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 5.14 mJ 10.28 mJ 12.50 mJ 2.57 mJ 5.14 mJ 10.28 mJ 12.50 mJ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Replace the underlined phrase in braket grammatically and conceptually with the help of the given options. If the given sentence is correct then select the option 'The given sentence is correct'.The new client had to pay money to initiates a account and activate their service had to pay money to initiate an The given sentence is correct have to pay money to initiate an had to pay money to initiate a had to pay money to initiate an The given sentence is correct have to pay money to initiate an had to pay money to initiate a 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
MGVCL Exam Paper (30-07-2021 Shift 1) An alternator is supplying load of 300 kW at a power factor of 0.5 lagging. If the power factor is raised to unity, how many more kilowatts can alternator supply for the same kVA loading? 300 kW 600 kW 450 kW 150 kW 300 kW 600 kW 450 kW 150 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = S*cosφS = P/cosφS = 300/0.5S = 600 kVAFor same kVA rating and unity power factor,P = 600*1P = 600 kWPextra = 600 - 300Pextra = 300 kW
MGVCL Exam Paper (30-07-2021 Shift 1) The current flowing to a balanced delta connected load through line 'a' is 15 A when the conductor of line 'b' is open. With the current in line 'a' as reference, compute the symmetrical components of the line currents. Assume the phase sequence of 'abc'. Ia₀ = 0 AIa₁ = (7.5 - j13) AIa₂ = (7.5 + j13) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) A Ia₀ = 0 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 - j13) AIa₂ = (7.5 + j13) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) A Ia₀ = 0 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(15 + 15∟-120°)= 7.5 + j*4.33 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(15 + 15∟120°)= 7.5 - j*4.33 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0
MGVCL Exam Paper (30-07-2021 Shift 1) મારુતારુંનો ઝગડો હજી ખતમ થયો નથી, ચારપાંચ વર્ષ થઇ ગયા અને રાતદિવસ સાથે ઉઠવા-બેસવાનું તો ચાલુ જ છે. ઉપરોક્ત વાક્યમાં ક્યા ક્યા વ્યાકરણીય સમાસનો ઉપયોગ કરાયો છે? અવ્યવીભાવ, મધ્યમપદલોપી દ્વન્દ્વ તત્પુરુષ, દ્વન્દ્વ, મધ્યમપદલોપી અવ્યવીભાવ, મધ્યમપદલોપી દ્વન્દ્વ તત્પુરુષ, દ્વન્દ્વ, મધ્યમપદલોપી ANSWER DOWNLOAD EXAMIANS APP
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