For a given voltage and overall diameter of a single core cable there is a certain value of the conductor radius (or core) diameter that gives a minimum potential gradient at the surface. A small conductor radius allows a greater insulation thickness, but on the other hand, the smaller radius of curvature tends to increase the stress; while the effect of too large a diameter of core is to cause an increase of the stress through reduction of the total thickness of the insulation.

**For fixed values of the voltage V and overall diameter D, the g _{max} will be minimum when d log_{e} D/d is maximum and d log_{e} D/d will be maximum if: **

For the low and medium voltage cables, the diameter of the conductor determined from the above consideration, i.e. d = 2V/g_{max} is somewhat less than that determined from the consideration of the safe current density, therefore, the main criterion for determining conductor diameter for such cables is the current carrying capacity.

ADVERTISEMENTS:

On the other hand for high voltage cables the value of core diameter d determined from the above expression would, in general, give a conductor of cross-section much larger than required from the consideration of current carrying capacity.

The desired overall diameter of the conductor without increasing the cross-sectional area can be had in two ways, by making the conductor hollow or by building it with a lead sheath. The latter method has the advantage of eliminating at the same time the increase of stress due to stranding which may be as high as 20%.

**Example: **

ADVERTISEMENTS:

Show that for a given voltage V and maximum stress E_{max} in the dielectric of a single core cable, the sheath diameter D is a minimum when D: d = e where d is conductor diameter and e is the base of natural logarithms. Determine D and d for V = 10 kV and E_{max} = 23 kV/cm.

**Solution: **

Maximum stress, g_{max} = E_{max} = 23 kV/cm = 23 × 10^{5} V/m (assume rms)

Applied voltage, V = 10,000 V

ADVERTISEMENTS:

∴ Diameter of conductor, d = 2V/g_{max} = (2 × 10,000) / (23 × 10^{5}) = 0.0087 m or 8.7 mm Ans.

Internal diameter of sheath, D = e × d = 2.71828 × 8.7 = 23.64 mm Ans.