MGVCL Exam Paper (30-07-2021 Shift 3) The most commonly used method for the protection of three phase feeder is Differential protection Reverse power protection Time graded protection None of these Differential protection Reverse power protection Time graded protection None of these ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Differential Pilot Wire Protection is simply a differential protection scheme applied to feeders.Several differential schemes are applied for protection of line but Mess Price Voltage balance system and Translay Scheme are most popularly used.
MGVCL Exam Paper (30-07-2021 Shift 3) સારું:નરસું::શુક્લ: ? અવિશાની કૃષ્ણ પરમ અમર અવિશાની કૃષ્ણ પરમ અમર ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A conductor having surface density is embedded in a dielectric medium of permittivity. The electric field in the medium is E. If it is known that the pressure p on the conductor surface is equal to the electric energy density in the medium, then p (in SI unit)is given by σ²/(2πε) σ²/(2ε) σ²/(4π) σ/(4πε) σ²/(2πε) σ²/(2ε) σ²/(4π) σ/(4πε) ANSWER EXPLANATION DOWNLOAD EXAMIANS APP As the point is near to the conductor will act as infinite sheetElectric field (E) is given by:E = σ²/(2ϵ)
MGVCL Exam Paper (30-07-2021 Shift 3) A 3 pF capacitor is charged by a constant current of 2 pA for six seconds. The voltage across the capacitor at the end of charging will be 2 V 8 V 6 V 4 V 2 V 8 V 6 V 4 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation know that the charge stored in a capacitor is given as:Q = C*VQ = I*tV = (I*t)/C= (2μ*6)/(3μ)= 4 V
MGVCL Exam Paper (30-07-2021 Shift 3) When was the Constitution of India adopted by the Constituent Assembly? 13th January 1950 15th August 1947 26th November 1949 January 25 1930 13th January 1950 15th August 1947 26th November 1949 January 25 1930 ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The current in the coil of a large electromagnet falls from 6 A to 2 A in 10 ms. The induced emf across the coil is 100 V. Find the self-inductance of the coil. 1.25 H 0.25 H 0.5 H 0.75 H 1.25 H 0.25 H 0.5 H 0.75 H ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation of self inductance:V = Ls*(di/dt)Ls = V/(di/dt)= 100/((6 - 2)/(0.01))= 0.250 H= 250 mH
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