MGVCL Exam Paper (30-07-2021 Shift 2) Which one of the following is the largest windfarm located in India? Muppandal Mundra Vindhayachal Talcher Muppandal Mundra Vindhayachal Talcher ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rank - Windfarm name - Capacity1 - Muppandal windfarm - 1500 MW2 - Jaisalmer Wind Park - 1064 MW3 - Brahmanvel Wind Farm Dhule - 964 MW4 - Damanjodi Wind Farm - 99 MW5 - Tuppadahalli Wind Farm - 56.1 MW
MGVCL Exam Paper (30-07-2021 Shift 2) In the circuit of figure, the switch is closed to position-1 for a long time. At t = 0 the switch position is changed from postion-1 to postion-2. Find an expression for the current i(t). i(t) = (E/R) sin√(LC) t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) sin√(LC) t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) cos[1/√(LC)]t A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Standard eqution of current for this condition,i(t) = i(0+)*cos(wt)i(0+) = E/RFor LC circuit,w = 1/√(LC)i(t) = (E/R)*cos((1/√(LC))*t) A
MGVCL Exam Paper (30-07-2021 Shift 2) In the context of smart meters, the term 'SMETS' stand for Smart Metering Equipment Technical Specifications Standard Metering Equipment Technical Specifications Standard Monitoring Equipment Technical Specifications Smart Monitoring Equipment Technical Specifications Smart Metering Equipment Technical Specifications Standard Metering Equipment Technical Specifications Standard Monitoring Equipment Technical Specifications Smart Monitoring Equipment Technical Specifications ANSWER DOWNLOAD EXAMIANS APP
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. 96.32 µF 86.32 µF 66.32 µF 76.32 µF 96.32 µF 86.32 µF 66.32 µF 76.32 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) A circular loop has its radius increasing at a rate of 2 m/s. The loop is placed perpendicular to a constant magnetic field of 0.1 Wb/m². When the radius of the loop is 2 m, the emf induced in the loop will be 0.2π V 0.4π V zero 0.8π V 0.2π V 0.4π V zero 0.8π V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP emf = dΨ/dt = change of flux linkageemf = dф/dtemf = d(B*Area)/dtemf = B*d(area)/dtemf = B*d(πr²)/dtemf = Bπ*2r*dr/dtemf = Bπ*2r*change of radius per unit timeemf = 0.1*π*2*2emf = 4π
MGVCL Exam Paper (30-07-2021 Shift 2) In a single phase energy meter, the shunt electromagnet is energized by a ____of ____ pressure coil, thin wire of more turns current coil, thin wire of more turns pressure coil, thick wire of few turns current coil, thick wire of few turns pressure coil, thin wire of more turns current coil, thin wire of more turns pressure coil, thick wire of few turns current coil, thick wire of few turns ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In single phase energy meter,Current coil is made up from thick wire having few no. of turns.Pressure coil is made up from thin wire having more no. of turns.
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