Transmission Line Capacitance

Set 6 — Transmission Line Capacitance

Twenty worked problems on the parameter that makes a long line behave differently from a short one. Capacitance follows from the same geometric-mean construction as inductance and produces a formula of almost identical shape — but with the actual radius in place of the GMR, and with the earth acting as a mirror that inductance calculations never had to consider. The consequences are large: a charging current that flows whether or not there is a load, and the voltage rise that Set 14 will call the Ferranti effect.

Textbook Chapter 7 · 20 solved · 12 practice · 3 challenge · 12 MCQs

i Method Recap

The master relation. \(C_n = \dfrac{2\pi\varepsilon_0}{\ln(D_m/r)}\) F/m to neutral, with \(\varepsilon_0 = 8.854\times10^{-12}\) F/m so that \(2\pi\varepsilon_0 = 5.563\times10^{-11}\). Typical overhead values are 0.008–0.012 µF/km.
The actual radius, not the GMR. This is the single most important difference from Set 5. Charge resides on the conductor surface, so there is no internal electric field to absorb and no factor of 0.7788. Using \(D_s\) instead of \(r\) overstates the capacitance by about 4%.
Line-to-line and line-to-neutral differ by two. For a single-phase line \(C_{ab} = \pi\varepsilon_0/\ln(D/r)\) and \(C_n = 2C_{ab}\), because the neutral plane sits midway and each conductor sees half the spacing in potential terms. Quoting one where the other is meant is a factor-of-two error.
The earth is a mirror. A conductor at height \(h\) above a perfectly conducting earth behaves as though a conductor of opposite charge sat at depth \(h\) below it. The image raises the capacitance, by a factor \(\ln\!\big[(D/r)\cdot 2h/\sqrt{D^2+4h^2}\big]\) in the denominator — an effect of well under 1% at normal line heights.
Bundling raises capacitance where it lowered inductance. The bundle radius \(r_b = \sqrt{rd}\) enters the denominator, so a two-conductor bundle raises \(C\) by about 35% while cutting \(X_L\) by 26%. Both changes raise the line's power-transfer capability.
Charging current flows with no load connected. \(I_c = 2\pi f C_n V_{ph}\) per unit length, and the charging MVA is \(\sqrt3\,V_LI_c\). On a long line or any cable this is a first-order quantity, not a correction.
Susceptance and reactance move oppositely with length. \(B = 2\pi fC\) adds along the line while \(X_c = 1/B\) falls. Quoting a capacitive reactance without saying whether it is per kilometre or for the whole line is the commonest source of confusion in this set.

VideoWalkthrough

Problem 1Warm-upSingle-Phase Line

A single-phase line operating at 50 Hz has conductors of diameter 1.608 cm spaced 6 m between centres. Find the line-to-line capacitance and the capacitance to neutral, both per kilometre.

Solution

Conductor radius:

\[ r = \frac{1.608}{2} = 0.804\ \text{cm} = 0.00804\ \text{m} \]

Line-to-line capacitance:

\[ C_{ab} = \frac{\pi\varepsilon_0}{\ln(D/r)} = \frac{\pi(8.854\times10^{-12})}{\ln(6/0.00804)} \]

\[ = \frac{2.7816\times10^{-11}}{\ln(746.3)} = \frac{2.7816\times10^{-11}}{6.6152} = 4.205\times10^{-12}\ \text{F/m} \]

Capacitance to neutral is twice this, because the neutral plane lies midway between the conductors and each sees only half the potential difference:

\[ C_n = 2C_{ab} = 8.410\times10^{-12}\ \text{F/m} = 0.00841\ \mu\text{F/km} \]

Equivalently, and more directly:

\[ C_n = \frac{2\pi\varepsilon_0}{\ln(D/r)} = \frac{5.5632\times10^{-11}}{6.6152} = 8.410\times10^{-12}\ \text{F/m}\ \checkmark \]

Two capacitances, one geometry, and a factor of two between them. \(C_{ab}\) is what a bridge measured between the two conductors would read; \(C_n\) is what each conductor presents to the neutral plane, and it is the quantity that goes into a per-phase equivalent circuit. Since every line model from Set 11 onwards is per-phase, \(C_n\) is almost always the one wanted — but data sheets are not always explicit, and a factor of two in the shunt branch is enough to move a Ferranti-effect calculation substantially.

Answer\(C_{ab} = 4.205\) pF/m, \(C_n = 8.410\) pF/m \(= 0.00841\ \mu\text{F}\)/km

Problem 2Challenge-liteRadius or GMR?

Explain why the capacitance formula uses the actual conductor radius \(r\) while the inductance formula uses the geometric mean radius \(D_s = 0.7788r\). Quantify the error made by using the GMR in a capacitance calculation on the line of Problem 1.

Solution

The physical reason. Inductance is computed from magnetic flux linkage, and magnetic flux penetrates the conductor: some of it links only part of the current. That internal flux is real, contributes \(\mu_0/8\pi\) per metre, and the factor 0.7788 exists solely to absorb it into the logarithm.

Capacitance is computed from charge, and in electrostatic equilibrium all the charge on a conductor resides on its surface. The electric field inside is identically zero, so there is nothing internal to absorb:

\[ E_{\text{inside}} = 0 \quad\Rightarrow\quad \text{no internal contribution} \quad\Rightarrow\quad \text{use } r \]

The error if the GMR were used. With \(D_s = 0.7788(0.00804) = 0.006261\) m:

\[ C_n = \frac{5.5632\times10^{-11}}{\ln(6/0.006261)} = \frac{5.5632\times10^{-11}}{\ln(958.3)} = \frac{5.5632\times10^{-11}}{6.8653} = 8.103\times10^{-12}\ \text{F/m} \]

Against the correct 8.410 pF/m:

\[ \frac{8.410 - 8.103}{8.410}\times100 = 3.65\% \]

An understatement of about 3.7% — the same order as the internal-inductance fraction of Set 5 Problem 3, which is no coincidence.

The two formulas look alike and are not alike. The similarity is genuine — both come from a logarithmic potential in two dimensions — but they describe different fields obeying different boundary conditions at the conductor surface. A magnetic field passes through a conductor; an electrostatic field stops at it. Anyone who has memorised "GMD over GMR" as a single incantation will apply it to capacitance and be 4% wrong, in the direction of underestimating the charging current — and 4% is exactly the size of effect that makes a Ferranti calculation disagree with a measurement without being obviously broken.

AnswerCharge lives on the surface, so \(r\) is correct; using \(D_s\) understates \(C_n\) by 3.65%

Problem 3Exam levelThree-Phase Line

Calculate the capacitance to neutral per km of a three-phase line whose conductors are 2 cm in diameter, placed at the corners of a triangle with sides 5 m, 6 m and 7 m. The line is fully transposed and carries a balanced load.

Solution

Conductor radius — and note it is the radius itself, not the GMR:

\[ r = \frac{2}{2} = 1\ \text{cm} = 0.01\ \text{m} \]

Because the line is transposed, the unequal spacings are replaced by their geometric mean, exactly as in Set 5:

\[ D_m = \sqrt[3]{(5)(6)(7)} = \sqrt[3]{210} = 5.944\ \text{m} \]

Capacitance to neutral:

\[ C_n = \frac{2\pi\varepsilon_0}{\ln(D_m/r)} = \frac{5.5632\times10^{-11}}{\ln(5.944/0.01)} = \frac{5.5632\times10^{-11}}{\ln(594.4)} \]

\[ = \frac{5.5632\times10^{-11}}{6.3875} = 8.71\times10^{-12}\ \text{F/m} = 0.00871\ \mu\text{F/km} \]

The value sits inside the usual 0.008–0.012 µF/km band for an overhead line, so no error in \(r\) is indicated.

Transposition serves capacitance exactly as it served inductance, and for the same reason. An untransposed line has three unequal capacitances to neutral, so a balanced set of voltages drives an unbalanced set of charging currents — a standing zero-sequence current that flows even with the line on open circuit. Since the geometric mean appears in both parameters through the same averaging over three sections, one transposition scheme fixes both, which is a small piece of good fortune in an otherwise inconvenient piece of construction.

Answer\(D_m = 5.944\) m, \(C_n = 8.71\) pF/m \(= 0.00871\ \mu\text{F}\)/km

Problem 4Warm-upNeutral Capacitance

Find the capacitance to neutral per km of a single-phase line composed of single-strand conductors of radius 0.328 cm, spaced 3 m apart, neglecting the effect of the earth.

Solution

Working directly with the capacitance-to-neutral form:

\[ C_n = \frac{2\pi\varepsilon_0}{\ln(D/r)} = \frac{2\pi(8.854\times10^{-12})}{\ln\!\left(\dfrac{3}{0.328\times10^{-2}}\right)} \]

Evaluating the logarithm:

\[ \frac{3}{0.00328} = 914.6, \qquad \ln(914.6) = 6.8186 \]

Hence:

\[ C_n = \frac{5.5632\times10^{-11}}{6.8186} = 8.159\times10^{-12}\ \text{F/m} = 0.008159\ \mu\text{F/km} \]

This value is used again in Problem 5, where the earth is restored, and in Problem 6, where the height is varied.

The spacing is 914 times the radius, and only its logarithm matters. That ratio could be 500 or 2000 and the capacitance would change by only \(\pm 20\%\) — which is why overhead-line capacitance clusters so tightly around 0.01 µF/km across every voltage class from 33 to 400 kV. Cables break the pattern completely, because there the "spacing" is millimetres of insulation rather than metres of air, as Problem 14 shows.

Answer\(C_n = 8.159\) pF/m \(= 0.008159\ \mu\text{F}\)/km

Problem 5Exam levelEarth Effect

Repeat Problem 4 with the conductors 7.5 m above the ground, taking the earth into account by the method of images. Compare the two results.

Solution

The method of images. A perfectly conducting earth is an equipotential surface. Its effect on the field above it is reproduced exactly by removing the earth and placing, at depth \(h\) below the original position, an image conductor carrying the opposite charge. The two conductors of the line therefore acquire two images, at distances \(2h\) and \(\sqrt{D^2 + (2h)^2}\) from the originals.

The images modify the potential and hence the capacitance:

\[ C_n = \frac{2\pi\varepsilon_0}{\ln\!\left[\dfrac{D}{r}\times\dfrac{2h}{\sqrt{D^2+(2h)^2}}\right]} \]

Evaluating the image factor with \(D = 3\) m and \(h = 7.5\) m:

\[ \frac{2h}{\sqrt{D^2+(2h)^2}} = \frac{15}{\sqrt{9 + 225}} = \frac{15}{15.297} = 0.9806 \]

The argument of the logarithm is reduced by this factor, so the logarithm shrinks and the capacitance grows:

\[ 914.6 \times 0.9806 = 896.9, \qquad \ln(896.9) = 6.7989 \]

\[ C_n = \frac{5.5632\times10^{-11}}{6.7989} = 8.183\times10^{-12}\ \text{F/m} \]

Comparing with the 8.159 pF/m of Problem 4:

\[ \frac{8.183 - 8.159}{8.159}\times100 = 0.29\% \]

The earth raises the capacitance, always, and by almost nothing. It must raise it: the image is an opposite charge, which draws additional charge onto the conductor for the same potential. But at 7.5 m with 3 m spacing the effect is under a third of a per cent, far smaller than the tolerance on the conductor radius itself, and every practical calculation ignores it. The images matter enormously elsewhere — in the zero-sequence capacitance, where the earth is the return path, and in the surge impedance seen by lightning — but for positive-sequence charging current they are a curiosity.

Answer\(C_n = 8.183\) pF/m with earth against \(8.159\) without — an increase of 0.29%

Problem 6Warm-upHeight Above Ground

For the line of Problem 4, compute the capacitance to neutral at heights of 4 m, 7.5 m and 15 m, and state at what height the earth effect may safely be ignored.

Solution

The image factor \(k = 2h/\sqrt{D^2+(2h)^2}\) at each height, with \(D = 3\) m:

\[ h = 4: \ k = \frac{8}{\sqrt{9+64}} = 0.9363; \qquad h = 7.5: \ k = 0.9806; \qquad h = 15: \ k = \frac{30}{\sqrt{9+900}} = 0.9950 \]

Each reduces the argument 914.6 and hence raises the capacitance:

\[ \begin{array}{cccc} h\ (\text{m}) & k & \ln(914.6k) & C_n\ (\text{pF/m}) \\ \hline 4 & 0.9363 & 6.7528 & 8.238 \\ 7.5 & 0.9806 & 6.7989 & 8.183 \\ 15 & 0.9950 & 6.8135 & 8.165 \\ \infty & 1 & 6.8186 & 8.159 \end{array} \]

As percentages above the no-earth value:

\[ h = 4\ \text{m}: +0.97\%, \qquad h = 7.5\ \text{m}: +0.29\%, \qquad h = 15\ \text{m}: +0.07\% \]

The effect falls away rapidly once \(2h \gg D\), since then \(k \to 1\). For any transmission line — where clearances are several times the phase spacing is not required, but heights of 10 m and more are universal — the earth may be neglected without hesitation.

The effect vanishes as \(D^2/8h^2\), which is why it disappears so fast. Expanding \(k = (1 + D^2/4h^2)^{-1/2} \approx 1 - D^2/8h^2\) shows the correction is second order in the ratio of spacing to height. Doubling the height quarters it. This is a useful thing to know in reverse: for a low-voltage distribution line strung at 6 m with 1 m spacing the correction is 0.35%, while for a busbar a metre above an earthed floor it would be dominant — the formula's neglect is a statement about overhead line geometry, not a general licence.

Answer8.238, 8.183 and 8.165 pF/m at 4, 7.5 and 15 m — negligible above about 10 m

Problem 7Warm-upCapacitive Reactance

A three-phase 50 Hz line has conductors 2.772 cm in diameter spaced 6 m equilaterally. Find the capacitance to neutral per km and the capacitive reactance to neutral per km.

Solution

Radius and capacitance:

\[ r = \frac{2.772}{2} = 1.386\ \text{cm} = 0.01386\ \text{m} \]

\[ C_n = \frac{5.5632\times10^{-11}}{\ln(6/0.01386)} = \frac{5.5632\times10^{-11}}{\ln(432.9)} = \frac{5.5632\times10^{-11}}{6.0706} = 9.164\times10^{-12}\ \text{F/m} \]

\[ C_n = 0.009164\ \mu\text{F/km} \]

Capacitive reactance per kilometre — the capacitance of one kilometre is \(9.164\times10^{-9}\) F:

\[ X_c = \frac{1}{2\pi fC_n} = \frac{1}{314.16(9.164\times10^{-9})} = \frac{1}{2.879\times10^{-6}} = 3.473\times10^{5}\ \Omega\cdot\text{km} \]

The units require care. This is \(3.473\times10^5\) ohm-kilometres: for a line of length \(l\) km the actual reactance to neutral is \(X_c/l\), because capacitances in parallel add and reactances therefore fall.

Series and shunt parameters scale oppositely with length, and the units record it. Series reactance is quoted in \(\Omega\)/km and multiplied by the length; shunt capacitive reactance is quoted in \(\Omega\cdot\)km and divided by it. A 280 km line therefore has 280 times the series reactance and one 280th of the shunt reactance of a single kilometre. Losing track of this is the single most frequent arithmetic error in line calculations, and the surest guard is to work in susceptance, which simply adds.

Answer\(C_n = 0.009164\ \mu\text{F}\)/km, \(X_c = 3.473\times10^5\ \Omega\cdot\text{km}\)

Problem 8Warm-upSusceptance

For the line of Problem 7, find the susceptance to neutral per km, and the total susceptance and capacitive reactance of a 280 km line.

Solution

Susceptance per kilometre:

\[ B = 2\pi fC_n = 314.16(9.164\times10^{-9}) = 2.879\times10^{-6}\ \text{S/km} \]

Susceptances in parallel add, so for the whole line they simply multiply by the length:

\[ B_{\text{total}} = 2.879\times10^{-6} \times 280 = 8.061\times10^{-4}\ \text{S} \]

The corresponding capacitive reactance of the whole line:

\[ X_c = \frac{1}{B_{\text{total}}} = \frac{1}{8.061\times10^{-4}} = 1240\ \Omega \]

Checking against Problem 7's per-km figure, which must be divided by the length:

\[ \frac{3.473\times10^{5}}{280} = 1240\ \Omega\ \checkmark \]

Work in susceptance and the bookkeeping takes care of itself. \(B\) per km multiplied by kilometres gives \(B\) — the same rule as for series reactance, with no inversion anywhere. Converting to \(X_c\) only at the very end removes every opportunity for the length to be applied the wrong way round. This is also why the nominal-\(\pi\) model of Set 11 is written with \(Y/2\) at each end rather than \(2X_c\): admittance is the natural currency for shunt elements.

Answer\(B = 2.879\times10^{-6}\) S/km; for 280 km, \(B = 8.061\times10^{-4}\) S and \(X_c = 1240\ \Omega\)

Problem 9Exam levelCharging Current

The line of Problems 7 and 8 operates at 220 kV. Find the charging current per km and the total charging current of the 280 km line.

Solution

The charging current is driven by the phase voltage against the shunt susceptance:

\[ V_{ph} = \frac{220\,000}{\sqrt3} = 127\,017\ \text{V} \]

Per kilometre:

\[ I_c = V_{ph}B = 127\,017(2.879\times10^{-6}) = 0.3657\ \text{A/km} \]

For the full length:

\[ I_c = 0.3657 \times 280 = 102.4\ \text{A} \]

Equivalently, using the total susceptance of Problem 8:

\[ I_c = 127\,017(8.061\times10^{-4}) = 102.4\ \text{A}\ \checkmark \]

This current flows with the far end of the line open-circuited and no load whatever connected.

A hundred amperes flowing into a line that is delivering nothing. It is purely reactive, so it consumes no real power beyond the \(I^2R\) it causes in the conductors, but it occupies conductor capacity and it must be supplied by the source. At light load it can exceed the load current entirely, in which case the line is a net generator of reactive power — which is the condition that produces the Ferranti voltage rise of Set 14, and the reason Problem 19 finds long lines fitted with shunt reactors to absorb it.

Answer\(0.366\) A/km; \(102.4\) A for the 280 km line

Problem 10Exam levelCharging MVA

Find the total charging MVA of the 280 km, 220 kV line of Problem 9, and express it as a fraction of a 100 MVA base.

Solution

Charging MVA is three-phase reactive power, computed from line quantities:

\[ Q_c = \sqrt3\,V_LI_c = \sqrt3(220\,000)(102.4) = 39.0\times10^{6} = 39.0\ \text{MVAr} \]

Equivalently, and more directly from the susceptance:

\[ Q_c = V_L^{2}B_{\text{total}} = (220\,000)^{2}(8.061\times10^{-4}) = 39.0\ \text{MVAr}\ \checkmark \]

The second form is worth remembering — \(Q = V_L^2B\) needs neither \(\sqrt3\) nor a current, because the \(\sqrt3\) factors cancel between the phase voltage and the three-phase summation.

On a 100 MVA base:

\[ Q_{c,pu} = \frac{39.0}{100} = 0.390\ \text{p.u.} \]

Per hundred kilometres this line generates about 14 MVAr — a figure worth carrying for 220 kV lines generally, since the parameters vary so little from line to line.

Charging MVA rises as the square of the voltage and linearly with length, so it dominates at EHV. The same line at 400 kV would generate \((400/220)^2 = 3.3\) times as much, around 129 MVAr — a substantial power station's worth of reactive output from a piece of wire. This is why EHV lines are commissioned with shunt reactors already installed rather than added later, and why the reactive-power balance of Set 34 treats transmission lines as reactive sources at light load and sinks at heavy load.

Answer\(Q_c = 39.0\) MVAr \(= 0.390\) p.u. on 100 MVA

Problem 11Exam levelBundled Line

A line of equivalent spacing 12 m uses two-conductor bundles at 0.45 m spacing, each sub-conductor having a radius of 0.0125 m. Find the capacitance to neutral per km, and compare with a single conductor of the same radius.

Solution

The bundle radius for capacitance uses the actual radius of the sub-conductor, not its GMR:

\[ r_b = \sqrt{r\,d} = \sqrt{(0.0125)(0.45)} = 0.075\ \text{m} \]

It happens to equal the inductive \(D_{sb}\) of Set 5 Problem 10 only because that problem was given a GMR of 0.0125 m while this one is given a radius of 0.0125 m. For a real conductor the two bundle figures differ by the usual factor.

Bundled:

\[ C_n = \frac{5.5632\times10^{-11}}{\ln(12/0.075)} = \frac{5.5632\times10^{-11}}{\ln(160)} = \frac{5.5632\times10^{-11}}{5.0752} = 1.0962\times10^{-11}\ \text{F/m} \]

\[ = 0.01096\ \mu\text{F/km} \]

Single conductor, same geometry:

\[ C_n = \frac{5.5632\times10^{-11}}{\ln(12/0.0125)} = \frac{5.5632\times10^{-11}}{6.8669} = 8.102\times10^{-12} = 0.00810\ \mu\text{F/km} \]

The increase:

\[ \frac{0.01096 - 0.00810}{0.00810}\times100 = 35.3\% \]

Bundling raises capacitance by 35% while cutting inductance by 26%, and both changes help. Lower \(X_L\) raises the power-transfer limit \(EV\sin\delta/X\) directly; higher \(C\) lowers the surge impedance \(\sqrt{L/C}\) and so raises the surge impedance loading, the natural power a line carries with a flat voltage profile. A bundled line is therefore better on both counts, which — together with the corona reduction of Set 8 — is why nothing above 220 kV is built any other way.

Answer\(r_b = 0.075\) m, \(C_n = 0.01096\ \mu\text{F}\)/km — 35.3% above the single conductor

Problem 12Exam levelDouble Circuit

A transposed double-circuit three-phase line has the two conductors of each phase 21 m apart, each of radius 1.38 cm, and an equivalent GMD between phases of 12.7 m. Find the capacitance to neutral per km and the 50 Hz susceptance.

Solution

The two conductors of one phase form a group of two, so the effective radius follows the two-conductor bundle rule with the very large spacing of 21 m:

\[ r_b = \sqrt{r\,D_{aa'}} = \sqrt{(0.0138)(21)} = \sqrt{0.2898} = 0.538\ \text{m} \]

Capacitance to neutral:

\[ C_n = \frac{2\pi\varepsilon_0}{\ln(D_m/r_b)} = \frac{5.5632\times10^{-11}}{\ln(12.7/0.538)} = \frac{5.5632\times10^{-11}}{\ln(23.61)} \]

\[ = \frac{5.5632\times10^{-11}}{3.1617} = 1.759\times10^{-11}\ \text{F/m} = 0.0176\ \mu\text{F/km} \]

Susceptance to neutral per km:

\[ B = 2\pi fC_n = 314.16(1.759\times10^{-8}) = 5.53\times10^{-6}\ \text{S/km} \]

Roughly double the single-circuit value of Problem 7 — as expected, since there are twice as many conductors, though the widened effective radius makes it slightly more than double.

A double circuit doubles the charging current as well as halving the reactance. Set 5 Problem 13 found the inductance falling to 47% of a single circuit; here the capacitance rises to about 190% of it. Both effects lower the surge impedance \(\sqrt{L/C}\) substantially, which raises the line's natural loading — but they also mean a double-circuit line at light load generates twice the reactive power, and the shunt compensation must be sized for it.

Answer\(r_b = 0.538\) m, \(C_n = 0.0176\ \mu\text{F}\)/km, \(B = 5.53\times10^{-6}\) S/km

Problem 13Exam levelCharging vs Load

The 280 km, 220 kV line of Problem 9 has a thermal rating of 400 A. Express the charging current as a percentage of that rating, and find the load current at which the charging current becomes negligible — say, below 5% of the load.

Solution

From Problem 9 the total charging current is 102.4 A. As a fraction of the thermal rating:

\[ \frac{102.4}{400}\times100 = 25.6\% \]

The charging current is in quadrature with the load current, so they combine as phasors rather than arithmetically. For the charging component to be 5% of the load:

\[ I_{\text{load}} = \frac{102.4}{0.05} = 2048\ \text{A} \]

That is five times the thermal rating and therefore unattainable. The conclusion is unavoidable:

\[ \text{The charging current is never negligible on this line.} \]

Even at full thermal load the charging current is a quarter of it, and the resultant magnitude is \(\sqrt{400^2 + 102.4^2} = 413\) A at the sending end — over the rating, though the two components do not both flow at the same point along the line.

This is precisely why a 280 km line cannot be modelled as a series impedance. The short-line model of Set 9 discards the shunt branch entirely, which would discard a current a quarter the size of the rated load. Somewhere between "negligible" and "dominant" lies the boundary between the short-line and medium-line models, and it falls at around 80 km — the criterion being exactly this comparison. Set 11 takes up the nominal-\(\pi\) model that restores the shunt branch, and Set 12 the distributed model that stops lumping it at all.

Answer25.6% of the thermal rating; the charging current is never negligible on a line of this length

Problem 14Challenge-liteCables

A 132 kV three-phase cable has a capacitance to neutral of 0.25 µF/km and a current rating of 500 A. Find the charging current per km, and the length at which the charging current alone equals the full rating. Compare with an overhead line of 0.009 µF/km.

Solution

Phase voltage and susceptance per km:

\[ V_{ph} = \frac{132\,000}{\sqrt3} = 76\,210\ \text{V}, \qquad B = 314.16(0.25\times10^{-6}) = 7.854\times10^{-5}\ \text{S/km} \]

Charging current per km:

\[ I_c = 76\,210(7.854\times10^{-5}) = 5.99\ \text{A/km} \]

The length at which this alone reaches the rating:

\[ l_{\text{crit}} = \frac{500}{5.99} = 83.5\ \text{km} \]

Beyond this length the cable cannot deliver any load at all — its entire capacity is consumed charging itself.

The overhead comparison. At 0.009 µF/km the charging current would be

\[ I_c = 76\,210(314.16)(0.009\times10^{-6}) = 0.216\ \text{A/km} \]

\[ l_{\text{crit}} = \frac{500}{0.216} = 2320\ \text{km} \]

A ratio of nearly 28 in the capacitance, and therefore in the critical length.

This is the reason submarine and long underground links are direct current. The cable's dielectric is millimetres thick where an overhead line has metres of air, so the logarithm collapses and the capacitance rises by more than an order of magnitude. Beyond roughly 80 km an a.c. cable is transmitting nothing but its own charging current — and unlike an overhead line it cannot be compensated at intervals, since there is nowhere at sea to put a reactor. Converting to d.c. removes the charging current entirely, because a d.c. cable's capacitance draws current only when the voltage changes.

Answer5.99 A/km; critical length \(83.5\) km against \(2320\) km overhead

Problem 15Challenge-liteDuality

Compare the inductance and capacitance formulas of Sets 5 and 6. Identify every structural similarity and every difference, and explain the origin of each difference.

Solution

Setting them side by side:

\[ L = \frac{\mu_0}{2\pi}\ln\frac{D_m}{D_s}\ \text{H/m}, \qquad C_n = \frac{2\pi\varepsilon_0}{\ln(D_m/r)}\ \text{F/m} \]

The similarities, and why they exist:

\[ \begin{array}{ll} \text{Both logarithmic in } D_m & \text{Both fields fall as } 1/x \text{ in two dimensions} \\ \text{Both use the same } D_m & \text{The same conductor geometry produces both} \\ \text{Both use a permeability/permittivity of free space} & \text{Air is neither magnetic nor dielectric to any useful degree} \end{array} \]

The differences, and their origins:

\[ \begin{array}{ll} D_s \text{ against } r & \text{Magnetic flux enters the conductor; electric flux does not} \\ \ln \text{ in numerator against denominator} & L \text{ is flux per current; } C \text{ is charge per volt} \\ \text{Earth ignored against earth by images} & \text{Earth is a magnetic non-entity but an electrostatic mirror} \end{array} \]

The third difference deserves a note. The earth affects capacitance because it is a conductor and therefore an equipotential — it constrains the electric field. It barely affects positive-sequence inductance because at 50 Hz its permeability is that of free space and the balanced phase currents sum to zero, leaving little net field to interact with it. Under zero-sequence conditions, where the currents do not sum to zero, the earth becomes the return path and dominates the inductance entirely.

The reciprocal placement of the logarithm is the deepest of the differences. Widening the spacing increases inductance and decreases capacitance, so the two always move oppositely. That is what makes the surge impedance \(\sqrt{L/C}\) so much more sensitive to geometry than either parameter alone — and, conversely, what makes the product \(LC\) almost independent of geometry, as Problem 16 shows with a result that is one of the most striking in the whole subject.

AnswerSame \(D_m\) and logarithmic form; differ in \(D_s\) vs \(r\), reciprocal placement, and the earth's role

Problem 16Challenge-liteVelocity of Propagation

Neglecting the internal inductance (that is, taking \(D_s = r\)), form the product \(LC\) for an overhead line and evaluate \(1/\sqrt{LC}\). Interpret the result.

Solution

With \(D_s = r\) the two logarithms are identical and cancel in the product:

\[ LC = \left[\frac{\mu_0}{2\pi}\ln\frac{D_m}{r}\right]\left[\frac{2\pi\varepsilon_0}{\ln(D_m/r)}\right] = \mu_0\varepsilon_0 \]

Every trace of the geometry has disappeared. The spacing, the conductor size, the arrangement of the phases — none of it survives.

Evaluating numerically:

\[ LC = (4\pi\times10^{-7})(8.854\times10^{-12}) = 1.1127\times10^{-17}\ \text{s}^{2}/\text{m}^{2} \]

Hence:

\[ \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1.1127\times10^{-17}}} = \frac{1}{3.336\times10^{-9}} = 2.998\times10^{8}\ \text{m/s} \]

Interpretation. This is the speed of light in vacuum. Disturbances travel along an overhead line at essentially \(c\), because the field between the conductors is in air and it is the field, not the conductor, that carries the wave.

\[ v = \frac{1}{\sqrt{\mu_0\varepsilon_0}} = c \]

Restoring the internal inductance makes \(L\) about 4% larger, so the real velocity is roughly 2% below \(c\) — around \(2.94\times10^8\) m/s. For a cable, where the dielectric has \(\varepsilon_r \approx 2.3\), the velocity falls to about \(c/\sqrt{2.3} = 0.66c\).

This is the result that turns a line from a circuit element into a waveguide. At 50 Hz the wavelength is \(c/f = 6000\) km, so a 300 km line is a twentieth of a wavelength and a lumped model is defensible; a 1000 km line is a sixth of one and it is not. The whole distinction between the short, medium and long line models of Sets 9 to 12 is a statement about what fraction of 6000 km the line occupies — and this problem is where that 6000 km comes from. It is also why travelling-wave protection can locate a fault by timing, at roughly 300 m per microsecond.

Answer\(LC = \mu_0\varepsilon_0\), independent of geometry; \(1/\sqrt{LC} = 2.998\times10^8\) m/s \(= c\)

Problem 17Exam levelPer-Unit Admittance

Express the total shunt susceptance of the 280 km, 220 kV line of Problem 8 in per-unit on a base of 100 MVA, 220 kV.

Solution

The base admittance is the reciprocal of the base impedance:

\[ Y_B = \frac{1}{Z_B} = \frac{\text{MVA}_B}{\text{kV}_B^{2}} = \frac{100}{(220)^{2}} = \frac{100}{48\,400} = 2.066\times10^{-3}\ \text{S} \]

The total susceptance from Problem 8:

\[ B_{\text{total}} = 8.061\times10^{-4}\ \text{S} \]

Hence:

\[ B_{pu} = \frac{8.061\times10^{-4}}{2.066\times10^{-3}} = 0.390\ \text{p.u.} \]

Cross-checking against the charging MVA of Problem 10, since at 1.0 p.u. voltage the per-unit susceptance and the per-unit reactive power are numerically equal:

\[ Q_{c,pu} = V_{pu}^{2}B_{pu} = (1.0)^{2}(0.390) = 0.390\ \text{p.u.}\ \checkmark \]

Susceptance in per-unit and charging MVA in per-unit are the same number at rated voltage, and that is not a coincidence. \(Q = V^2B\) in any units, and per-unit is built so that \(V = 1\) at rated voltage. This makes the per-unit susceptance immediately meaningful as a reactive-power figure — 0.390 means "this line generates 39% of base MVA in vars when unloaded" — and it is why load-flow programs report line charging in per-unit without further comment.

Answer\(B_{pu} = 0.390\) p.u. on 100 MVA, 220 kV

Problem 18Warm-upConductor Size

For a line of fixed 6 m spacing, find the capacitance to neutral for conductor radii of 0.5 cm, 1 cm and 2 cm, and comment on the sensitivity.

Solution

Applying \(C_n = 2\pi\varepsilon_0/\ln(6/r)\) at each radius:

\[ \begin{array}{cccc} r\ (\text{cm}) & 6/r & \ln(6/r) & C_n\ (\text{pF/m}) \\ \hline 0.5 & 1200 & 7.090 & 7.847 \\ 1.0 & 600 & 6.397 & 8.696 \\ 2.0 & 300 & 5.704 & 9.753 \end{array} \]

Quadrupling the radius from 0.5 to 2 cm raises the capacitance by

\[ \frac{9.753 - 7.847}{7.847}\times100 = 24.3\% \]

Each doubling of the radius adds a fixed amount to the denominator's reciprocal — the increments are 10.8% and 12.2%, growing slowly because the logarithm is shrinking.

Sixteen times the metal buys a quarter more capacitance, which is why nobody sizes a conductor for its capacitance. Conductor size is decided by current-carrying capacity, by short-circuit withstand and — at EHV — by corona, and the resulting capacitance is simply accepted. Where capacitance genuinely needs to be raised, bundling does it far more cheaply, as Problem 11 showed: two sub-conductors gave 35% for twice the metal, against 24% for sixteen times.

Answer7.85, 8.70 and 9.75 pF/m — a 24% rise for a fourfold radius

Problem 19Exam levelShunt Compensation

The 280 km, 220 kV line of Problem 10 generates 39 MVAr of charging. A shunt reactor is to be installed to absorb 60% of it. Find the reactor rating, its reactance and its inductance.

Solution

Required reactor rating:

\[ Q_L = 0.60 \times 39.0 = 23.4\ \text{MVAr} \]

A three-phase shunt reactor connected across the line absorbs \(Q = V_L^2/X_L\), so:

\[ X_L = \frac{V_L^{2}}{Q_L} = \frac{(220\times10^{3})^{2}}{23.4\times10^{6}} = \frac{4.84\times10^{10}}{2.34\times10^{7}} = 2068\ \Omega\ \text{per phase} \]

Its inductance:

\[ L = \frac{X_L}{2\pi f} = \frac{2068}{314.16} = 6.58\ \text{H} \]

The remaining uncompensated charging:

\[ 39.0 - 23.4 = 15.6\ \text{MVAr} \]

Compensation is deliberately partial. Full compensation would leave the line with no reactive generation at all at light load, and would then require reactive support the moment load appears.

A 6.6 henry reactor is a substantial machine — several metres tall, oil-filled, and costing a significant fraction of a substation bay. It exists solely to cancel the effect of the geometry computed in Problem 3, and it is switched in at light load and out at heavy load, because the line's reactive balance reverses as the loading passes surge impedance loading. The economics of that switching, and the choice of how much to compensate, is the subject of Set 34; the parameter that forces the question is the \(0.00871\ \mu\text{F}\)/km with which this set began.

Answer\(Q_L = 23.4\) MVAr, \(X_L = 2068\ \Omega\)/phase, \(L = 6.58\) H

Problem 20Exam levelGeometry to Charging

A 400 kV, 300 km line uses two-conductor bundles at 0.45 m spacing with sub-conductor radius 0.0125 m, the bundles arranged horizontally at 10 m between adjacent phases and transposed. Carry the calculation from geometry through to charging MVA and per-unit susceptance on a 100 MVA base.

Solution

Step 1 — bundle radius:

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

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

\[ D_m = \sqrt[3]{(10)(10)(20)} = 12.60\ \text{m} \]

Step 3 — capacitance to neutral:

\[ C_n = \frac{5.5632\times10^{-11}}{\ln(12.60/0.075)} = \frac{5.5632\times10^{-11}}{\ln(168)} = \frac{5.5632\times10^{-11}}{5.1240} = 1.0857\times10^{-11}\ \text{F/m} \]

\[ = 0.01086\ \mu\text{F/km} \]

Step 4 — susceptance:

\[ B = 314.16(1.0857\times10^{-8}) = 3.411\times10^{-6}\ \text{S/km}, \qquad B_{\text{total}} = 300(3.411\times10^{-6}) = 1.023\times10^{-3}\ \text{S} \]

Step 5 — charging MVA:

\[ Q_c = V_L^{2}B_{\text{total}} = (400\times10^{3})^{2}(1.023\times10^{-3}) = 164\ \text{MVAr} \]

Step 6 — per-unit on 100 MVA, 400 kV:

\[ Y_B = \frac{100}{(400)^{2}} = 6.25\times10^{-4}\ \text{S}, \qquad B_{pu} = \frac{1.023\times10^{-3}}{6.25\times10^{-4}} = 1.637\ \text{p.u.} \]

164 MVAr of reactive generation from 300 km of wire — more than the base MVA of the study. Compare Set 5 Problem 20, where the same line's series reactance came to only 0.060 p.u. A 400 kV line is electrically short in series terms and enormous in shunt terms, and that asymmetry is the defining feature of EHV transmission. It is why such lines are commissioned with reactors, why they must never be energised from one end without them, and why the Ferranti effect of Set 14 is a design constraint at 400 kV where at 33 kV it is a curiosity.

Answer\(C_n = 0.01086\ \mu\text{F}\)/km, \(B = 1.023\times10^{-3}\) S, \(Q_c = 164\) MVAr, \(B_{pu} = 1.637\)

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 three-phase line has conductors of radius 1 cm spaced 4 m equilaterally. Find \(C_n\) per km.

Show answer
\(5.5632\times10^{-11}/\ln(400) = 5.5632\times10^{-11}/5.9915 = \mathbf{9.29}\) pF/m \(= 0.00929\ \mu\)F/km.
P2. The same line at 132 kV. Find the charging current per km.

Show answer
\(V_{ph} = 76\,210\) V; \(B = 314.16(9.29\times10^{-9}) = 2.919\times10^{-6}\) S/km; \(I_c = \mathbf{0.222}\) A/km.
P3. Find the charging MVA of 150 km of that line.

Show answer
\(B_{\text{tot}} = 4.379\times10^{-4}\) S; \(Q = (132\,000)^2(4.379\times10^{-4}) = \mathbf{7.63}\) MVAr.
P4. A single-phase line has \(C_{ab} = 5\) pF/m. What is \(C_n\)?

Show answer
\(C_n = 2C_{ab} = \mathbf{10}\) pF/m. The neutral plane halves the potential each conductor works against.
P5. Why does using the GMR instead of the radius understate capacitance, and by roughly how much?

Show answer
\(D_s < r\), so \(\ln(D_m/D_s) > \ln(D_m/r)\) and \(C\) comes out smaller — by about \(\mathbf{4\%}\). There is no internal electric field to absorb.
P6. A conductor of radius 1 cm sits 8 m above ground with 3 m spacing. Find the image factor.

Show answer
\(2h/\sqrt{D^2+4h^2} = 16/\sqrt{9+256} = 16/16.279 = \mathbf{0.983}\) — a capacitance increase of about 0.25%.
P7. A line has \(B = 3\times10^{-6}\) S/km. Find the capacitive reactance to neutral of 200 km.

Show answer
\(B_{\text{tot}} = 6\times10^{-4}\) S, so \(X_c = \mathbf{1667}\ \Omega\). Add susceptances, then invert once.
P8. A two-conductor bundle has sub-conductor radius 0.015 m at 0.4 m spacing. Find the bundle radius for capacitance.

Show answer
\(\sqrt{(0.015)(0.4)} = \mathbf{0.0775}\) m — using the actual radius, unlike the inductance case.
P9. A 220 kV cable has \(C_n = 0.20\ \mu\)F/km and a rating of 600 A. Find its critical length.

Show answer
\(I_c = 127\,017(314.16)(0.2\times10^{-6}) = 7.98\) A/km, so \(l_{\text{crit}} = 600/7.98 = \mathbf{75}\) km.
P10. Why is \(1/\sqrt{LC}\) independent of the line's geometry?

Show answer
The logarithms cancel between \(L\) and \(C\), leaving \(LC = \mu_0\varepsilon_0\). The velocity is \(\mathbf{c}\), because the wave travels in the air, not the metal.
P11. A line generates 50 MVAr of charging. Size a shunt reactor for 70% compensation at 400 kV.

Show answer
\(Q_L = 35\) MVAr; \(X_L = (400\times10^3)^2/35\times10^6 = \mathbf{4571}\ \Omega\); \(L = \mathbf{14.6}\) H.
P12. A calculation gives \(C_n = 0.9\ \mu\)F/km for an overhead line. What is wrong?

Show answer
A hundred times the usual 0.008–0.012 band. Almost certainly a unit error in the radius — metres used where centimetres were meant, collapsing the logarithm.

Challenge

Challenge Problems

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

C1. Derive the capacitance to neutral of a single-phase line from first principles, and show precisely where the boundary condition on the conductor surface forces the use of \(r\) rather than any effective radius.

Show answer
Setup. Two long parallel conductors of radius \(r\), spacing \(D \gg r\), carrying charges \(+q\) and \(-q\) per metre.

Field of one conductor. By Gauss's law on a coaxial cylinder of radius \(x\) outside the conductor, \(E = q/2\pi\varepsilon_0 x\). Inside the conductor \(E = 0\) identically, since the charge is a surface charge and the metal is an equipotential — this is the boundary condition that decides everything.

Potential difference. Integrating from the surface of conductor 1 to the surface of conductor 2 along the line joining them, and superposing both charges:
\[ V_{12} = \frac{q}{2\pi\varepsilon_0}\ln\frac{D-r}{r} + \frac{q}{2\pi\varepsilon_0}\ln\frac{D-r}{r} \approx \frac{q}{\pi\varepsilon_0}\ln\frac{D}{r} \]
Capacitance:
\[ C_{ab} = \frac{q}{V_{12}} = \frac{\pi\varepsilon_0}{\ln(D/r)}, \qquad C_n = 2C_{ab} = \frac{2\pi\varepsilon_0}{\ln(D/r)} \]
Where \(r\) is forced. The integration begins at \(x = r\), the conductor surface, because that is where the field begins to exist. There is no contribution from \(x < r\) to include, so no effective radius can arise. Contrast the magnetic case, where the integration must begin at \(x = 0\) and the internal region contributes the \(\mu_0/8\pi\) that the 0.7788 absorbs.

The \(D \gg r\) approximation replaces \(D - r\) by \(D\) and assumes the charge is uniformly distributed round each conductor. The second assumption is the weaker one: proximity draws charge preferentially towards the facing sides. For \(D/r > 20\) the error is under 0.1%, and overhead lines have \(D/r\) in the hundreds — but a busbar pair at \(D/r = 4\) would need the exact treatment.
C2. Charging current flows whether or not a line is loaded. Trace its consequences through voltage profile, reactive balance, protection and switching, and explain why the same phenomenon is a nuisance at 400 kV and irrelevant at 11 kV.

Show answer
Voltage profile. The charging current is capacitive, so it leads and flows through the line's series inductance to produce a voltage rise towards the open end — the Ferranti effect of Set 14. On a 400 kV line of a few hundred kilometres the rise can exceed 10%, taking equipment beyond its rated insulation.

Reactive balance. A line generates \(V^2B\) and absorbs \(I^2X_L\). At light load the first dominates and the line is a var source; at heavy load the second dominates and it is a var sink. The crossover is surge impedance loading. Consequently the same line needs reactors at night and capacitors at peak, and the switching schedule of Set 34 exists to manage exactly this reversal.

Protection. Charging current is a standing current with no fault present, so differential protection must be compensated for it or it will see a permanent apparent differential. Distance relays see the line's apparent impedance shifted by it. Both are routine but neither is free.

Switching. Energising an unloaded line means switching a large capacitance, which draws a high-frequency inrush and can produce restrikes in the breaker; de-energising traps charge on the line, so the conductor can sit at peak voltage until it bleeds away, and reclosing onto trapped charge doubles the stress. Line breakers are specified for capacitive switching duty for this reason.

Why 400 kV and not 11 kV. Charging MVA is \(V^2B\) while load MVA is \(\sqrt3 VI\) — so the ratio of charging to load rises linearly with voltage. It also rises with length, and high-voltage lines are long while 11 kV feeders are short. Problem 20 found 164 MVAr from 300 km at 400 kV; a 5 km, 11 kV feeder of the same construction would generate about 2 kVAr, which nothing in the system would notice.
C3. The earth is treated as a perfect conductor in the method of images. Examine that assumption: what does real soil resistivity do, when does it matter, and why does it matter far more for zero-sequence quantities than for the positive-sequence capacitance computed in this set?

Show answer
What the assumption does. A perfect conductor forces the earth's surface to be an exact equipotential, so the image sits at depth exactly \(h\) and carries exactly \(-q\). Real soil has finite conductivity, so the equipotential is imperfect and the effective image sits deeper than \(h\), at a complex depth that depends on frequency and resistivity.

Carson's depth. The standard result gives an equivalent depth \(D_e \approx 658.5\sqrt{\rho/f}\) metres, with \(\rho\) in \(\Omega\)m. For \(\rho = 100\) and \(f = 50\) this is about 930 m — three orders of magnitude below the conductor rather than the 7.5 m of a perfect mirror.

Why capacitance barely cares. Two reasons. First, the capacitive effect of the earth was already tiny — 0.29% in Problem 5 — so even a large fractional change in it is negligible. Second, and more fundamentally, capacitance is an electrostatic quantity, and at 50 Hz even poor soil is an excellent electrostatic conductor: the relaxation time \(\varepsilon\rho\) is microseconds, far shorter than a cycle. The perfect-mirror assumption is genuinely good for the electric field.

Why zero-sequence inductance cares enormously. Under zero-sequence conditions the earth carries real current, and the loop area is set by that 930 m equivalent depth rather than by the metres between conductors. The result is a zero-sequence inductance two to three times the positive-sequence value, and one that varies with soil resistivity by a factor of two or more between wet clay and dry rock. Earth-fault current calculations and earth-fault relay settings therefore depend on a soil survey in a way that nothing in this set does — and that dependence, together with the loss of \(\sum I_k = 0\) noted in Set 5 Challenge C3, is why Set 22 treats the zero-sequence network as a wholly separate construction rather than a variation on this one.

Self-Test

Multiple-Choice Questions

MCQ 1. In the capacitance formula the denominator contains:
(a) \(\ln(D_m/D_s)\) (b) \(\ln(D_m/r)\) (c) \(\ln(r/D_m)\) (d) \(\ln(D_s/D_m)\)

Show answer
(b). The actual radius, because charge resides on the surface and there is no internal electric field to absorb.
MCQ 2. Using the GMR instead of the radius in a capacitance calculation:
(a) overstates \(C\) by 4% (b) understates \(C\) by 4% (c) has no effect (d) doubles \(C\)

Show answer
(b). \(D_s < r\) makes the logarithm larger and the capacitance smaller — Problem 2.
MCQ 3. For a single-phase line, \(C_n\) relates to \(C_{ab}\) as:
(a) \(C_n = C_{ab}\) (b) \(C_n = 2C_{ab}\) (c) \(C_n = C_{ab}/2\) (d) \(C_n = \sqrt3 C_{ab}\)

Show answer
(b). The neutral plane lies midway, so each conductor works against half the potential difference.
MCQ 4. Accounting for the earth by the method of images:
(a) reduces \(C\) (b) increases \(C\) (c) leaves \(C\) unchanged (d) may do either

Show answer
(b) increases — the image is an opposite charge, drawing more charge onto the conductor for the same potential. The effect is under 1% at normal heights.
MCQ 5. A typical overhead line has a capacitance to neutral of about:
(a) 0.001 µF/km (b) 0.01 µF/km (c) 0.1 µF/km (d) 1 µF/km

Show answer
(b). The band 0.008–0.012 is remarkably universal. Option (c) is the order for a cable.
MCQ 6. Susceptance and capacitive reactance scale with line length as:
(a) both increase (b) both decrease (c) \(B\) increases, \(X_c\) decreases (d) \(B\) decreases, \(X_c\) increases

Show answer
(c). Capacitances in parallel add, so \(B\) adds and \(X_c = 1/B\) falls — Problem 8.
MCQ 7. Bundling a line changes its parameters by:
(a) raising both \(L\) and \(C\) (b) lowering both (c) lowering \(L\), raising \(C\) (d) raising \(L\), lowering \(C\)

Show answer
(c). Both changes lower the surge impedance and raise the transfer capability — Problem 11.
MCQ 8. The charging current of a line flows:
(a) only under load (b) only under fault (c) whenever the line is energised (d) only at the sending end

Show answer
(c). It is driven by the voltage, not the load, and flows with the far end open — Problem 9.
MCQ 9. The charging MVA of a line is most directly computed as:
(a) \(\sqrt3 V_LI_c\) (b) \(V_L^2B\) (c) \(V_{ph}^2B\) (d) \(3V_L^2B\)

Show answer
(b). Option (a) is also correct but needs the current first; \(V_L^2B\) needs neither a current nor a \(\sqrt3\) — Problem 10.
MCQ 10. Long a.c. cables are impractical chiefly because:
(a) their resistance is too high (b) their charging current consumes the rating (c) they overheat (d) their inductance is too high

Show answer
(b). The critical length is around 80 km, beyond which no load can be delivered at all — Problem 14.
MCQ 11. The product \(LC\) for an overhead line, neglecting internal inductance, equals:
(a) \(\mu_0\varepsilon_0\) (b) \(\mu_0/\varepsilon_0\) (c) depends on spacing (d) depends on conductor radius

Show answer
(a). The logarithms cancel and every trace of geometry disappears, giving a propagation velocity of \(c\) — Problem 16.
MCQ 12. A shunt reactor is fitted to a long EHV line in order to:
(a) improve the power factor of the load (b) absorb the line's charging vars (c) reduce the series reactance (d) limit the fault current

Show answer
(b). It counteracts the reactive generation of the line's own capacitance and the voltage rise it causes — Problem 19.

Reference

Key Formulas

Quantity	Relation	Notes
Permittivity of free space	\(\varepsilon_0 = 8.854\times10^{-12}\) F/m	\(2\pi\varepsilon_0 = 5.5632\times10^{-11}\)
Capacitance to neutral	\(C_n = 2\pi\varepsilon_0/\ln(D_m/r)\) F/m	Actual radius, not GMR
Line-to-line	\(C_{ab} = \pi\varepsilon_0/\ln(D/r)\)	\(C_n = 2C_{ab}\)
Transposed line	\(D_m = \sqrt[3]{D_{12}D_{23}D_{31}}\)	Same \(D_m\) as for inductance
Earth effect	\(C_n = \dfrac{2\pi\varepsilon_0}{\ln\!\big[\frac{D}{r}\cdot\frac{2h}{\sqrt{D^2+4h^2}}\big]}\)	Raises \(C\); under 1% above 10 m
Image factor	\(k = 2h/\sqrt{D^2+4h^2} \approx 1 - D^2/8h^2\)	Second order in \(D/h\)
Two-conductor bundle	\(r_b = \sqrt{rd}\)	Actual radius \(r\)
Three-conductor bundle	\(r_b = \sqrt[3]{rd^2}\)
Four-conductor bundle	\(r_b = 1.09\sqrt[4]{rd^3}\)
Susceptance	\(B = 2\pi fC_n = 314.16C_n\) at 50 Hz	S per unit length; adds with length
Capacitive reactance	\(X_c = 1/B\)	\(\Omega\cdot\)km; divide by length
Charging current	\(I_c = V_{ph}B\)	Flows with the line unloaded
Charging MVA	\(Q_c = V_L^2B = \sqrt3 V_LI_c\)	First form needs no current
Per-unit susceptance	\(B_{pu} = B/Y_B\), \(Y_B = \text{MVA}_B/\text{kV}_B^2\)	Equals \(Q_{c,pu}\) at 1.0 p.u. volts
Shunt reactor	\(X_L = V_L^2/Q_L\) per phase	Sized as a fraction of \(Q_c\)
Propagation velocity	\(v = 1/\sqrt{LC} = c\) for an overhead line	Wavelength 6000 km at 50 Hz
Typical values	\(C_n \approx 0.008\)–\(0.012\ \mu\text{F}\)/km	Cables: 0.2–0.4 µF/km

Diagnostics

Common Mistakes

Using the GMR instead of the radius. The single most frequent error in this set, carried across by habit from Set 5. It understates \(C\) by about 4% — Problem 2.
Confusing \(C_n\) with \(C_{ab}\). A factor of two, and per-phase models always want \(C_n\) — Problem 1.
Multiplying the capacitive reactance by the line length. It must be divided. Work in susceptance, which adds, and invert only at the end — Problems 7 and 8.
Using the line voltage where the phase voltage is meant in \(I_c = V B\). A factor of \(\sqrt3\) — but note that \(Q_c = V_L^2B\) correctly uses the line voltage.
Assuming the earth effect is significant. It is under 0.3% at 7.5 m and falls as \(D^2/8h^2\) — Problems 5 and 6.
Assuming the earth effect reduces capacitance. It always increases it, because the image carries opposite charge.
Forgetting that transposition applies to capacitance too. An untransposed line has three unequal shunt capacitances and a standing zero-sequence charging current — Problem 3.
Neglecting the shunt branch on a line of any length. At 280 km the charging current was a quarter of the thermal rating — Problem 13.
Applying overhead-line capacitance figures to cables. Cables are 20 to 40 times higher, which is what makes their critical length so short — Problem 14.
Sizing a conductor to change its capacitance. Sixteen times the metal buys 24% — bundling is far cheaper — Problem 18.
Quoting \(C\) in F/m when µF/km was wanted, or the reverse. They differ by \(10^{-9}\): \(1\) pF/m \(= 10^{-3}\ \mu\)F/km.
Accepting a capacitance far outside 0.008–0.012 µF/km for an overhead line. Almost always a unit error in the radius or the spacing.

Looking Ahead

Both line parameters are now derived from the same geometry: a series reactance around 0.4 \(\Omega\)/km and a shunt susceptance around \(3\times10^{-6}\) S/km, from a conductor radius and three spacings. The line model is complete.

Set 7 applies the same machinery to bundled and multi-circuit arrangements in more detail, and Set 8 finds the limit that geometry finally imposes — the surface voltage gradient at which the air ionises, which decides the minimum conductor size at EHV and therefore feeds back into both parameters computed here. From Set 9 the parameters stop being derived and start being used: first in the short-line model that discards this set's capacitance entirely, then in the medium and long models that restore it, and finally in the wave equation of Set 12, where the 6000 km wavelength found in Problem 16 becomes the governing length scale.