MGVCL Exam Paper (30-07-2021 Shift 2) Find the word which is correctly spelt from the given options. Socialy Abolishid Oscillate seriuous Socialy Abolishid Oscillate seriuous ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A wire pilot is generally economical for distances up to 20 to 25 km 30 km to 60 km 10 km to 15 km 75 km to 100 km 20 to 25 km 30 km to 60 km 10 km to 15 km 75 km to 100 km ANSWER EXPLANATION DOWNLOAD EXAMIANS APP A wire pilot is generally economical for distances up to 5 or 10 miles, beyond which a carrier-current pilot usually becomes more economical.
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. 3% 4% 6% 5% 3% 4% 6% 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 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. 86.32 µF 76.32 µF 96.32 µF 66.32 µF 86.32 µF 76.32 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) Magnetic tape is not practical for applications where data must be quickly recalled because tape is a/an____. Random access medium expensive storage medium None of these Sequential access medium Random access medium expensive storage medium None of these Sequential access medium ANSWER DOWNLOAD EXAMIANS APP
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√(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 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
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