MGVCL Exam Paper (30-07-2021 Shift 3) Choose the word which expresses nearly the opposite meaning of the given word "DILIGENT " Lasy Active Careful Busy Lasy Active Careful Busy ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Match the following shown in table 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) A = (ii), B = (i), C = (iii) A = (iii), B = (i), C = (ii) 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) The current in the coil of a large electromagnet falls from 6 A to 2 A in 10 ms. The induced emf across the coil is 100 V. Find the self-inductance of the coil. 1.25 H 0.75 H 0.5 H 0.25 H 1.25 H 0.75 H 0.5 H 0.25 H ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation of self inductance:V = Ls*(di/dt)Ls = V/(di/dt)= 100/((6 - 2)/(0.01))= 0.250 H= 250 mH
MGVCL Exam Paper (30-07-2021 Shift 3) When was the Constitution of India adopted by the Constituent Assembly? 15th August 1947 26th November 1949 13th January 1950 January 25 1930 15th August 1947 26th November 1949 13th January 1950 January 25 1930 ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A single phase motor connected to 400 V, 50 Hz supply takes 25 A at a power factor of 0.75 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.95 lagging. 62.55 pF 82.55 pF 72.55 pF 92.55 pF 62.55 pF 82.55 pF 72.55 pF 92.55 pF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For single phase induction motor,P = V*I*cosφ= 400*25*0.75= 7500 WQ1 = P*tan(φ) = 7500*0.80= 6015 VARNow power factor change to 0.95So, φ' = 25.84 degreeQ = 7500*tan(18.19)= 2464.424Qnet = Q - Q'= 6015 - 2464.424= 4500.8C = V²/(2*ᴨ*f*Q)= (400*400)/(2*ᴨ*50*3575.76)= 8.255*10^(-5)= 82.55 μF
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 2 V 6 V 8 V 4 V 2 V 6 V 8 V 4 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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