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
Equivalent resistance of the circuit, Req = (12||18) + 8 Req = 7.2 + 8 Req = 15.2 ohm Source current = current through 8 ohm resistor, I = Vr/R8 I = 20/8 I = 2.5 A Supply votage V = I*Req V = 2.5*15.2 V = 38 V
Where, Pmp is maximum power in watt Vmp maximum voltage at maximum power in volt Imp is maximum current at maximum power in ampere Voc open circuit voltage in volt Isc is short circuit current in ampere
Fill Factor = Pmp/(Vosc*Isc) = (Vmp*Imp)/(Voc*Isc) = (2.65*0.5)/(0.7*3) = 0.63
EXAMIANS