MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is linux operating system? HP-UX MacOS Fedora IBM-AIX HP-UX MacOS Fedora IBM-AIX ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) If the velocity of electromagnetic wave in free space is 3 × 10⁸ m/s, then the velocity in a medium with εr of 4.5 and μr of 2 would be 9 × 10⁸ m/s 27 × 10⁸ m/s 1 × 10⁸ m/s 3 × 10⁸ m/s 9 × 10⁸ m/s 27 × 10⁸ m/s 1 × 10⁸ m/s 3 × 10⁸ m/s ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Speed of EM wave in medium = C/√(εr*μr)= (300000000)/√9= 300000000/3= 100000000= 1*10⁸ m/s
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 nH 50 nH 50 mH 25 mH 25 nH 50 nH 50 mH 25 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) 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) cos[1/√(LC)]t A i(t) = (E/R) cos√(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[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) sin√(LC) t A i(t) = (E/R) sin[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) A three-phase, 12 kV, 50 Hz and 300 kW load is operating at 0.8 pf lag. If the pf is to be improved to unity, using star connected capacitor, the per phase value of the reactive VAR required is: 100 kVAR 50 kVAR 25 kVAR 75 kVAR 100 kVAR 50 kVAR 25 kVAR 75 kVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The reactive VAR required = kW*(tanφ1 -tanφ2)Cosφ1 = 0.8φ1 = 36.87 deg.Cosφ2 = 0φ2 = 0 deg.kVAR required = kW*(tan(36.87) - tan(0))= 300*1000(0.75)= 225.00 kVARPer phase value = 225.00/3= 75 kVAR
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 76.32 µF 86.32 µF 96.32 µF 66.32 µF 76.32 µF 86.32 µF 96.32 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
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