MGVCL Exam Paper (30-07-2021 Shift 3) In a power network, 375 kV is recorded at a 400 kV bus. A 45 MVAR, 400 kV shunt reactor is connected to the bus. What is the reactive power absorbed by the shunt reactor? 49.55 MVAR 39.55 MVAR 59.55 MVAR 69.55 MVAR 49.55 MVAR 39.55 MVAR 59.55 MVAR 69.55 MVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Q = V²/XX = 400²/45= 3555.55 ohmFor 375 kV,Q = 375²/3555.55= 39.55 kVAR
MGVCL Exam Paper (30-07-2021 Shift 3) The International boundary 'Radcliffe Line’ lies between India and Japan China Afghanistan Pakistan Japan China Afghanistan Pakistan ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) નીચે પૈકી કેટલી કહેવતો વિરુદ્ધાથીઁ છે?.1. પાંચ બોલે તે પરમેશ્વર - ગામને મોઢે ગળણું ન બંધાય.2. ચોરની ચાર અને જોનારની બે - વિશ્વાસે વહાણ ચાલે.3. ખાલી ચણો વાગે ઘણો - અધુરો ઘડો છલકાય.4. હાથના કર્યા હૈયે વાગ્યા - દીવો લઈને કૂવામાં પડવું એક પણ નહિ કુલ 3 બધી કહેવતો સમાનર્થી છે કુલ 1 એક પણ નહિ કુલ 3 બધી કહેવતો સમાનર્થી છે કુલ 1 ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Murray loop test is performed on a faulty cable 300 m long. At balance, the resistance connected to the faulty core was set at 20 Ω and the resistance of the resistor connected to the sound core was 40 Ω. Calculate the distance of the fault point from the test end. 400 m 300 m 100 m 200 m 400 m 300 m 100 m 200 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,R2 is the resistance connected to the faulty core in ohmR2 is the resistance of the resistor connected to the sound core in ohmR1 Distance of fault location (Lx) = R2/(R1 + R2)*(2*L)= (20/60)*600= 200 m.
MGVCL Exam Paper (30-07-2021 Shift 3) A star connected synchronous generator rated at 500 MVA, 50 kV has a reactance of 0.5 pu. Find the ohmic value of the reactance. 2.5 Ω 1 Ω 0.25 Ω 0.1 Ω 2.5 Ω 1 Ω 0.25 Ω 0.1 Ω ANSWER EXPLANATION DOWNLOAD EXAMIANS APP pu = actual/baserated phase consired at base value for alternator.ohmic pu = ohmic actual/ohmic basepu = ohmic actual*current base/voltage baseConsider all above phase valuesMVA (3phase) = 3 Vph IphIph = 500*10⁶/(3*28901.73) = 5766.66 Apu = ohmic actual*current base/voltage base0.5 = ohmic actual*5766.66/28901.73ohmic actual = 2.5 Ω
MGVCL Exam Paper (30-07-2021 Shift 3) Name the application under MS Office software bundle, that we use to create audio- visual presentation. MS Excel MS Access MS Power point MS Word MS Excel MS Access MS Power point MS Word ANSWER DOWNLOAD EXAMIANS APP
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