MGVCL Exam Paper (30-07-2021 Shift 2) Find the word which is correctly spelt from the given options. seriuous Abolishid Socialy Oscillate seriuous Abolishid Socialy Oscillate ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) What is the full form of the acronym 'AIDS' in diseases? Acquired Immune Deficiency Syndrome Acquire in Deficiency Syndrome Amino Intersect Deficiency Syndrome Assert in Dander Syndrome Acquired Immune Deficiency Syndrome Acquire in Deficiency Syndrome Amino Intersect Deficiency Syndrome Assert in Dander Syndrome ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In a short circuit test on circuit breaker, time to reach the peak re-striking voltage is 50 µs with the peak restriking voltage of 100 kV. Determine average RRRV (Rate of rise re-striking voltage)? 1 x 10⁶ kV/sec 2 x 10⁶ kV/sec 3 x 10⁶ kV/sec 4 x 10⁶ kV/sec 1 x 10⁶ kV/sec 2 x 10⁶ kV/sec 3 x 10⁶ kV/sec 4 x 10⁶ kV/sec ANSWER EXPLANATION DOWNLOAD EXAMIANS APP RRRV = Peak value of restriking voltage/time to reach the peak re-striking voltageRRRV = 100*1000/(50 μs)RRRV = 2*10⁶ kV/sec
MGVCL Exam Paper (30-07-2021 Shift 2) The line currents in amperes in phases a, b and c respectively are 500 +j150, 100 – j600 and – 300 + j600 referred to the same reference vector with phase sequence of abc. Find the symmetrical component of currents. Ia₀ = (100+j50) AIa₁ = (200+ j107.75) AIa₂ = (200-j107.75) A Ia₀ = (100+j50) AIa₁ = (546.41+ j165.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (546.41+j165.47) AIa₂ = (-146.41-j65.47) A Ia₀ = (100+j50) AIa₁ = (146.41+ j65.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (200+ j107.75) AIa₂ = (200-j107.75) A Ia₀ = (100+j50) AIa₁ = (546.41+ j165.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (546.41+j165.47) AIa₂ = (-146.41-j65.47) A Ia₀ = (100+j50) AIa₁ = (146.41+ j65.47) AIa₂ = (546.41-j165.47) A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Calculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 500 + j150 + (- 0.5 + j*0.866)*(100 – j6000) + (-0.5 - j*0.866)*(– 300 + j600)= (546.41 + j*156.47) AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= 500 + j150 + (- 0.5 - j*0.866)*(100 – j6000) + (-0.5 + j*0.866)*(– 300 + j600)= (-146.41 - j*56.47) AIao = 1/3*(Ia + Ib + Ic)= 1/3*(500 + j150 + 100 – j6000 – 300 + j600)= (100 + j*50) A
MGVCL Exam Paper (30-07-2021 Shift 2) A single phase motor connected to 415 V, 50 Hz supply takes 30 A at a power factor of 0.7 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.9 lagging. 76.32 µF 96.32 µF 66.32 µF 86.32 µF 76.32 µF 96.32 µF 66.32 µF 86.32 µF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = VIcosφ = 8715 wattQ = VAR required to improve PF = P(tanφ₁ - tanφ₂) = 4670 VARCapacitive Reactance, Xc = V²/Q = 36.8790 ohm.C = (2πfXc)⁻¹ = 86.314 μF
MGVCL Exam Paper (30-07-2021 Shift 2) In the RLC series circuit shown in figure, resonance occurs when the value of C is 20 µF. The supply voltages is v(t) = 20 sin (500t) V. Find the value of L. 5 mH 0.2 H 0.2 mH 5 H 5 mH 0.2 H 0.2 mH 5 H ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,ω = 500 rad/sec = 2πf.At Resonance condition,ωL = 1/(ωC)L 0.2 HAlso at series resonace1. fr = 1/[2π√(LC)]2. PF = 13. Maximum current4. Source voltage = voltage across resistor.
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