MGVCL Exam Paper (30-07-2021 Shift 1) Dena Bank and Vijaya Bank recenlty got merged with which of the following banks? Bank of Baroda Punjab National Bank State Bank of India Indian Overseas Bank Bank of Baroda Punjab National Bank State Bank of India Indian Overseas Bank 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 125 V 200 V 150 V 100 V 125 V 200 V 150 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) 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, મધ્યમપદલોપી દ્વન્દ્વ તત્પુરુષ, અવ્યવીભાવ, મધ્યમપદલોપી દ્વન્દ્વ, મધ્યમપદલોપી દ્વન્દ્વ તત્પુરુષ, ANSWER DOWNLOAD EXAMIANS APP
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 + 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 + j13) A Ia₀ = 0 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) 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 + j13) 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) 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 150 kW 450 kW 600 kW 300 kW 150 kW 450 kW 600 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) Match the following A = (ii), B = (i), C = (iii) A = (iii), B = (i), C = (ii) A = (ii), B = (iii), C = (i) A = (iii), B = (ii), C = (i) A = (ii), B = (i), C = (iii) A = (iii), B = (i), C = (ii) A = (ii), B = (iii), C = (i) A = (iii), B = (ii), C = (i) ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Bus Type - Known Parameter - Unknown ParameterLoad Bus -P, Q - V, phase angleGenerator Bus - P, V (magnitude) - Q, Voltage phase angleSlack Bus Voltage - magnitude and phase angle - P, Q
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