MGVCL Exam Paper (30-07-2021 Shift 2) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 4 hours. What is the permissible overload for a duration of four hours. Assume the permissible load kVA as a fraction of rated kVA is 1.24. 930 kVA 620 kVA 1240 kVA 2480 kVA 930 kVA 620 kVA 1240 kVA 2480 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
MGVCL Exam Paper (30-07-2021 Shift 2) A MOSFET rated for 15 A, carries a periodic current as shown in figure. The ON state resistance of the MOSFET is 0.15 Ω. The average ON state loss in the MOSFET is 15 W 33.8 W 3.8 W 7.5 W 15 W 33.8 W 3.8 W 7.5 W ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Ip = 10 ATon = πT = 2πIavg = Ip*√(Ton/T)Iavg = 10*√(1/2)= 7.07 AThe average ON state loss in MOSFET, P = Iavg²*RP = 7.07²*0.15P = 7.5 W
MGVCL Exam Paper (30-07-2021 Shift 2) A coil having an inductance of 100 mH is magnetically coupled to another coil having an inductance of 900 mH. The coefficient of coupling between the coils is 0.45. Calculate the equivalent inductance if the two coils are connected in series opposing. 1270 mH 730 mH 1135 mH 932 mH 1270 mH 730 mH 1135 mH 932 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP M = k√(L₁L₂)M= 0.45√(100*900)M = 135 mHL_eq for opposite connection of coils = L₁ + L₂ - 2ML_eq = 900 + 100 - 2*135L_eq = 730 mH
MGVCL Exam Paper (30-07-2021 Shift 2) In the circuit of figure, the switch is closed to position-1 for a long time. At t = 0 the switch position is changed from postion-1 to postion-2. Find an expression for the current i(t). i(t) = (E/R) sin√(LC) t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) sin√(LC) t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos[1/√(LC)]t A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Standard eqution of current for this condition,i(t) = i(0+)*cos(wt)i(0+) = E/RFor LC circuit,w = 1/√(LC)i(t) = (E/R)*cos((1/√(LC))*t) A
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. 100 m 200 m 400 m 300 m 100 m 200 m 400 m 300 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,d is distance in mQ is value of variable resistor in ohmP is variable resistance in ohmHere Q/P = 1/3 is given,d = Q/(P + Q)*Loop length= 1/4*(2*600)= 300 m
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