MGVCL Exam Paper (30-07-2021 Shift 2) Which one of the following is the largest windfarm located in India? Vindhayachal Talcher Mundra Muppandal Vindhayachal Talcher Mundra Muppandal ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rank - Windfarm name - Capacity1 - Muppandal windfarm - 1500 MW2 - Jaisalmer Wind Park - 1064 MW3 - Brahmanvel Wind Farm Dhule - 964 MW4 - Damanjodi Wind Farm - 99 MW5 - Tuppadahalli Wind Farm - 56.1 MW
MGVCL Exam Paper (30-07-2021 Shift 2) Recently, India and Benin signed MoUs on education, health and ____ Against Corruption Trading e-visa facilities Defence Against Corruption Trading e-visa facilities Defence ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In the circuit of figure, the switch is closed to position-1 for a long time. At t = 0 the switch position is changed from postion-1 to postion-2. Find an expression for the current i(t). i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) sin√(LC) t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) sin√(LC) t A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Standard eqution of current for this condition,i(t) = i(0+)*cos(wt)i(0+) = E/RFor LC circuit,w = 1/√(LC)i(t) = (E/R)*cos((1/√(LC))*t) A
MGVCL Exam Paper (30-07-2021 Shift 2) Calculate the inductance per meter of a 50 Ω load cable that has an capacitance of 20 pF/m. 50 nH 25 nH 25 mH 50 mH 50 nH 25 nH 25 mH 50 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Characteristic impedance, Z = √(L/C)50 = √(L/20*10⁻¹²)L = (50*50)*(20*10⁻¹²)L = 50000*10⁻¹²L = 50 nH
MGVCL Exam Paper (30-07-2021 Shift 2) પાણી : પાણીગ્રહણ :: હસ્ત : ??? છુટાછેડા હસ્તમેળાપ પ્રેમાલાપ વિવાહ છુટાછેડા હસ્તમેળાપ પ્રેમાલાપ વિવાહ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 3-phase motor load has a p.f. of 0.397 lagging. The two wattmeter method used to measure power show the input as 30 kW. Find the reading on each wattmeter. W1 = -10 kW and W2 = 40 kW W1 = 10 kW and W2 = 20 kW W1 = 5 kW and W2 = 25 kW W1 = -5 kW and W2 = 35 kW W1 = -10 kW and W2 = 40 kW W1 = 10 kW and W2 = 20 kW W1 = 5 kW and W2 = 25 kW W1 = -5 kW and W2 = 35 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,Cosφ = 0.397P = 30 kW = √3*V_L*I_L*cosφV_L*I_L = 30kW/(√3*0.397) = 43628.48 VAReadings of wattmeter W1 = V_L*I_L*cos(30+φ)Readings of wattmeter W2 = V_L*I_L*cos(30-φ)
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