Current Rating of Cable

       Once all the thermal resistances are known then the current carrying capacity of a cable can be determined. When the cables carry an excessive current the cable get heated up. It is not advisable to operate the cable at excessively high temperature because of the following reasons :
1. Due to high temperature, oil in the oil filled cables get expanded and this may lead to bursting of the sheath.
2. High temperature can cause unequal expansion which leads to the information of voids. Such voids may lead to ionisation and finally lead to insulation failure.
3. The dielectric losses increase with temperature which also can lead to breakdown of insulation.
       Hence the cable must be operated at a current less than the maximum current carrying capacity of the cable.
       The current rating of a cable is dependent on the following factors,
1. The maximum permissible temperature at which conductor insulation can be operated.
2. Heat dissipation arrangement through the cable.
3. The ambient conditions as well as conditions at the time of installation.
       Let us see how to determine current carrying capacity of a cable.
       Let          n = Number of phases
                      R = Conductor resistance at 65 in /m
                      I = R.M.S. value of current in each core
      Hence the total core loss is given by n I2R.
      The heat generated in the core of the cable passes through the dielectric medium to the sheath.
Let                 θm = Maximum permissible of the core
                      θ= Sheath temperature
      Then we can write,
                           nI2R = (θms )/S1                                            .................(1)
      Where          S1= Thermal resistance of the dielectric
       Let                λ = Sheath loss/ Core loss
       Sheath loss = λ core loss
       Total loss = Core loss + Sheath loss
                        = (1+λ) core loss = (1+λ) nI2R
       This is the heat flowing through bedding, serving and ground. While the total thermal resistance of bedding, serving and ground is S4 + S5 +G. And the corresponding temperature difference is difference between sheath temperature θ and ambient temperature . Hence we can write,
        (1+λ) nI2R = (θ- θa )/(S4 + S5 +G)                      ..................(2)
       As θ is generally not known, eliminate θ from (1) and (2).
                    θ- θ= nI2RS1
      and          θ- θa   = (1+λ) nI2R (S4 + S5 +G)
                      θ- θa   = nI2R {S1+(1+λ)(S4 + S5 +G)}

       This is the required current carrying capacity of a cable.
       The allowable temperature values for the various types of cables are given in the Table 1.

       Practically the current carrying capacity also depends on the factors like grouping and proximately of the other cables, load factor, load cycle, ambient soil temperatures in which cable is to lay etc.
       In practice S4, S5 and λ can be neglected hence current carrying capacity can be expressed as,

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