MGVCL Exam Paper (30-07-2021 Shift 3) Find the word which is correctly spelt from the given options. Abolishid Seriuous Syndrome Obecity Abolishid Seriuous Syndrome Obecity ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Which of the following medium is used between CPU and RAM to speed up the processing power of a CPU? Cache memory DRAM Virtual memory Flash memory Cache memory DRAM Virtual memory Flash memory ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A rectangular conductor is 1.6 inches wide and 0.25 inch thick. What is its area in square mils? 400000 square mils 4000 square mils 40000 square mils 400 square mils 400000 square mils 4000 square mils 40000 square mils 400 square mils ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 1.6 inches = 1.6 inches x 1,000 mils per inch = 1,600 mils0.25 inch = 0.25 inch x 1,000 mils per inch = 250 milsArea = 1,600 x 250 = 400,000 square mils
MGVCL Exam Paper (30-07-2021 Shift 3) A three phase four pole 50 Hz induction motor has a rotor resistance of 0.02 Ω/phase and stand-still reactance of 0.5 Ω/phase. Calculate the speed at which the maximum torque is developed. 1440 rpm 1525 rpm 1500 rpm 1475 rpm 1440 rpm 1525 rpm 1500 rpm 1475 rpm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Condition for maximum torque,Sm = R2/X2= 0.04Ns = (120*f)/P= (120*50)/4= 1500 rpmS = (Ns - Nr)/NsNr = Ns(1 - Sm)= 1500*(1 - 0.04)= 1440 rpm
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 0.1 Ω 1 Ω 0.25 Ω 2.5 Ω 0.1 Ω 1 Ω 0.25 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) A 3 pF capacitor is charged by a constant current of 2 pA for six seconds. The voltage across the capacitor at the end of charging will be 4 V 8 V 2 V 6 V 4 V 8 V 2 V 6 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation know that the charge stored in a capacitor is given as:Q = C*VQ = I*tV = (I*t)/C= (2μ*6)/(3μ)= 4 V
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