MGVCL Exam Paper (30-07-2021 Shift 1) દુર્ગતિ' શબ્દનો વિરુદ્ધાર્થી શબ્દ ક્યો છે? અદુર્ગતી સળંગગતિ કુગતિ સુગતિ અદુર્ગતી સળંગગતિ કુગતિ સુગતિ ANSWER DOWNLOAD EXAMIANS APP
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. 325√3 VAR 425√3 VAR 425/√3 VAR 325/√3 VAR 325√3 VAR 425√3 VAR 425/√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
MGVCL Exam Paper (30-07-2021 Shift 1) Rearrange the following to form a meaningful sentence and find the most logical order from the given options.P: 60 countries using dance as a tool forQ: building social cohesion and resolvingR: conflict throughout the worldS: battery Dance has travelled far and wide across SPQR PQRS PQSR SPRQ SPQR PQRS PQSR SPRQ 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. 200 V 150 V 125 V 100 V 200 V 150 V 125 V 100 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) In computer networks, the nodes are the computer that originates data all the mentioned above the computer that routes data the computer that terminates data the computer that originates data all the mentioned above the computer that routes data the computer that terminates data 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. 18.75 MVA 12 MVA 9.6 MVA 15 MVA 18.75 MVA 12 MVA 9.6 MVA 15 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
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