It is usual, when single-conductor cables are used, to transpose them at regular intervals in order to minimize the disturbance to other circuits.If the cables are arranged in any other formation than at the corners of an equilateral triangle, it is desirable to transpose regularly for the additional reason of equalizing the impedance-drops in each line, which will be different without transposition, due to the asymmetrical arrangement of the cables. It is sometimes convenient to lay the cables in a plane with the middle cable equidistant from the two outer cables, and the impedance in this case will be considered, when the cables are not transposed and when the sheaths are earthed at one point only.It is shown in an Appendix that earthing the sheaths at both ends, so that part of the sheath currents can return through the earth, will not appreciably alter the impedance-drops. The method is readily applicable to any irregular arrangement of cables, but the formulae become rather lengthy and it is only proposed to deal here with the arrangement of the cables in a plane.An example is worked out fully to show the method, and the results of experimental work carried out to verify the theory are given at the end of the paper.In the example chosen, it is shown that in order to deliver a balanced 3-phase load of 500 amperes at a power factor of 0.8 (lagging) and with voltages between lines of 6 000 volts, when the cables are arranged in a plane without transposition, it is necessary at the sending end(5 miles distant) to have a voltage between one outer line (A) and the middle line (B) of 7260 volts, between the other outer line (C) and the middle line (B) a voltage of 7 000 volts, and between the two outer lines a voltage of 7 340 volts, when the sheaths are insulated from each other.If the sheaths are bonded together, but earthed at one end only, the voltages necessary are 7 290 volts between the outer line (A) and the middle line (B); 7 i70 volts between the outer line (C) and the middle line (B); and 7 500 volts between the two outer lines.The unbalance in both cases may be seen to be considerable.If the sheaths are earthed at both ends, and the earth resistance is zero, the voltages required at the sending end become 7 270, 7 170 and 7 500 volts respectively. These figures are very little different from the figures given above, namely, 7 290, 7 170 and 7 500 volts, for the case of bonded sheaths with only one earth point.Results of example worked out in Section (3).Phase sequence of lines is A, B, C, where A and C are the two outer lines.At receiving end, current = 500 amperes, power factor = 0.8. Voltage between lines = 6 000 volts.If, with the cables arranged in a plane, regular transposition of the cables is effected, then the impedance-drop in each of the three lines is the same over an integral multiple of three transpositions, and is less than the average of the three impedance-drops in the untransposed lines. The difference is usually small, but may amount to 10 per cent when the cables are laid close together. The difference diminishes as the cables are spaced further apart.When the cables are regularly transposed, the estimation of the impedance-drops presents no difficulties, for the system is then equivalent to a symmetrically arranged system with axes of cables3√2 times as far apart as the axes of adjacent cables in the plane arrangement.It follows from this that if the cables are laid close together the impedance-drops will always be greater for the plane arrangement with regular transposition than for the symmetrical arrangement. The difference is about 20 per cent when the cables are in contact. The difference diminishes as the cables are spaced further apart. The sheath currents modify these results slightly, but, unless the sheath losses are very large, no appreciable error will be introduced by assuming an equivalent symmetrical arrangement.