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
Slip, s = (Ns - Nr)/Ns Ns = 120f/P 0.05 = (1000 - Nr)/1000 Nr = 950 rpm Speed of rotor with respect to stator magnetic field = 950 - 1000 = -50 rpm
Speed of the stator magnetic field with respect to stator core = 0 Speed of the rotor magnetic field with respect to rotor = sNs Speed of the stator magnetic field with respect to rotor magnetic field = 0 Speed of the rotor magnetic field with respect to stator core = Ns
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)
EXAMIANS