A single core cable is equivalent to two long co-axial cylinders. The inner cylinder is the conductor itself while the outer cylinder is the lead sheath. The lead sheath is always at earth potential.

Let d = Conductor diameter

D = Total diameter with sheath

The co-axial cylindrical from of cable and its section are shown in the Fig. 1(a) and (b).

Fig. 1 Capacitance of single core cable |

Let Q = Charge per meter length of conductor in coulombs

ε = Permittivity of material between core and sheath

Now ε = ε

_{o}ε_{r} Where ε

_{o}= permittivity of free space = 8.854 x 10^{-12}F/m and ε

_{r }= Relative permittivity of the medium Consider an elementary cylinder with radius x and axial length of 1 m. The thickness of the cylinder is dx.

According to Gauss's theorem, the lines of flux emanating due to charge Q on the conductor are in parallel direction and total flux line are equal to the total charge possessed i.e. Q lines. As lines are in radial direction, the cross-sectional area through which lines pass is surface area. For a cylinder with radius x, the surface area is (2πx x axial length) m

The electric field intensity at any point P on the elementary cylinder is given by,

^{2}. As axial length considered is 1 m, the surface area is 2 π x m^{2}.The electric field intensity at any point P on the elementary cylinder is given by,

g

Note that as length considered is 1 m, the capacitance is F/m.

If required for length 'l' multiply c by 'l'.

Substituting value of ε

If the length l of cable is known then the total capacitance of cable is,

X

Then the charging current of the cable is given by,

Where V

The electrical stress in insulation is the electric field intensity acting at any point P in insulation.

The stress is maximum at the surface of the conductor i.e. when x = r.

Similar the minimum stress will be at the length i.e. x = R hence

The variation of stress in dielectric material is shown ion the Fig. 2.

The ratio of maximum and minimum stress is,

In practice, the maximum stress value should be as low as possible. When the voltage V and sheath diameter D are fixed, the only parameter to be selected is the core diameter d. So d should be selected for which value is minimum.

The value of will be minimum when ∂g

Now the value of ∂g

The value of minimum g

3. Using stranded copper conductors around a lead tube instead of hemp.

_{x }= Dx/ε where Dx = Electric flux density Hence the work done in moving a unit charge through a distance dx in the direction of an electric field is g

_{x }dx. Therefore the work done in moving a unit charge from the conductor to sheath is the potential difference between the conductor and the sheath given by,

The capacitance of a cable is given by,Note that as length considered is 1 m, the capacitance is F/m.

If required for length 'l' multiply c by 'l'.

Substituting value of ε

_{o},If the length l of cable is known then the total capacitance of cable is,

**Note**: To avoid the confusion of units, students can use the expression given by equation (1), to calculate capacitance while solving the problem.

**Charging current**: When the capacitance C of a cable is known then its reactance is given by,X

_{c }= 1/(ωC) = 1/(2πfC) ΩThen the charging current of the cable is given by,

Where V

_{ph }= Phase voltage between core and sheath = V_{line}/√3__1.1 Stress in Insulation__

The stress is maximum at the surface of the conductor i.e. when x = r.

Similar the minimum stress will be at the length i.e. x = R hence

The variation of stress in dielectric material is shown ion the Fig. 2.

Fig. 2 |

The ratio of maximum and minimum stress is,

**Key Point**: If value of voltage used is r.m.s. we get r.m.s. values of stresses and if value of voltage used is peak, we get peak values of stresses.__1.2 Economical core Diameter__

The value of will be minimum when ∂g

_{max}/∂d = 0Now the value of ∂g

_{max}/∂d must be zero to get minimum g_{max}.**Key Point**: The core diameter must be 1/2.718 times the sheath diameter D so as to give the minimum value of g_{max}.The value of minimum g

_{max}is,**.**Minimum g^{.}._{max }= V/r ..... As r = d/2 For high voltage cable, for a required if d is determined by the expression (5), it gives very large values of d than required for current carrying capacity. And such extra copper required can increase the cost tremendously. Hence ti increase d without the use of an extra copper following methods are used :

1. Aluminium is used instead of copper as the aluminium size is more than copper for the same current carrying capacity.

2. Using stranded copper conductors around a dummy core of tube instead of hemp.3. Using stranded copper conductors around a lead tube instead of hemp.

**Read examples on cables**
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