MGVCL Exam Paper (30-07-2021 Shift 3) A rectangular conductor is 1.6 inches wide and 0.25 inch thick. What is its area in square mils? 40000 square mils 400000 square mils 400 square mils 4000 square mils 40000 square mils 400000 square mils 400 square mils 4000 square mils ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 1.6 inches = 1.6 inches x 1,000 mils per inch = 1,600 mils0.25 inch = 0.25 inch x 1,000 mils per inch = 250 milsArea = 1,600 x 250 = 400,000 square mils
MGVCL Exam Paper (30-07-2021 Shift 3) The sending end and receiving end voltages of the short transmission line are 150 kV and 120 kV respectively. Calculate its percentage voltage regulation. 20% 40% 30% 25% 20% 40% 30% 25% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,Vs - sending end voltage in kVVr - Receiving end voltage in kV.Voltage regulation = (Vs - Vr)/Vr= (150 - 120)/120= 0.25% R = 25 %
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 કુલ 1 કુલ 3 એક પણ નહિ બધી કહેવતો સમાનર્થી છે કુલ 1 કુલ 3 એક પણ નહિ બધી કહેવતો સમાનર્થી છે ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Message ____ means that the data must arrive at the receiver exactly as sent. Confidentiality Integrity Authentication None ot these Confidentiality Integrity Authentication None ot these ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Fill in the blanks with suitable Article from the given alternatives.Everyone is ready to accept ____ new rules framed by the executive body No article the an a No article the an a 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 = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - 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 = (-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 = 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
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