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
Frequency of natural oscillation is given by, fn = {((dPe/dδ)at(δo))/M)} dPe/dδ = ((V1*V2)/X*(cosδ)) = (11/05)*cosδ = (11/0.5)*0.5 M = (H*s)/(πf) = 4/(50π)
The capacity of reactor = 50 MVAr when it is operated at 415 kV. Impedance of reactor = V2/MVAr = (415*415)2/50 = 3444.5 Ω For 400 kV, Reactive power observed = (400*400)/3444.5 = 46.45 MVAR
Or
VAr capacity of capacitor or reactor is proportional to V²
EXAMIANS