MGVCL Exam Paper (30-07-2021 Shift 1) 1 terabyte is equal to None of these 1024 kilobytes 1024 bytes 1024 gigabytes None of these 1024 kilobytes 1024 bytes 1024 gigabytes ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Case 1: The PV connected to the DC bus was injecting 20 kW of power constituting of 1 kW of lossCase 2: The same PV is connected to the AC bus constituting to 1.5 kW loss. Find out the change in efficiency from Case 1 to Case 2? 5% increase 5% decrease 2.5% increase 2.5% decrease 5% increase 5% decrease 2.5% increase 2.5% decrease ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In DC microgrid system,Efficiency α 1/PinFor 1 kW loss injecting power is 20 kWFor 1.5 kW loss injecting power will be proportional to 30 kWSo,Efficiency will be reduced 0.025
MGVCL Exam Paper (30-07-2021 Shift 1) In the following question, one part of the sentence may have an error. Find out which part of the sentence has an error and select the option corresponding to it. If the sentence contains no error, Select "No error'' option. (Avoid punctuation errors)(A) Bubin together with / (B) her parents are coming / (C) to attend the party. / (D) NO ERROR A D B C A D B C ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) The operating system is the most common type of ___ Firmware Communication System Software Application Software Firmware Communication System Software Application Software ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) A fully controlled natural commutated 3-phase bridge rectifier is operating with a firing angle α = 30°, The peak to peak voltage ripple expressed as a ratio of the peak output dc voltage at the output of the converter bridge is 0.5 1-(√3/2) √3-1 √3 -1 0.5 1-(√3/2) √3-1 √3 -1 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,φ is firing angleVoltage ripple for 3-phase bridge rectifier = cos(φ) = Cos(30)= 1/2
MGVCL Exam Paper (30-07-2021 Shift 1) Two-wattmeter method is used to measure the power taken by a 3-phase induction motor on no load. The wattmeter readings are 375 W and –50 W. Calculate the reactive power taken by the load. 425√3 VAR 425/√3 VAR 325√3 VAR 325/√3 VAR 425√3 VAR 425/√3 VAR 325√3 VAR 325/√3 VAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Reactive power taken, Q = √3*(W1 - W2)Q = √3*(375 + 50)Q = √3*425 kVAR
EXAMIANS