MGVCL Exam Paper (30-07-2021 Shift 1) Calculate the capacitance per meter of a 50 Ω load cable that has an inductance of 50 nH/m. 10 pF 20 pF 40 pF 30 pF 10 pF 20 pF 40 pF 30 pF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Z = √(L/C)C = L/Z²C = 50*10⁻⁹/(50*50)C = 0.02 nFC = 20 pF
MGVCL Exam Paper (30-07-2021 Shift 1) A given solar cell has the following specifications: Isc = 4 A and Voc = 0.7 V. Consider that 20 identical cells with the above specifications are to be interconnected to create a PV module. What is the open-circuit voltage of the PV module if all the solar cells are connected in a series configuration? 12.6 V 14 V 11.2 V 0.7 V 12.6 V 14 V 11.2 V 0.7 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For M no. of series configuration of solar cell,Open circuit voltage of PV module= M*Voc= 20*0.7= 14 VShort circuit current of PV module = M*Isc= 20*4= 80 A
MGVCL Exam Paper (30-07-2021 Shift 1) Choose the best option from the given alternatives which can be substituted for the given word/sentenceA disease prevailing in a locality Host Habitat Endemic Epic Host Habitat Endemic Epic ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Which of the following is an independent malicious program that spread copies of itself from computer to computer? Virus Trap door Worm Trojan horse Virus Trap door Worm Trojan horse ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Murray loop test is performed on a faulty cable 300 m long. At balance, the resistance connected to the faulty core was set at 15 Ω and the resistance of the resistor connected to the sound core was 45 Ω. Calculate the distance of the fault point from the test end. 150 m 225 m 75 m 450 m 150 m 225 m 75 m 450 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,R2 is the resistance connected to the faulty core in ohmR2 is the resistance of the resistor connected to the sound core in ohmR1 Distance of fault location (Lx) = R2/(R1 + R2)*(2*L)= (15/60)*600= 150 m
MGVCL Exam Paper (30-07-2021 Shift 1) An alternator is supplying load of 300 kW at a power factor of 0.5 lagging. If the power factor is raised to unity, how many more kilowatts can alternator supply for the same kVA loading? 300 kW 150 kW 450 kW 600 kW 300 kW 150 kW 450 kW 600 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = S*cosφS = P/cosφS = 300/0.5S = 600 kVAFor same kVA rating and unity power factor,P = 600*1P = 600 kWPextra = 600 - 300Pextra = 300 kW
EXAMIANS