Fig. 1 Single core cable |
In such cables,
the leakage current flows radially from centre towards the surface as
shown in the Fig.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
consider an elementary section of the cylindrical cable of radius x and
the thickness dx as shown in the Fig. 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
Fig. 2 Elementary ring |
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 of l of the cable.
Cross-section area = Surface area for length l of cable
= (2 π x) x l
Hence the resistance of this elementary cylindrical shell is,
where ρ = 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.
This can be expressed interms of diameters as,
The value of Ri
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 insulation resistance 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 resistance get connected in series.
Read this example
Read this example
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