MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is true regarding access lists applied to an interface? You can apply only one access list on any interface You can apply two access lists to any interface You can place as many access lists as you want on any interface until you run out of memory. One access list may be configured, per direction, for each layer 3 protocol configured on an interface You can apply only one access list on any interface You can apply two access lists to any interface You can place as many access lists as you want on any interface until you run out of memory. One access list may be configured, per direction, for each layer 3 protocol configured on an interface ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In transformer, bushing failure occurs due to(i) transformer vibrations(ii) dielectric losses(ii) partial discharge (i) and (iii) only (i), (ii) and (iii) (ii) and (iii) only (i) and (ii) only (i) and (iii) only (i), (ii) and (iii) (ii) and (iii) only (i) and (ii) only ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Choose the word which expresses nearly the opposite meaning of the given word "WHIMSICAL" Fantastic Regular Funny Curious Fantastic Regular Funny Curious 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) A delta-connected balanced three-phase load is supplied from a three-phase, 400 V supply. The line current is 20 A and the power taken by the load is 10 kW. Find the power factor. 0.82 0.72 0.52 0.62 0.82 0.72 0.52 0.62 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = 10 kWVL = 400 VIL = 20 A3- phse power, P = √3*VL*IL*cosφcosφ = 10*1000/(√3*400*20)= 0.7216
MGVCL Exam Paper (30-07-2021 Shift 2) In the context of smart meters, the term 'SMETS' stand for Standard Monitoring Equipment Technical Specifications Standard Metering Equipment Technical Specifications Smart Monitoring Equipment Technical Specifications Smart Metering Equipment Technical Specifications Standard Monitoring Equipment Technical Specifications Standard Metering Equipment Technical Specifications Smart Monitoring Equipment Technical Specifications Smart Metering Equipment Technical Specifications ANSWER DOWNLOAD EXAMIANS APP
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