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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
    Lol... that's the C&G for you. Looks like a question from the 2391 or 2395 exam sheet pre 18th ed as there's no adjustment for supply voltage.. The answer is D (0.2) but it assumes that the given  tabulated value for the final circuit resistance is the same as the measured resistance with respect to the ambient temperature.

    Therefore Zs = Ze +(R1+R2)*1.2

    Ze = Zs*1/1.2 - (R1+R2)

    A real answer = 0.19 when using 0.8 and 0.25 when using 1/1.2

    Legh

Reply
  • 0
    Lol... that's the C&G for you. Looks like a question from the 2391 or 2395 exam sheet pre 18th ed as there's no adjustment for supply voltage.. The answer is D (0.2) but it assumes that the given  tabulated value for the final circuit resistance is the same as the measured resistance with respect to the ambient temperature.

    Therefore Zs = Ze +(R1+R2)*1.2

    Ze = Zs*1/1.2 - (R1+R2)

    A real answer = 0.19 when using 0.8 and 0.25 when using 1/1.2

    Legh

Children
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