MGVCL Exam Paper (30-07-2021 Shift 2) Operating system is a program that runs for short duration a program that runs during log off a layer between the user and the hardware a program that runs during log off a program that runs for short duration a program that runs during log off a layer between the user and the hardware a program that runs during log off ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Fill in the blanks with suitable Preposition from the given alternatives.England skipper Joe Root is excited to bat ____ number three in the Ashes cricket series on at over in on at over in ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A sudden, non power frequency change in the steady state condition of voltage or current that is bidirectional in polarity is called as Distortion Harmonic Impulsive transient Oscillatory transients Distortion Harmonic Impulsive transient Oscillatory transients ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Oscillatory transient is the a sudden, non power frequency change in the steady state condition of voltage or current that is bidirectional in polarity.
MGVCL Exam Paper (30-07-2021 Shift 2) In a short circuit test on circuit breaker, time to reach the peak re-striking voltage is 50 µs with the peak restriking voltage of 100 kV. Determine average RRRV (Rate of rise re-striking voltage)? 3 x 10⁶ kV/sec 1 x 10⁶ kV/sec 2 x 10⁶ kV/sec 4 x 10⁶ kV/sec 3 x 10⁶ kV/sec 1 x 10⁶ kV/sec 2 x 10⁶ kV/sec 4 x 10⁶ kV/sec ANSWER EXPLANATION DOWNLOAD EXAMIANS APP RRRV = Peak value of restriking voltage/time to reach the peak re-striking voltageRRRV = 100*1000/(50 μs)RRRV = 2*10⁶ kV/sec
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. 620 kVA 2480 kVA 930 kVA 1240 kVA 620 kVA 2480 kVA 930 kVA 1240 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) If all the devices are connected to a central hub, then the topology is called Star Topology Tree Topology Bus Topology Tree Topology Star Topology Tree Topology Bus Topology Tree Topology ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS