MGVCL Exam Paper (30-07-2021 Shift 3) Transistors were used in First Generation Computers Third Generation Computers Fourth Generation Computers Second Generation Computers First Generation Computers Third Generation Computers Fourth Generation Computers Second Generation Computers ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Nairobi negotiations are related to which of the following Organisation? World Economic Forum The United Nations Children's Fund UNESCO World Trade Organisation World Economic Forum The United Nations Children's Fund UNESCO World Trade Organisation ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) નીચે પૈકી કેટલી કહેવતો વિરુદ્ધાથીઁ છે?.1. પાંચ બોલે તે પરમેશ્વર - ગામને મોઢે ગળણું ન બંધાય.2. ચોરની ચાર અને જોનારની બે - વિશ્વાસે વહાણ ચાલે.3. ખાલી ચણો વાગે ઘણો - અધુરો ઘડો છલકાય.4. હાથના કર્યા હૈયે વાગ્યા - દીવો લઈને કૂવામાં પડવું એક પણ નહિ કુલ 3 કુલ 1 બધી કહેવતો સમાનર્થી છે એક પણ નહિ કુલ 3 કુલ 1 બધી કહેવતો સમાનર્થી છે ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) સારું:નરસું::શુક્લ: ? પરમ અવિશાની અમર કૃષ્ણ પરમ અવિશાની અમર કૃષ્ણ 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 = (-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. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - 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) The current in the coil of a large electromagnet falls from 6 A to 2 A in 10 ms. The induced emf across the coil is 100 V. Find the self-inductance of the coil. 1.25 H 0.75 H 0.25 H 0.5 H 1.25 H 0.75 H 0.25 H 0.5 H ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation of self inductance:V = Ls*(di/dt)Ls = V/(di/dt)= 100/((6 - 2)/(0.01))= 0.250 H= 250 mH
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