MGVCL Exam Paper (30-07-2021 Shift 1) ___ is the process of actuating equipment operation at remote locations. Supervisory control Supplementary control Data processing Data acquisition Supervisory control Supplementary control Data processing Data acquisition ANSWER DOWNLOAD EXAMIANS APP
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. 40 pF 20 pF 30 pF 10 pF 40 pF 20 pF 30 pF 10 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) By inserting a plate of dielectric material between the plates of a parallel plate capacitor at constant potential, the energy stored in the capacitor is increased five times. The dielectric constant of the material is 25 1/25 5 1/5 25 1/25 5 1/5 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Energy stored by capacitor, Q = ½CV²Dielectric constant is directly proportional to C and capacitance is proportional to the dielectrical constant
MGVCL Exam Paper (30-07-2021 Shift 1) The operating system is the most common type of ___ Communication Application Software Firmware System Software Communication Application Software Firmware System Software ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) For protection of inter-turn fault in transformer, the relays preferred are Overcurrent relay and differential relay Overcurrent relay and Buchholz's relay Buchholz's relay, overcurrent relay and differential relay Buchholz's relay and differential relay Overcurrent relay and differential relay Overcurrent relay and Buchholz's relay Buchholz's relay, overcurrent relay and differential relay Buchholz's relay and differential relay ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Intern turn winding fault can be easily detect by differential protection relay.Also detect by buchholz relay (gas operated relay), cause buchholz relay is used to detect the internal fault of transformer.
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? 150 kW 450 kW 300 kW 600 kW 150 kW 450 kW 300 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
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