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Examoogle • User • info@examoogle.com • EE-2024-8821
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The leakage resistance of a 50 km long cable is 1 MΩ. For a 100 km long cable it will be
1 MΩ
2 MΩ
0.66 MΩ
None of the above
None of the above
D is correct because the leakage resistance for 100 km is 0.5 MΩ, which is not listed in options A, B, or C.
For a cable of length l1 = 50 km, the leakage resistance R1 = 1 MΩ.
R=2πlρiln(RinnerRouter)=Lρs
Identify the relationship
Leakage resistance of a cable is inversely proportional to its length (L). If the cable length increases, the effective insulation path area decreases, thus resistance decreases.
R∝L1
Set up the ratio
Using the inverse proportionality, the product of resistance and length is constant: R1×L1=R2×L2.
R2=R1×L2L1
Calculate the resistance for 100 km
Substitute R1=1 MΩ, L1=50 km, and L2=100 km into the formula.
R2=1×10050=0.5 MΩ
D is correct because the leakage resistance for 100 km is 0.5 MΩ, which is not listed in options A, B, or C.
Note that while conductor resistance increases with length, leakage (insulation) resistance decreases with length because leakage paths act as parallel resistors.