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 86.32 µF 96.32 µF 66.32 µF 76.32 µF 86.32 µF 96.32 µF 66.32 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) સોહેલ ખાન : મલાયકા અરોર :: સૈફ અલી ખાન : ?? માધુરી દિક્ષિત પરવીન બાબી કિરણ ખેર અમૃતા સિંહ માધુરી દિક્ષિત પરવીન બાબી કિરણ ખેર અમૃતા સિંહ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) સ્વાધીન' નો વિરોધીભાસી શબ્દ કયો છે? પરાધીન કુવાધીન અસ્વાધીન નીસ્વધીન પરાધીન કુવાધીન અસ્વાધીન નીસ્વધીન ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In a typical DC micro grid, hybrid storage system is a combination of Lead acid battery and Capacitor None of these Lead acid battery and ultra capacitor / super capacitor Lead acid battery and Lithium ion battery Lead acid battery and Capacitor None of these Lead acid battery and ultra capacitor / super capacitor Lead acid battery and Lithium ion battery ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Hybrid storage system is a combination of microturbines, fuel cells, photovoltaics (PV), Lead acid battery and ultra or super capacitors etc.
MGVCL Exam Paper (30-07-2021 Shift 2) Find the word which is correctly spelt from the given options. Wrinkel Sentimant Maximum Absurrd Wrinkel Sentimant Maximum Absurrd ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 3-phase motor load has a p.f. of 0.397 lagging. The two wattmeter method used to measure power show the input as 30 kW. Find the reading on each wattmeter. W1 = -10 kW and W2 = 40 kW W1 = -5 kW and W2 = 35 kW W1 = 10 kW and W2 = 20 kW W1 = 5 kW and W2 = 25 kW W1 = -10 kW and W2 = 40 kW W1 = -5 kW and W2 = 35 kW W1 = 10 kW and W2 = 20 kW W1 = 5 kW and W2 = 25 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,Cosφ = 0.397P = 30 kW = √3*V_L*I_L*cosφV_L*I_L = 30kW/(√3*0.397) = 43628.48 VAReadings of wattmeter W1 = V_L*I_L*cos(30+φ)Readings of wattmeter W2 = V_L*I_L*cos(30-φ)
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