The maximum and minimum stress in the dielectric of a single core cable are 80 kV/cm (rms) and 20 kV/cm (rms), respectively. If the conductor diameter is 4 cm, then find the thickness of insulation.

The correct answer is option 2):(6 cm)

Concept:

Maximum dielectric stress (at the surface of the conductor) is given by

\({{g}_{mx}}=\frac{V}{rln\left( \frac{R}{r} \right)}\)

Minimum dielectric stress (at the inner sheath) is given by

\({{g}_{mn}}=\frac{V}{Rln\left( \frac{R}{r} \right)}~\)

Condition for minimizing dielectric stress (economical)

\(\frac{R}{r}=e\)

The ratio between the maximum and minimum stress in the dielectric of a single-core cable is given as

 \(g_{mx} \over g_{mn} \) = \(D \over d\)

Insulation thickness =  \(D-d \over 2\)

Where

g = electric field intensity

d = Diameter of the conductor 

D = Diameter of the inner sheath

r = conductor radius

R = inner sheath radius

V = peak value of the potential of conductor w.r.t sheath

Calculation:

Giveng_mx =  80 kV/cm

g_mn =  20 kV/cm

d = 4 cm

D = \(g_{mx} \over g_{mn} \) ×  4

=  \(80 \over 20 \) ×  4

= 16

Insulation thickness =  \(D-d \over 2\)

= \(16 -4 \over 2\)

= 6 cm