MGVCL Exam Paper (30-07-2021 Shift 2) In the circuit shown in figure, the voltage across 8 Ω resistor is 20 V. Calculate the supply voltage. 18 V 47 V 38 V 27 V 18 V 47 V 38 V 27 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equivalent resistance of the circuit,Req = (12||18) + 8Req = 7.2 + 8Req = 15.2 ohmSource current = current through 8 ohm resistor, I = Vr/R8I = 20/8I = 2.5 ASupply votage V = I*ReqV = 2.5*15.2V = 38 V
MGVCL Exam Paper (30-07-2021 Shift 2) A generating station having 'n' section busbars each rated at Q kVA with ×% reactance is connected on the tie-bars systems through busbar reactances of b%. Determine the short circuit kVA if a 3-phase fault takes place on one section. Q[(1/x) + (n-1)/(bn+x)] Q[(1/x) + (n+1)/(bn+x)] Q[(1/x) + (n+1)/(bn-x)] Q[(1/x) - (n+1)/(bn-x)] Q[(1/x) + (n-1)/(bn+x)] Q[(1/x) + (n+1)/(bn+x)] Q[(1/x) + (n+1)/(bn-x)] Q[(1/x) - (n+1)/(bn-x)] ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Short circuit kVA of any bus is given by:Skva = Sbase*(100/% X)Where,Sbase - base rating in kVA% X - value of reactance in %
MGVCL Exam Paper (30-07-2021 Shift 2) A wire pilot is generally economical for distances up to 30 km to 60 km 10 km to 15 km 20 to 25 km 75 km to 100 km 30 km to 60 km 10 km to 15 km 20 to 25 km 75 km to 100 km ANSWER EXPLANATION DOWNLOAD EXAMIANS APP A wire pilot is generally economical for distances up to 5 or 10 miles, beyond which a carrier-current pilot usually becomes more economical.
MGVCL Exam Paper (30-07-2021 Shift 2) સોહેલ ખાન : મલાયકા અરોર :: સૈફ અલી ખાન : ?? અમૃતા સિંહ પરવીન બાબી કિરણ ખેર માધુરી દિક્ષિત અમૃતા સિંહ પરવીન બાબી કિરણ ખેર માધુરી દિક્ષિત ANSWER DOWNLOAD EXAMIANS APP
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 7.5 W 3.8 W 33.8 W 15 W 7.5 W 3.8 W 33.8 W 15 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 circular loop has its radius increasing at a rate of 2 m/s. The loop is placed perpendicular to a constant magnetic field of 0.1 Wb/m². When the radius of the loop is 2 m, the emf induced in the loop will be 0.8π V zero 0.4π V 0.2π V 0.8π V zero 0.4π V 0.2π V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP emf = dΨ/dt = change of flux linkageemf = dф/dtemf = d(B*Area)/dtemf = B*d(area)/dtemf = B*d(πr²)/dtemf = Bπ*2r*dr/dtemf = Bπ*2r*change of radius per unit timeemf = 0.1*π*2*2emf = 4π
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