MGVCL Exam Paper (30-07-2021 Shift 2) Choose the word which expresses nearly the opposite meaning of the given word "WHIMSICAL" Funny Curious Fantastic Regular Funny Curious Fantastic Regular ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A wire pilot is generally economical for distances up to 30 km to 60 km 10 km to 15 km 20 to 25 km 75 km to 100 km 30 km to 60 km 10 km to 15 km 20 to 25 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) 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) sin√(LC) t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) sin√(LC) t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos√(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
MGVCL Exam Paper (30-07-2021 Shift 2) Fill in the blanks with suitable Article from the given alternatives.I heard ____interesting story yesterday a an the No article a an the No article ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 50 Hz synchronous generator is connected to an infinite bus through a line. The p.u. reactances of generator and the line are j0.3 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. 1.42 Hz 1.62 Hz 1.52 Hz 1.32 Hz 1.42 Hz 1.62 Hz 1.52 Hz 1.32 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))= (11/05)*cosδ= (11/0.5)*0.5M = (H*s)/(πf)= 4/(50π)fn = (1.76*(3/50π))= 9.4 rad/sec = 1.32 Hz
MGVCL Exam Paper (30-07-2021 Shift 2) હુ ગીત ગાઈશ' ને કર્મણીમાં ફેરવો. હુ ગીતને ગાઈશ મારાથી ગીત ગવાશે ગીતને હુ જ ગાઇશ ગીતને હુ ગાઈશ હુ ગીતને ગાઈશ મારાથી ગીત ગવાશે ગીતને હુ જ ગાઇશ ગીતને હુ ગાઈશ ANSWER DOWNLOAD EXAMIANS APP