MGVCL Exam Paper (30-07-2021 Shift 2) Encrypted security payload extension header is new in IPv4 IP IPv6 IPv5 IPv4 IP IPv6 IPv5 ANSWER DOWNLOAD EXAMIANS APP
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. 930 kVA 620 kVA 2480 kVA 1240 kVA 930 kVA 620 kVA 2480 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) Which of the following is linux operating system? IBM-AIX MacOS HP-UX Fedora IBM-AIX MacOS HP-UX Fedora ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Rearrange the following to form a meaningful sentence and find the most logical order from the given options.P: dancers make the companyQ: and they do not haveR: a big team of professionalS: a ‘guru shishya’ hierarchy RPSQ RPQS PQSR PQRS RPSQ RPQS PQSR PQRS ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) The relation between electric intensity E, voltage applied V and the distance d between the plates of a parallel plate condenser is E = V/d E = V/(d)² E = Vx(d)² E = V x d E = V/d E = V/(d)² E = Vx(d)² E = V x d ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The relation between electric intensity E, voltage applied V and the distance d between the plates of a parallel plate condenser is given by:V = E*LE = V/L
MGVCL Exam Paper (30-07-2021 Shift 2) A three-phase, 12 kV, 50 Hz and 300 kW load is operating at 0.8 pf lag. If the pf is to be improved to unity, using star connected capacitor, the per phase value of the reactive VAR required is: 50 kVAR 100 kVAR 25 kVAR 75 kVAR 50 kVAR 100 kVAR 25 kVAR 75 kVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The reactive VAR required = kW*(tanφ1 -tanφ2)Cosφ1 = 0.8φ1 = 36.87 deg.Cosφ2 = 0φ2 = 0 deg.kVAR required = kW*(tan(36.87) - tan(0))= 300*1000(0.75)= 225.00 kVARPer phase value = 225.00/3= 75 kVAR
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