MGVCL Exam Paper (30-07-2021 Shift 3) Tiger Woods is associated with which of the following fields? Sports Literature Cinema Politics Sports Literature Cinema Politics ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) વિષમ' ની સંધી છૂટી પાડો. - કયો વિકલ્પ સાચો છે? વિસ્ + અમ વિ + સમ વીસ + અમ વી + સમ વિસ્ + અમ વિ + સમ વીસ + અમ વી + સમ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Two-wattmeter method is used to measure the power taken by a 3-phase induction motor on no load. The wattmeter readings are 400 W and -50 W. Calculate the reactive power taken by the load 350√3 VAR 350/√3 VAR 450√3 VAR 450/√3 VAR 350√3 VAR 350/√3 VAR 450√3 VAR 450/√3 VAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Reactive power taken (Q) = √3*(W1 - W2)= √3*(450 + 50)= √3*450 kVAR
MGVCL Exam Paper (30-07-2021 Shift 3) દ્રશ્ય' શબ્દનો વિરુદ્ધાર્થ શબ્દ આપેલો વિકલ્પોમાંથી કયો છે? દ્રષ્ટિ દ્રશ્યમાન દ્રષ્ટિવંત અદ્રશ્ય દ્રષ્ટિ દ્રશ્યમાન દ્રષ્ટિવંત અદ્રશ્ય 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 = 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 = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) 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 = (-5 + j2.44) A.Ia₂ = (5 - j2.44) 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) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 2 hours. What is the permissible overload for a duration of two hours. Assume the permissible load kVA as a fraction of rated kVA is 1.43. 715 kVA 1430 kVA 700 kVA 1050 kVA 715 kVA 1430 kVA 700 kVA 1050 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload for transformer = Full load kVA*factor for permissible overload= 1000*1.43= 1430 kVA
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