MGVCL Exam Paper (30-07-2021 Shift 2)
In a test by Murray loop for ground fault on 600 m of cable having a resistance of 1.5 Ω/km, the faulty cable is looped with a sound cable of the same length and area of cross section. If the ratio of the other two arms of the testing network at balance is 3 : 1, find the distance of the fault from the testing end of cables.
Where, d is distance in m Q is value of variable resistor in ohm P is variable resistance in ohm Here Q/P = 1/3 is given, d = Q/(P + Q)*Loop length = 1/4*(2*600) = 300 m
Here, ω = 500 rad/sec = 2πf. At Resonance condition, ωL = 1/(ωC) L 0.2 H Also at series resonace 1. fr = 1/[2π√(LC)] 2. PF = 1 3. Maximum current 4. Source voltage = voltage across resistor.
EXAMIANS