MGVCL Exam Paper (30-07-2021 Shift 3) Match the following shown in table A = (iii), B = (i), C = (ii) A = (ii), B = (iii), C = (i) A = (ii), B = (i), C = (iii) A = (iii), B = (ii), C = (i) A = (iii), B = (i), C = (ii) A = (ii), B = (iii), C = (i) A = (ii), B = (i), C = (iii) 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
MGVCL Exam Paper (30-07-2021 Shift 3) What is the value of the characteristic impedance of a transmission line with impedance and admittance of 15 and 5? 1.732 1.414 0.577 0.7.7 1.732 1.414 0.577 0.7.7 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Characteristic impedance = √(impedance/admittance) = 1.73
MGVCL Exam Paper (30-07-2021 Shift 3) Fill in the blanks with suitable Preposition from the given alternatives.BCCI receives____1000 applications for Team India head coach position since for over against since for over against 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. 100 m 200 m 300 m 400 m 100 m 200 m 300 m 400 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. 0.25 Ω 0.1 Ω 1 Ω 2.5 Ω 0.25 Ω 0.1 Ω 1 Ω 2.5 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) What is the value of capacitance of a capacitor which has a voltage of 4 V and has 16 C of charge? 16 F 8 F 2 F 4 F 16 F 8 F 2 F 4 F ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation is given by:C = Q/V= 16/4= 4 F
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