MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is true regarding access lists applied to an interface? You can apply two access lists to any interface One access list may be configured, per direction, for each layer 3 protocol configured on an interface You can place as many access lists as you want on any interface until you run out of memory. You can apply only one access list on any interface You can apply two access lists to any interface One access list may be configured, per direction, for each layer 3 protocol configured on an interface You can place as many access lists as you want on any interface until you run out of memory. You can apply only one access list on any interface ANSWER DOWNLOAD EXAMIANS APP
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 Sequential access medium None of these expensive storage medium Random access medium Sequential access medium None of these expensive storage medium ANSWER DOWNLOAD EXAMIANS APP
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 96.32 µF 76.32 µF 66.32 µF 86.32 µF 96.32 µF 76.32 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) Which one of the following is the committee on poverty estimation? Ad hoc committe S.S.Tarapore committee Tendulkar committee Karve committee Ad hoc committe S.S.Tarapore committee Tendulkar committee Karve committee 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 = -5 kW and W2 = 35 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 kW W1 = -5 kW and W2 = 35 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 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)
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) sin√(LC) t A i(t) = (E/R) cos√(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) sin√(LC) t A i(t) = (E/R) cos√(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
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