MGVCL Exam Paper (30-07-2021 Shift 3) In which of the following situations, there is no need to provide directional overcurrent protection. double end fed, single feeder ring main single end fed, parallel feeder single end fed, single feeder double end fed, single feeder ring main single end fed, parallel feeder single end fed, single feeder ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Directional overcurrent relay protection is used for parallel feeder and ring main system.It is not used for the protection of radial system.
MGVCL Exam Paper (30-07-2021 Shift 3) મેં મીઠાઈ બનાવી'. વાક્યને કેવળ ક્રિયાપદની પ્રેરક રચના કઈ છે? મેં મીઠાઈ બનાવડાવી. મીઠાઈ તો હું જ બનવું ને! હું મીઠાઈ બનાવવા લાગી. મેં મીઠાઈ બનાવી લીધી. મેં મીઠાઈ બનાવડાવી. મીઠાઈ તો હું જ બનવું ને! હું મીઠાઈ બનાવવા લાગી. મેં મીઠાઈ બનાવી લીધી. ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) સાબરમતી: અમદાવાદ:: મૂસી: ગોવા લંડન વેનિસ હૈદરાબાદ ગોવા લંડન વેનિસ હૈદરાબાદ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A 50 Hz synchronous generator is connected to an infinite bus through a line. The p.u. reactances of generator and the line are j0.4 p.u. and j0.2 p.u. respectively. The generator no load voltage is 1.1 p.u. and that of infinite bus is 1.0 p.u. The inertia constant of the generator is 4 MW-sec/MVA. Determine the frequency of natural oscillations if the generator is loaded to 60% of its maximum power transfer capacity and small perturbation in power is given. 0.8 Hz 1.6 Hz 1.2 Hz 1.8 Hz 0.8 Hz 1.6 Hz 1.2 Hz 1.8 Hz ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Frequency of natural oscillation is given by,fn = {((dPe/dδ)at(δo))/M)}dPe/dδ = ((V1*V2)/X*(cosδ))= (1.1/0.6)*cosδ= (1.1/0.6)*0.5= 0.91M = (H*s)/(πf)= 4/(50π)fn = (0..91/(4/50π))= 8.9 rad/sec= 1.2 Hz
MGVCL Exam Paper (30-07-2021 Shift 3) Rearrange the following to form a meaningful sentence and find the most logical order from the given options.P: globally, out of which over 200 millionQ: are now in India, andR: flipkart has 700 million usersS: growing exponentially RPQS PQRS RPSQ RSQP RPQS PQRS RPSQ RSQP 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/(2ε) σ²/(2πε) σ²/(4π) σ/(4πε) σ²/(2ε) σ²/(2πε) 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)
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