimes quote \(X\) at an assumed frequency and sometimes \(L\) — and a 50 Hz value used in a 60 Hz study understates the reactance by 20%. It matters more for the harmonic studies of Set 39, where the fifth harmonic sees five times this reactance and the network's resonances are decided by it.

Answer\(0.402\ \Omega\)/km at 50 Hz, \(0.483\ \Omega\)/km at 60 Hz; \(80.4\ \Omega\) for 200 km

Problem 16Challenge-liteFlux Linkage

Derive the flux linkage of one conductor of a group of \(n\) parallel conductors carrying currents that sum to zero, and show how the inductance formula \(L = 2\times10^{-7}\ln(D_m/D_s)\) follows for a balanced three-phase line.

Solution

For conductors carrying \(I_1, I_2, \dots, I_n\) with \(\sum I_k = 0\), the flux linking conductor 1 per metre is

\[ \lambda_1 = 2\times10^{-7}\left[I_1\ln\frac{1}{D_{s1}} + I_2\ln\frac{1}{D_{12}} + \cdots + I_n\ln\frac{1}{D_{1n}}\right] \]

Each current contributes a term in the logarithm of the reciprocal of its distance from conductor 1, with the conductor's own contribution using \(D_{s1}\) in place of a distance to itself. The condition \(\sum I_k = 0\) is what makes the expression finite — otherwise the flux would diverge logarithmically with distance.

Apply to a balanced three-phase line. Take conductor \(a\), with \(I_b + I_c = -I_a\), and equilateral spacing \(D\):

\[ \lambda_a = 2\times10^{-7}\left[I_a\ln\frac{1}{D_s} + I_b\ln\frac{1}{D} + I_c\ln\frac{1}{D}\right] \]

Collecting the last two terms:

\[ \lambda_a = 2\times10^{-7}\left[I_a\ln\frac{1}{D_s} + (I_b + I_c)\ln\frac{1}{D}\right] = 2\times10^{-7}\left[I_a\ln\frac{1}{D_s} - I_a\ln\frac{1}{D}\right] \]

Hence

\[ \lambda_a = 2\times10^{-7}I_a\ln\frac{D}{D_s}, \qquad L_a = \frac{\lambda_a}{I_a} = 2\times10^{-7}\ln\frac{D}{D_s} \]

For unsymmetrical spacing the same steps apply to each of the three transposition sections, and averaging the three results replaces \(D\) by the geometric mean \(D_{eq}\) — which is the derivation promised in Problem 9.

The condition \(\sum I_k = 0\) is doing essential work here, and it is exactly KCL. A single isolated conductor has no well-defined inductance per unit length, because its flux integral diverges; inductance is always a property of a complete circuit. What makes a per-phase inductance meaningful for a three-phase line is that the other two phases constitute the return, and their currents sum with the first to zero. The moment that fails — under an earth fault, where zero-sequence current returns through the ground — the formula fails too, and the zero-sequence inductance of Set 22 has to be derived quite differently.

Answer\(\lambda_1 = 2\times10^{-7}\sum_k I_k\ln(1/D_{1k})\), giving \(L = 2\times10^{-7}\ln(D_m/D_s)\) when \(\sum I_k = 0\)

Problem 17Challenge-liteBundling vs Size

A line at 12 m equivalent spacing uses a single conductor of GMR 0.0125 m. To reduce its reactance by 26%, compare (a) increasing the conductor GMR and (b) using a two-conductor bundle at 0.45 m spacing. Quantify the metal required by each.

Solution

From Problem 12, the single conductor gives \(L = 1.373\) mH/km and the target is \(0.74 \times 1.373 = 1.016\) mH/km.

aBy enlarging the conductor. Solving for the GMR that achieves the target:

\[ 1.016\times10^{-3} = 0.2\ln\frac{12}{D_s'} \quad\Rightarrow\quad \ln\frac{12}{D_s'} = 5.08 \quad\Rightarrow\quad D_s' = \frac{12}{160.8} = 0.0746\ \text{m} \]

The GMR must rise by a factor of \(0.0746/0.0125 = 5.97\), and since GMR scales with radius, so must the radius. The cross-sectional area therefore scales as the square:

\[ \frac{a'}{a} = (5.97)^{2} = 35.6 \]

Thirty-six times the metal — and a conductor of that size would be unstringable, unspoolable and would need towers designed for its weight.

bBy bundling. Problem 12 already showed that two sub-conductors at 0.45 m achieve exactly this reduction. The metal required:

\[ \frac{a'}{a} = 2 \]

The comparison:

\[ \begin{array}{lcc} \text{Route} & \text{Metal} & \text{Practicable?} \\ \hline \text{Enlarge conductor} & 35.6\times & \text{No} \\ \text{Two-conductor bundle} & 2\times & \text{Yes} \end{array} \]

Bundling wins by a factor of eighteen, and the reason is that GMR is geometric. A solid conductor's GMR is tied to its radius and therefore to the square root of its metal; a bundle's GMR is tied to the spacing between sub-conductors, which costs nothing but air. The same argument explains why the four-conductor bundle of Problem 11 reaches \(D_{sb} = 0.2\) m on four times the metal, where a solid conductor of that GMR would need 256 times. Air, arranged correctly, is the cheapest conductor in the system.

AnswerEnlarging needs \(35.6\times\) the metal; bundling needs \(2\times\) — an eighteenfold advantage

Problem 18Warm-upWhole-Line Reactance

A transposed 220 kV line 200 km long has conductors of GMR 0.0125 m arranged horizontally with adjacent spacings of 6 m. Find the equivalent spacing, the inductance per km, and the total reactance of the line at 50 Hz.

Solution

Horizontal arrangement with 6 m between adjacent conductors makes the outer pair 12 m apart:

\[ D_{eq} = \sqrt[3]{(6)(6)(12)} = \sqrt[3]{432} = 7.56\ \text{m} \]

Inductance per phase:

\[ L = 0.2\ln\frac{7.56}{0.0125} = 0.2\ln(604.8) = 0.2(6.4050) = 1.281\ \text{mH/km} \]

Reactance per km:

\[ X = 314.16(1.281\times10^{-3}) = 0.402\ \Omega/\text{km} \]

Total for the line:

\[ X_{\text{total}} = 200(0.402) = 80.5\ \Omega \]

Notice how insensitive the answer is to the geometry. Problem 7 had 3 m equilateral spacing and gave 0.344 \(\Omega\)/km; this line has 7.56 m equivalent spacing — two and a half times larger — and gives 0.402. Almost every overhead line ever built lands between 0.3 and 0.45 \(\Omega\)/km per phase for exactly this reason, and that band is worth carrying as a check: a computed value outside it means an error in \(D_s\) far more often than an unusual line.

Answer\(D_{eq} = 7.56\) m, \(L = 1.281\) mH/km, \(X_{\text{total}} = 80.5\ \Omega\)

Problem 19Exam levelPer-Unit

Express the reactance of the line of Problem 18 in per-unit on a base of 100 MVA, 220 kV, and check the result against the expected band.

Solution

Base impedance of the 220 kV zone:

\[ Z_B = \frac{(220)^{2}}{100} = 484\ \Omega \]

Per-unit reactance:

\[ X_{pu} = \frac{80.5}{484} = 0.166\ \text{p.u.} \]

Against the band of Set 3 — transmission lines run from about 0.05 to 0.5 p.u. on a 100 MVA base — this is comfortably inside, and the line may be entered on a reactance diagram with confidence.

A useful rule follows from combining the two steps. For a 220 kV line on 100 MVA:

\[ X_{pu} \approx \frac{0.4\,l}{484} = 8.3\times10^{-4}\,l \quad (l\ \text{in km}) \]

so roughly 0.083 p.u. per hundred kilometres at this voltage — a figure worth carrying for quick estimates.

This is the point at which the geometry of Sets 5 to 8 hands over to the network analysis of Set 16 onwards. Everything from here on treats 0.166 as given. It is worth remembering that behind that single number lie a conductor diameter, three spacings, a transposition scheme and a base declaration — and that an error in any of them arrives at the load flow disguised as a perfectly plausible per-unit reactance.

Answer\(X_{pu} = 0.166\) p.u. on 100 MVA, 220 kV

Problem 20Exam levelGeometry to Per-Unit

A 400 kV, 300 km transposed line uses two-conductor bundles at 0.45 m spacing, the sub-conductors having GMR 0.0125 m. The bundle centres are arranged horizontally with 10 m between adjacent phases. Carry the calculation from geometry through to per-unit reactance on a 100 MVA base.

Solution

Step 1 — bundle GMR:

\[ D_{sb} = \sqrt{(0.0125)(0.45)} = 0.075\ \text{m} \]

Step 2 — equivalent spacing. Horizontal at 10 m adjacent makes the outer pair 20 m:

\[ D_{eq} = \sqrt[3]{(10)(10)(20)} = \sqrt[3]{2000} = 12.60\ \text{m} \]

Step 3 — inductance per phase:

\[ L = 0.2\ln\frac{12.60}{0.075} = 0.2\ln(168) = 0.2(5.1240) = 1.025\ \text{mH/km} \]

Step 4 — reactance:

\[ X = 314.16(1.025\times10^{-3}) = 0.322\ \Omega/\text{km}, \qquad X_{\text{total}} = 300(0.322) = 96.6\ \Omega \]

Step 5 — per-unit on 100 MVA, 400 kV:

\[ Z_B = \frac{(400)^{2}}{100} = 1600\ \Omega, \qquad X_{pu} = \frac{96.6}{1600} = 0.060\ \text{p.u.} \]

Worth comparing with Problem 19: this line is half as long again and yet has barely a third of the per-unit reactance, because \(Z_B\) rises as the square of the voltage.

Five steps, one logarithm, and a 300 km line has become the number 0.060. That is what every set from 16 onwards consumes. Notice what dominated: not the length, not the spacing, but the base voltage — the same \(1/V^2\) that governed conductor volume in Set 2 now governs per-unit reactance, and it is why a 400 kV corridor can carry power across a subcontinent while a 132 kV one of the same construction cannot reach a third as far before its reactance dominates.

Answer\(D_{sb} = 0.075\) m, \(D_{eq} = 12.60\) m, \(L = 1.025\) mH/km, \(X = 96.6\ \Omega = 0.060\) p.u.

Practice

Practice Problems

Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not.

P1. A solid conductor has a diameter of 1.2 cm. Find its GMR.

Show answer
\(r = 0.6\) cm, so \(D_s = 0.7788(0.6) = \mathbf{0.467}\) cm.
P2. A single-phase line has conductors of GMR 0.4 cm spaced 1.5 m apart. Find the loop inductance per km.

Show answer
\(4\times10^{-4}\ln(150/0.4) = 4\times10^{-4}(5.926) = \mathbf{2.37}\) mH/km.
P3. A three-phase line has spacings 2 m, 2.5 m and 4.5 m. Find the equivalent equilateral spacing.

Show answer
\(\sqrt[3]{(2)(2.5)(4.5)} = \sqrt[3]{22.5} = \mathbf{2.82}\) m.
P4. Conductors of GMR 0.011 m are spaced equilaterally at 4 m. Find \(L\) and \(X\) per km at 50 Hz.

Show answer
\(0.2\ln(4/0.011) = 0.2(5.894) = \mathbf{1.179}\) mH/km; \(X = \mathbf{0.370}\ \Omega\)/km.
P5. A two-conductor bundle has sub-conductors of GMR 0.015 m spaced 0.40 m. Find the bundle GMR.

Show answer
\(\sqrt{(0.015)(0.40)} = \sqrt{0.006} = \mathbf{0.0775}\) m.
P6. A three-conductor bundle has sub-conductors of GMR 0.015 m at the corners of a 0.40 m triangle. Find the bundle GMR.

Show answer
\(\sqrt[3]{(0.015)(0.40)^2} = \sqrt[3]{0.0024} = \mathbf{0.134}\) m.
P7. By how much does inductance change when the equivalent spacing is trebled?

Show answer
It increases by \(0.2\ln3 = \mathbf{0.220}\) mH/km, independent of the starting spacing.
P8. A line has \(L = 1.15\) mH/km. Find the total reactance of 150 km at 50 Hz, and its per-unit value on 100 MVA, 132 kV.

Show answer
\(X = 314.16(1.15\times10^{-3}) = 0.361\ \Omega\)/km, total \(\mathbf{54.2}\ \Omega\); \(Z_B = 174.24\ \Omega\), so \(\mathbf{0.311}\) p.u.
P9. Why is the coefficient \(4\times10^{-7}\) for a single-phase line but \(2\times10^{-7}\) per phase for a three-phase line?

Show answer
The single-phase figure is for the loop — both conductors in series. The three-phase figure is per phase, the return being shared among the other two phases.
P10. A four-conductor bundle has \(D_s = 0.012\) m on a 0.5 m square. Find its GMR.

Show answer
\(1.09\sqrt[4]{(0.012)(0.5)^3} = 1.09\sqrt[4]{0.0015} = 1.09(0.1968) = \mathbf{0.215}\) m.
P11. A line's computed reactance comes out at 3.2 \(\Omega\)/km. What is the most likely error?

Show answer
Nearly ten times the usual band, so \(D_s\) is far too small — most often a radius in cm used as metres, or a diameter used where a GMR was meant.
P12. Two conductors of a phase in a double-circuit line are 7 m apart, each of \(D_s = 0.014\) m. Find the effective GMR of the phase.

Show answer
\(\sqrt{(0.014)(7)} = \sqrt{0.098} = \mathbf{0.313}\) m — a very wide bundle, which is why double circuits have low reactance per phase.

Challenge

Challenge Problems

Each of these needs an idea rather than a formula. Decide what the governing principle is before opening the answer.

C1. Show that the internal inductance of a solid round conductor is \(\mu_0/8\pi\) per metre and is independent of the radius. Explain physically why the radius cancels, and identify the condition under which the result fails.

Show answer
Derivation. At radius \(x\) inside a conductor of radius \(r\) carrying \(I\) uniformly, the enclosed current is \(I_x = I(x^2/r^2)\) and Ampère's law gives \(H = I_x/2\pi x = Ix/2\pi r^2\).

The flux in an annulus \(dx\) is \(d\phi = \mu_0 H\,dx\), but it links only the fraction \(x^2/r^2\) of the current, so the flux linkage is
\[ d\lambda = \frac{x^2}{r^2}\,\mu_0\frac{Ix}{2\pi r^2}dx = \frac{\mu_0 I x^3}{2\pi r^4}dx \]
Integrating from 0 to \(r\):
\[ \lambda = \frac{\mu_0 I}{2\pi r^4}\cdot\frac{r^4}{4} = \frac{\mu_0 I}{8\pi} \quad\Rightarrow\quad L_{\text{int}} = \frac{\mu_0}{8\pi} = 0.5\times10^{-7}\ \text{H/m} \]
Why the radius cancels. Two competing effects scale oppositely. A larger conductor offers longer flux paths, which raises the flux; but for the same total current it has a lower current density, which lowers the field at every internal radius. The \(r^4\) from the integration exactly cancels the \(r^4\) in the denominator. The result is a statement about the shape of the internal field distribution, not about scale.

Where it fails. The derivation assumed uniform current density, which holds only at low frequency. As frequency rises the skin effect pushes current towards the surface, reducing the internal flux and hence \(L_{\text{int}}\) — at high enough frequency it tends to zero and the GMR tends to the physical radius \(r\) rather than \(0.7788r\). It also fails for ferromagnetic conductors, where \(\mu \ne \mu_0\) and the internal inductance is enormously larger — which is another reason the steel core of the ACSR in Problem 5 is kept out of the current path.
C2. Bundling reduces reactance by raising the GMR. Establish why this is so much cheaper than enlarging the conductor, derive the scaling law for both routes, and identify what eventually limits the number of sub-conductors.

Show answer
The two scaling laws. For a solid conductor, \(D_s = 0.7788r\) and metal \(\propto r^2\), so \(D_s \propto \sqrt{\text{metal}}\). For an \(n\)-conductor bundle, \(D_{sb} \approx (D_s d^{\,n-1})^{1/n}\) and metal \(\propto n\). The bundle's GMR is driven by \(d\), the spacing — which costs nothing.

Quantitatively (Problem 17): a 26% reactance cut needs 35.6 times the metal by enlargement and 2 times by bundling — an eighteenfold advantage. The gap widens for larger reductions, because the enlargement route is quadratic while the bundling route is linear.

What limits \(n\):
— Diminishing returns. From Problem 11, \(D_{sb}\) goes 1, 6, 10.9, 16 for \(n = 1,2,3,4\), and reactance depends on \(\ln D_{sb}\), so each addition buys less than the last.
— Mechanical cost. Spacers every 50–70 m along every span, and a wind and ice loading that grows with \(n\). Towers must be strengthened.
— Wind and galloping. Bundles are aerodynamically more troublesome than single conductors and require damping.
— Short-circuit forces. Sub-conductors attract violently during a fault and can clash; spacer design is governed by this.

What pushes it up: corona. As Set 8 shows, the surface gradient falls with more sub-conductors, and at 750 kV and above the corona and radio-interference limits demand four or six even where reactance alone would not justify them. The bundle number at EHV is therefore usually set by corona, and the reactance reduction is a welcome by-product — historically the reverse of how the technique is usually taught.
C3. The formula \(L = 2\times10^{-7}\ln(D_m/D_s)\) depends only on a ratio of distances, yet inductance plainly has dimensions. Resolve this, and explain what happens to the formula for a single isolated conductor.

Show answer
The resolution. The dimensions live entirely in the coefficient. \(2\times10^{-7}\) is \(\mu_0/2\pi\) in H/m, so the formula reads \(L = (\mu_0/2\pi)\ln(D_m/D_s)\) henries per metre of line. The logarithm is dimensionless, as it must be, and any consistent unit may be used for the two distances — which is why Problem 2 could work in centimetres throughout.

The single isolated conductor. Setting \(D_m \to \infty\) makes \(L \to \infty\): an isolated conductor has no finite inductance per unit length. This is not a defect of the formula but a correct statement about the physics — the flux from an infinitely long isolated current extends to infinity and its integral diverges logarithmically.

What rescues it is the condition \(\sum I_k = 0\) of Problem 16. When the currents sum to zero, the fields of the several conductors cancel at large distance faster than \(1/x\), the flux integral converges, and each conductor acquires a well-defined share of the circuit's inductance. Inductance is therefore always a property of a closed circuit, and "the inductance of a conductor" is meaningful only as shorthand for that conductor's contribution to a circuit whose return is specified.

Where this bites in practice: the zero-sequence network of Set 22. Under an earth fault the three phase currents do not sum to zero — the residue returns through the ground — so the derivation collapses and zero-sequence inductance must be computed with an explicit earth return path (Carson's equations), giving values two to three times the positive-sequence figure and dependent on soil resistivity. That dependence, absent from everything in this set, is a direct consequence of losing \(\sum I_k = 0\).

Self-Test

Multiple-Choice Questions

MCQ 1. The GMR of a solid round conductor of radius \(r\) is:
(a) \(r\) (b) \(0.7788r\) (c) \(1.09r\) (d) \(2.177r\)

Show answer
(b). Option (d) is the GMR of a seven-strand conductor in terms of the strand radius.
MCQ 2. The internal inductance of a solid round conductor:
(a) rises with radius (b) falls with radius (c) is independent of radius (d) is zero

Show answer
(c). It is \(\mu_0/8\pi = 0.5\times10^{-7}\) H/m always — see Challenge C1 for why the radius cancels.
MCQ 3. For a balanced three-phase line the inductance per phase is:
(a) \(2\times10^{-7}\ln(D_m/D_s)\) (b) \(4\times10^{-7}\ln(D_m/D_s)\) (c) \(2\times10^{-7}\ln(D_s/D_m)\) (d) \(\mu_0\ln(D_m/D_s)\)

Show answer
(a). The factor of four belongs to a single-phase loop, where both conductors contribute in series.
MCQ 4. A transposed line with spacings \(D_{12}, D_{23}, D_{31}\) behaves as one with equivalent spacing:
(a) \((D_{12}+D_{23}+D_{31})/3\) (b) \(\sqrt[3]{D_{12}D_{23}D_{31}}\) (c) \(\sqrt{D_{12}D_{23}}\) (d) the largest of the three

Show answer
(b) the geometric mean, because inductance depends on \(\ln D\) and the mean of logarithms is the log of the geometric mean.
MCQ 5. The GMR of a two-conductor bundle with sub-conductor GMR \(D_s\) and spacing \(d\) is:
(a) \(D_s + d\) (b) \(\sqrt{D_sd}\) (c) \(\sqrt[3]{D_sd^2}\) (d) \(1.09\sqrt[4]{D_sd^3}\)

Show answer
(b). Options (c) and (d) are the three- and four-conductor bundles.
MCQ 6. Doubling the equivalent spacing of a line increases its inductance by about:
(a) 100% (b) 50% (c) 13% (d) 0%

Show answer
(c) about 13% — the increment is \(0.2\ln2 = 0.139\) mH/km on a typical 1.1 mH/km.
MCQ 7. Bundling reduces line reactance because it:
(a) reduces the resistance (b) increases the GMD (c) increases the GMR (d) reduces the current

Show answer
(c). \(D_{sb}\) rises sixfold for a two-conductor bundle, and it sits in the denominator of the logarithm.
MCQ 8. A line is transposed principally to:
(a) reduce the total inductance (b) equalise the three phase inductances (c) reduce corona (d) improve the power factor

Show answer
(b). Transposition barely changes the average inductance; it removes the asymmetry between phases and the negative-sequence current it produces.
MCQ 9. The steel core of an ACSR conductor is neglected in inductance calculations because:
(a) it is not magnetic (b) it carries negligible current (c) it is outside the aluminium (d) it has zero resistance

Show answer
(b). Its low conductivity and high internal impedance keep the current in the aluminium; the steel's role is mechanical.
MCQ 10. A typical overhead line has a reactance per phase of about:
(a) 0.03 \(\Omega\)/km (b) 0.4 \(\Omega\)/km (c) 4 \(\Omega\)/km (d) 40 \(\Omega\)/km

Show answer
(b). The band 0.3–0.45 is remarkably universal and is the best available check on a computed \(D_s\).
MCQ 11. The flux-linkage formula for one conductor of a group requires:
(a) equal currents (b) currents summing to zero (c) equal spacings (d) a solid conductor

Show answer
(b). Without it the flux integral diverges and no finite inductance per unit length exists — Problem 16 and Challenge C3.
MCQ 12. To halve a line's reactance, it is cheapest to:
(a) double the conductor radius (b) halve the spacing (c) use a bundle (d) double the conductor area

Show answer
(c). Problem 17 finds bundling eighteen times cheaper in metal than enlargement for the same reduction; halving the spacing saves only 0.14 mH/km and is limited by clearance.

Reference

Key Formulas

Quantity	Relation	Notes
GMR, solid round	\(D_s = r' = 0.7788r\)	Absorbs the internal inductance
Internal inductance	\(\mu_0/8\pi = 0.5\times10^{-7}\) H/m	Independent of radius
Inductance, per phase	\(L = 2\times10^{-7}\ln(D_m/D_s)\) H/m	Three-phase, balanced
Inductance, loop	\(L = 4\times10^{-7}\ln(D/D_s)\) H/m	Single-phase two-wire
Working form	\(L = 0.2\ln(D_m/D_s)\) mH/km	\(= 0.4605\log_{10}(D_m/D_s)\)
GMD of two groups	\(D_m = \sqrt[mn]{\prod D_{ij}}\)	Every conductor of one to every conductor of the other
GMR of a group	\(D_s = \sqrt[n^2]{\prod D_{ij}}\)	Includes self-terms \(r'\)
Transposed line	\(D_{eq} = \sqrt[3]{D_{12}D_{23}D_{31}}\)	Geometric, not arithmetic, mean
Seven-strand GMR	\(D_s = 2.177r\)	\(r\) = strand radius
Six-strand (ACSR) GMR	\(D_s = 1.155d\)	\(d\) = strand diameter
Two-conductor bundle	\(D_{sb} = \sqrt{D_sd}\)
Three-conductor bundle	\(D_{sb} = \sqrt[3]{D_sd^2}\)	Equilateral
Four-conductor bundle	\(D_{sb} = 1.09\sqrt[4]{D_sd^3}\)	Square; \(1.09 = 2^{1/8}\)
Flux linkage of one conductor	\(\lambda_1 = 2\times10^{-7}\sum_k I_k\ln(1/D_{1k})\)	Requires \(\sum I_k = 0\)
Reactance	\(X = 2\pi fL = 314.16L\) at 50 Hz	\(L\) in H
Spacing sensitivity	\(\Delta L = 0.2\ln(D_2/D_1)\) mH/km	Depends on the ratio only
Typical values	\(L \approx 1.0\)–\(1.4\) mH/km, \(X \approx 0.3\)–\(0.45\ \Omega\)/km	The check worth carrying

Diagnostics

Common Mistakes

Using the radius where the GMR is meant. The 0.7788 factor is not optional; omitting it understates \(\ln(D_m/D_s)\) and hence the inductance by about 4%.
Using the diameter where the radius is meant. A factor of two inside a logarithm — small enough to look plausible and large enough to be wrong.
Applying the single-phase coefficient \(4\times10^{-7}\) to a three-phase line. Doubles the answer. The four belongs to a loop, the two to a phase — Problem 2.
Mixing units between \(D_m\) and \(D_s\). They appear only as a ratio, so any unit will do — but it must be the same unit. A spacing in metres over a GMR in centimetres is off by 100.
Using the arithmetic mean for the equivalent spacing. It must be the geometric mean, because the physics is logarithmic — Problem 8.
Forgetting the self-terms in a GMR calculation. A group of \(n\) conductors has \(n^2\) distances, of which \(n\) are the conductors to themselves at \(r'\) — Problems 4 and 6.
Confusing bundle spacing with phase spacing. \(d\) is centimetres to tens of centimetres within a bundle; \(D_m\) is metres between phases. Interchanging them gives an absurd answer.
Treating the 1.09 in the four-conductor formula as a typo. It is \(2^{1/8}\), arising from the square's diagonals — Problem 11.
Including the steel core of an ACSR conductor in the GMR. It carries negligible current and is excluded by construction — Problem 5.
Quoting \(L\) when \(X\) was asked, or vice versa. They differ by \(2\pi f\), and \(L\) is frequency-independent while \(X\) is not — Problem 15.
Accepting a reactance far outside 0.3–0.45 \(\Omega\)/km without investigating. Almost always a unit error in \(D_s\).
Applying these formulas to zero-sequence current. They assume \(\sum I_k = 0\), which an earth fault violates — Challenge C3.

Looking Ahead

The series reactance is now accounted for: a conductor diameter, three spacings and a logarithm produce the number that Set 4 simply assumed. Series resistance follows from the material and needs no geometry beyond the cross-section.

Sets 6 and 7 take the same geometry and extract the shunt parameter. Capacitance follows from an almost identical geometric-mean construction, with two differences that repeatedly catch people out: the actual radius replaces the GMR, since there is no internal electric field to absorb, and the earth acts as a mirror, requiring the method of images. Set 8 then finds the limit that geometry finally imposes — the surface voltage gradient at which the air itself begins to conduct.