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 86.32 µF 66.32 µF 76.32 µF 96.32 µF 86.32 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) Calculate the inductance per meter of a 50 Ω load cable that has an capacitance of 20 pF/m. 25 mH 50 nH 25 nH 50 mH 25 mH 50 nH 25 nH 50 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Characteristic impedance, Z = √(L/C)50 = √(L/20*10⁻¹²)L = (50*50)*(20*10⁻¹²)L = 50000*10⁻¹²L = 50 nH
MGVCL Exam Paper (30-07-2021 Shift 2) If all the devices are connected to a central hub, then the topology is called Star Topology Tree Topology Bus Topology Tree Topology Star Topology Tree Topology Bus Topology Tree Topology ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) સોહેલ ખાન : મલાયકા અરોર :: સૈફ અલી ખાન : ?? કિરણ ખેર પરવીન બાબી માધુરી દિક્ષિત અમૃતા સિંહ કિરણ ખેર પરવીન બાબી માધુરી દિક્ષિત અમૃતા સિંહ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A three-phase, 4-pole, 50 Hz induction motor runs at 1425 rpm. Find out the percentage slip of the induction motor. 4% 6% 3% 5% 4% 6% 3% 5% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Slip, s = (Ns - Nr)/NsNs = 120f/PNs = 1500 rpms = (1500 - 1425)/1500s = 0.05%s = 5%
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 = 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 W1 = -10 kW and W2 = 40 kW W1 = -5 kW and W2 = 35 kW W1 = 10 kW and W2 = 20 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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