MGVCL Exam Paper (30-07-2021 Shift 3) દ્રશ્ય' શબ્દનો વિરુદ્ધાર્થ શબ્દ આપેલો વિકલ્પોમાંથી કયો છે? દ્રષ્ટિ દ્રશ્યમાન અદ્રશ્ય દ્રષ્ટિવંત દ્રષ્ટિ દ્રશ્યમાન અદ્રશ્ય દ્રષ્ટિવંત ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The process of peeling of rocks into layers is called Brachans Delta Exfoliation Sublimation Brachans Delta Exfoliation Sublimation ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The current flowing in to a balanced delta connected load through line 'a' is 10 A when the conductor of line ’b’ is open. With the current in line 'a' as reference, compute the symmetrical components of the line currents. Assume the phase sequence of ’abc’. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(10 + 10∟-120°)= 5 + j*2.88 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(10 + 10∟120°)= 5 - j*2.88 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0
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 PQRS RPQS RSQP RPSQ PQRS RPQS RSQP RPSQ 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. 1.6 Hz 0.8 Hz 1.2 Hz 1.8 Hz 1.6 Hz 0.8 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) Fill in the blanks with suitable Preposition from the given alternatives.BCCI receives____1000 applications for Team India head coach position since over against for since over against for ANSWER DOWNLOAD EXAMIANS APP
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