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. 66.32 µF 96.32 µF 86.32 µF 76.32 µF 66.32 µF 96.32 µF 86.32 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) The wind speed increases with height because of enhancement of drag effect on the earth surface reduction in the gravitational force reduction of drag effect of the earth surface reduction in lift effect of the air enhancement of drag effect on the earth surface reduction in the gravitational force reduction of drag effect of the earth surface reduction in lift effect of the air ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Wind speed increases as the height from the ground increases mainly due to the decrease in the friction produced by land terrain.
MGVCL Exam Paper (30-07-2021 Shift 2) Choose the word which best expresses the similar meaning of the given word "LUCID" Puzzled Cloudy Clear Sensible Puzzled Cloudy Clear Sensible ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 4 hours. What is the permissible overload for a duration of four hours. Assume the permissible load kVA as a fraction of rated kVA is 1.24. 2480 kVA 620 kVA 1240 kVA 930 kVA 2480 kVA 620 kVA 1240 kVA 930 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
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 zero 0.4π V 0.8π V 0.2π V zero 0.4π V 0.8π V 0.2π 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) Fill in the blanks with suitable Article from the given alternatives.I heard ____interesting story yesterday the No article an a the No article an a ANSWER DOWNLOAD EXAMIANS APP
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