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Let S1 be the value of resistance for position 1 and S2 for position 2.
Therefore,[R+X]/S1=P/Q.
Hence the resistance of loop R+X=[P/Q]*S1=[5/10]*16=8 ohms
Resistance of each cable =8/2=4 ohms.
Length of each cable=4/0.4=10 km.
At postion 2 we have;R/[X+S2]=P/Q.
[R+X+S2]/[X+S2]=[P+Q]/Q OR [8+7]/[X+7]=[5+10]/10.
X=3.0 ohms.
Dstance of fault from test end=3/4=7.5 km
Kevinsenzoochieng answered the question on October 22, 2019 at 08:02
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