MGVCL Exam Paper (30-07-2021 Shift 3) Which is the function of Operating System? Devices Management Security Process Management All of these Devices Management Security Process Management All of these ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A 3 pF capacitor is charged by a constant current of 2 pA for six seconds. The voltage across the capacitor at the end of charging will be 6 V 8 V 2 V 4 V 6 V 8 V 2 V 4 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation know that the charge stored in a capacitor is given as:Q = C*VQ = I*tV = (I*t)/C= (2μ*6)/(3μ)= 4 V
MGVCL Exam Paper (30-07-2021 Shift 3) Match the following shown in table A = (ii), B = (iii), C = (i) A = (iii), B = (i), C = (ii) A = (iii), B = (ii), C = (i) A = (ii), B = (i), C = (iii) A = (ii), B = (iii), C = (i) A = (iii), B = (i), C = (ii) A = (iii), B = (ii), C = (i) A = (ii), B = (i), C = (iii) ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Bus Type - Known Parameter - Unknown ParameterLoad Bus -P, Q - V, phase angleGenerator Bus - P, V (magnitude) - Q, Voltage phase angleSlack Bus Voltage - magnitude and phase angle - P, Q
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 બધી કહેવતો સમાનર્થી છે કુલ 3 એક પણ નહિ કુલ 1 બધી કહેવતો સમાનર્થી છે કુલ 3 એક પણ નહિ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A single phase motor connected to 400 V, 50 Hz supply takes 25 A at a power factor of 0.75 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.95 lagging. 92.55 pF 72.55 pF 82.55 pF 62.55 pF 92.55 pF 72.55 pF 82.55 pF 62.55 pF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For single phase induction motor,P = V*I*cosφ= 400*25*0.75= 7500 WQ1 = P*tan(φ) = 7500*0.80= 6015 VARNow power factor change to 0.95So, φ' = 25.84 degreeQ = 7500*tan(18.19)= 2464.424Qnet = Q - Q'= 6015 - 2464.424= 4500.8C = V²/(2*ᴨ*f*Q)= (400*400)/(2*ᴨ*50*3575.76)= 8.255*10^(-5)= 82.55 μF
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. 25% 20% 30% 40% 25% 20% 30% 40% 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 %
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