Protection Engineering Basics and Schemes Protection Engineering | Page 58
= 0.0952 MΩ Now let suppose after next two or three days we need to conduct this test again, asked by client because of corrective maintenance and during these past two days, weather changes rain occurs, now outer surface resistance is reduces, because our insulator is now wet and always remember wet insulator resistance is always less.
R = 2 Mega ohms
R0 = 0.01 mega ohms
Now our IR tester gives us reading 0.01
= 2 ∗
2 + 0.01 = 0. 00995MΩ What happened resistance reduces by percentage?
= 0. 00995 = 10. 45 %
. 0952 This will influence client to replace insulator as resistance
decreases by significant percentage. Now if we employees the guard in our testing, our IR tester reads 2MegaΩ in