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The insulation resistance of a kilometre of the cable having a conductor diameter of 1.5 cm and an insulation thickness of 1.5 cm is 500 M Ω. What would be the insulation resistance if the thickness of the insulation were increased to 2.5 cm ? |
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Answer» The insulation resistance of a cable is First Case R = (2.3ρ/2πl)log10(r2/r1) r1 = 1.5/2 = 0.75 cm ; r2 = 0.75 + 1.5 = 2.25 cm ∴ r2/r1 = 2.25/0.75 = 3 ; log10 (3) = 0.4771 ∴ 500 = (2.3ρ/2πl) x 0.4771 ..... (i) Second Case r1 = 0.75 cm − as before r2 = 0.75 + 2.5 = 3.25 cm r2/r1 = 3.25/0.75 = 4.333 ; log10 (4.333) = 0.6368 ∴ R = (2.4ρ/2πl) x 0.6368 ......(ii) Dividing Eq. (ii) by Eq. (i), we get R/500 = 0.6368/0.4771; R = 500 x 0.6368/0.4771 = 667.4MΩ |
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