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
Simple logic of reactance diagram is that simplified equivalent circuit of power system in which the various components of power system are represented by their reactances. The reactance diagram can be obtained from impedance diagram if all the resistive components are neglected.
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