Insulation Resistance of a Cable

Insulation Resistance of a Cable

The Fig. 6.6.1 shows the section of a single core cable which is insulated with the help of layer of an insulating material.

In such cables, the leakage current flows radially from centre towards the surface as shown in the Fig. 6.6.1. Hence the cross-section of the path of such current is not constant but changes with its length. The resistance offered by cable to path of the leakage current is called an insulation resistance. So to calculate the insulation resistance consider an elementary section of the cylindrical cable of radius x and the thickness dx as shown in the Fig. 6.6.2. Let us find the resistance of this elementary ring. 

Let d = Diameter of core

r = d / 2 = Radius of core

D = Diameter with sheath

R = D / 2 = Radius of cable with sheath

As the leakage current flows radially outwards, the length along which the current flows in an elementary ring is dx. While the cross-sectional area perpendicular to the flow of current depends on the length I of the cable.

Cross-section area = Surface area for length l of cable

=  (2πx) × l

Hence the resistance of this elementary cylindrical shell is,

dR_i = ρ . dx / 2πx l   ... as R = ρl / A

where p = Resistivity of the insulating material

The total insulation resistance of the cable can be obtained by integrating the resistance of an elementary ring from inner radius upto the outer radius i.e. r to R.

The value of R_i is always very high. The expression shows that the insulation resistance is inversely proportional to its length. So as the cable length increases, the insulation resistance decreases.

This shows that if two cables are joined in series then total length increases and hence their conductor resistances are in series giving higher resistance but their insulation resistances are in parallel decreasing the effective insulation resistance. Thus if two cables are connected in parallel, conductor resistances get connected in parallel while the insulation resistances get connected in series. 

Example 6.6.1 Two underground cables A and B, each has a conductor resistance of 0.6 Ω and 0.8 Ω respectively while each fas on insulation resistance of 600 M Ω and 400 M Ω  respectively. If the cables are connected in, a) Series and b) Parallel, calculate the effective conductor and insulation resistance in each case.

Solution :

a) Cables are in series

When cables are in series, conductor resistances are in series while insulation resistances are in parallel.

Effective conductor resistance = 0.6 + 0.8 = 1.4 Ω

and Effective insulation resistance = 600 MΩ || 400 MΩ

600 × 400 / 600 + 400 = 240 MΩ

In series, length increases and insulation resistance is inversely proportional to length hence effective insulation resistance value decreases.

b) Cables are in parallel

In this case, the conductor resistances are in parallel while the insulation resistances are in series.

Effective conductor resistance = 0.6 || 0.8

= 0.6 × 0.8 / 0.6 + 0.8 = 0.3428 Ω

Effective insulation resistance = 600 M Ω + 400 M Ω = 1000 M Ω

Example 6.6.2 A single core cable has a conductor diameter of 1 cm and insulation thickness of 0.4 cm. The specific resistance of the insulation is 5 × 10¹⁴ Ω cm. Calculate the insulation resistance for a 2 km length of cable.

Solution : d = 1 cm, insulation thickness = 0.4 cm, ρ = 5 × 10¹⁴  Ω cm

Review Questions

1. Derive an expression for insulation resistance of a cable.

2. A single core cable has a conductor diameter of 1 cm and insulation thickness cf 0.5 cm. If the specific resistance of an insulation is 5 × 10¹⁴ Ω cm. Calculate the insulation resistance for 3 km length of the cable.        

[Ans.: 183.86 M Ω]