MGVCL Exam Paper (30-07-2021 Shift 1) માણસ: ઘર :: પશુ : ગુફા આકાશ અંતરિક્ષ પાતાળ ગુફા આકાશ અંતરિક્ષ પાતાળ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) The current flowing to a balanced delta connected load through line 'a' is 15 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 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 - j13) AIa₂ = (7.5 + j13) A Ia₀ = 0 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 - j13) AIa₂ = (7.5 + j13) A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(15 + 15∟-120°)= 7.5 + j*4.33 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(15 + 15∟120°)= 7.5 - j*4.33 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0
MGVCL Exam Paper (30-07-2021 Shift 1) Recently, Dr. Subir Vithal Gokarn passed away due to illness. He is a/an ___ Former IAS Officer Former Finance Minister Former RBI Deputy Governor Former Chief Minister of Gujarat Former IAS Officer Former Finance Minister Former RBI Deputy Governor Former Chief Minister of Gujarat ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) An eight-pole wave-connected armature has 600 conductors and is driven at 625 rev/min. If the flux per pole is 20 mWb, determine the generated emf. 500 V 250 V 125 V 325 V 500 V 250 V 125 V 325 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let,E→EMF in voltφ→Flux in weberA→Parallel path (for wave winding = 2)Z→No. of armature conductorsN→speed in rpmP = Number of polesE = PφNZ/(60A)E = 8*0.020*625/(60*2)E = 500 V
MGVCL Exam Paper (30-07-2021 Shift 1) Determine maximum permissible load which a 15 MVA, ONAN cooled transformer to IS : 2026 can take for 6 hr, if the initial load on it was 12 MVA. The weighted ambient temperature is 20˚C. Assume the permissible load kVA as a fraction of rated kVA is 1.25. 9.6 MVA 18.75 MVA 12 MVA 15 MVA 9.6 MVA 18.75 MVA 12 MVA 15 MVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Maximum permissible load = Rated MVA rating*fraction of rated kVAMaximum permissible load = 15*1.25Maximum permissible load = 18.75 MVA
MGVCL Exam Paper (30-07-2021 Shift 1) In computer networks, the nodes are the computer that routes data the computer that terminates data the computer that originates data all the mentioned above the computer that routes data the computer that terminates data the computer that originates data all the mentioned above ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS