In this method of grading, in between the core and the lead sheath number of metallic sheaths are placed which are called intersheaths. All these intersheaths are maintained at different potentials by connecting them to the tappings of the transformer secondary. These potentials are between the core potential and earth potential. Generally lead is used for these sheaths as it is flexible and corrosion resistance but as its mechanical strength is less, aluminium also can be used. Aluminium is low weight and mechanically strong but it is much costlier than lead.

Using the intersheaths, maintaining at different potential, uniform distribution of stress is obtained in the cables.

Consider a cable with core diameter d and overall diameter with lead sheath as D. Let two intersheaths are used having diameter d

_{1}and d_{2}which are kept at the potentials V_{1}and V_{2}respectively. The intersheaths and stress distribution is shown in the Fig. 1.

Fig. 1 |

Let V

_{1}= Voltage of intersheath 1 with respect to earth V

_{2}= Voltage of intersheath 2 with respect to earth It has been proved that stress at a point which is at a distance x is inversely proportional to distance x and given by,

Where k is constant.

So electric stress between he conductor and intersheath 1 is,

Now potential difference between core and the first intersheath is V-V

_{1}. Substituting in equation (2) we get,

Now this stress is maximum at x = d/2, on core surface.

Similarly potential difference between intersheath 1 and intersheath 2 is V

_{1}- V_{2}. Now g

_{1 }will be maximum at the surface of intersheath 1 i.e. x = d_{1}/2. The potential difference between intersheath 2 and outermost sheath V

_{2}is only as potential of intersheath is maintained at V_{2 }with respect to earth. This g

_{3 }will be maximum at x = d_{2}/2 Choosing proper values of V

_{1 }and V_{2}, g_{1max}, g_{2max}etc. can be made equal and hence uniform distribution of stress can be obtained. The stress can be made to vary between same maximum and minimum values as shown in the Fig. 1, by choosing d

_{1}and d_{2}such that, d

_{1}/d = d_{2}/d_{1}= D/d_{2}= α and g

_{1max }= g_{2max}= g_{3max} Let us try the express voltages V

_{1 }and V_{2 }interms of V and α. Substituting value of V

_{2 }from equation (11),**Read examples on cables**

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