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Exam question

Eclipse
Guys,

can anyone explain the how the following answer to this  question is derived ( its from an old test paper) which my son is doing, 

The maximum tabulated value of earth fault loop impedance for a circuit is: 1.65 ohms, if the resistance of the line to cpc of a new circuit is 1.13ohms  the maximum accepted external earth fault loop impedance is:

A. 0.52 OHMS

B. 0.45 OHMS

C. 0.35 OHMS

D.0.20 OHMS


The answer given to him is "D"  but no explanation as to how its worked out,

really appreciate any feedback!!
Parents
  • 0
    Well, the figure we`d be looking at to BS7671 is 0.52 so that is the answer cos that is what is tabulated in 7671.

    However the figure we`d be working to (designing and subsequent inspection )  is the 80% figure so we would be looking at 0.20 , if it is higher we`d headscratch and do proper sums


    Don`t know why some calls it the 3/4 rule as when I went to school that would be 75% not 80%. Mind you the speed of light has altered since I went to school (it used to be 186234 miles per second back in those days)
Reply
  • 0
    Well, the figure we`d be looking at to BS7671 is 0.52 so that is the answer cos that is what is tabulated in 7671.

    However the figure we`d be working to (designing and subsequent inspection )  is the 80% figure so we would be looking at 0.20 , if it is higher we`d headscratch and do proper sums


    Don`t know why some calls it the 3/4 rule as when I went to school that would be 75% not 80%. Mind you the speed of light has altered since I went to school (it used to be 186234 miles per second back in those days)
Children
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