in laying of new 4c x 120 sqmm LV cable, what is the minimum requirement value in megger?

Question

To charge the supply in a LV power cable, what is the maximum Mega ohms required in a new cable and what is the minimum value for old cable?

Answer 1

The minimum isulation in either case, tested with a 500volt DC 'megger' is 0.5 Megohm, but I would expect the new cable to be up in the hundreds or tens of hundreds! Test between each conductor and each conductor to earth.

16th 713-14 &Table 71A

Answer 2

Minimum, usable LV cable for 3 phase power is 2MOhm.
With 2,000MOhm Megger, it goes to infinitive.
With larger scale Megger, it can goes to 10,000 MOhm.

Answer 3

build a triangle with sides of length a, b, c to correspond with the radians ?/7, 2?/7, and 4?/7 respectively. Bisect 4?/7 to 2?/7 and bisect between the radians lower back to ?/7. Now we are complete with construction what we desire. it may be impossible to describe with the labeling, so label the vertex of radian ?/7 as A, the vertex of radian 2?/7 as B, and the vertex of 4?/7 as C. The bisector of radian C intersects the triangle at element D. the 2d bisector intersects are E. Triangle BCE is resembling triangle ABC, subsequently | BE | = a million/c, | CE | = b/c. Triangle ACE is isosceles, subsequently | AE | = b. Triangle CDE is isosceles subsequently | CD | = b/c. Triangle CDE is resembling triangle ACE subsequently | DE | = b/c². when you consider that Triangle BCD is isosceles, b/c = a million/c + b/c². when you consider that Triangle ACE is isosceles, c = b + a million/c. b/c = a million/c + b/c² bc = b + c c = b + a million/c c² = bc + a million c² = b + c + a million c² - c - a million = b bc = b + c c³ - c² - c = c² - c - a million + c c³ - c² - c = c² - a million c³ - 2c² - c + a million = 0 clean up for c employing the cubic formula. After fixing for c, you could clean up for b. by employing utilising the regulation of Sines, sin (2?/7) and sin (4?/7) could be expressed as sin (?/7). the only last step is fixing for sin (?/7). sin (?/7) = (2 sin (?/7) cos (?/7)) / b b/2 = cos (?/7) b/2 = a million - sin² (?/7) sin (?/7) = ?(a million - b/2) Plug b in and clean up for it. properly, you have sin (?/7) so which you apart from could have the real fee.

Answer 4

Work to 1 Megohm of insulation resistance per Kv of operating voltage

Answer 5

maximum mega ohms is infinite (or near to)
minimum is 0.5 M ohms 16th edition wiring regs

test is usually done with a 500v insulation resistance tester.

for larger cables high current insulation resistance testers can be used, although not specified in the regs.