MGVCL Exam Paper (30-07-2021 Shift 1) In computer networks, the nodes are all the mentioned above the computer that terminates data the computer that originates data the computer that routes data all the mentioned above the computer that terminates data the computer that originates data the computer that routes data ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Three resistances 750 Ω, 600 Ω and 200 Ω are connected in parallel. The total current is 1 A. Determine the voltage applied. 100 V 200 V 150 V 125 V 100 V 200 V 150 V 125 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equivalent resistance of the circuit, Req = (750|| 600 || 200)Req = 125 ΩCurrent from source = 1 ASource voltage = 1*125Source voltage = 125 V
MGVCL Exam Paper (30-07-2021 Shift 1) An electric potential field is produced in air by point 1 μC and 4 μC located at (-2, 1, 5) and (1, 3, -1) respectively. The energy stored in the field is 2.57 mJ 12.50 mJ 5.14 mJ 10.28 mJ 2.57 mJ 12.50 mJ 5.14 mJ 10.28 mJ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Determine maximum permissible load which a 15 MVA, ONAN cooled transformer to IS : 2026 can take for 6 hr, if the initial load on it was 12 MVA. The weighted ambient temperature is 20˚C. Assume the permissible load kVA as a fraction of rated kVA is 1.25. 12 MVA 15 MVA 9.6 MVA 18.75 MVA 12 MVA 15 MVA 9.6 MVA 18.75 MVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Maximum permissible load = Rated MVA rating*fraction of rated kVAMaximum permissible load = 15*1.25Maximum permissible load = 18.75 MVA
MGVCL Exam Paper (30-07-2021 Shift 1) 1 terabyte is equal to 1024 kilobytes 1024 gigabytes None of these 1024 bytes 1024 kilobytes 1024 gigabytes None of these 1024 bytes 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. 10 pF 40 pF 20 pF 30 pF 10 pF 40 pF 20 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
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