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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!!
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  • 0
    This could be a catch question. Remember the 80% "rule"? 1.65 * 0.8=1.32. 1.12 + 0.2 = 1.32!

    One needs to think quite hard in exams, they ARE out to get you. The key is the "maximum tabulated velue", which is not the same as the OSG for example. The 80% is also arbitrary of course.


    David CEng
Reply
  • 0
    This could be a catch question. Remember the 80% "rule"? 1.65 * 0.8=1.32. 1.12 + 0.2 = 1.32!

    One needs to think quite hard in exams, they ARE out to get you. The key is the "maximum tabulated velue", which is not the same as the OSG for example. The 80% is also arbitrary of course.


    David CEng
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