MGVCL Exam Paper (30-07-2021 Shift 2) Which one of the following is the largest windfarm located in India? Mundra Vindhayachal Talcher Muppandal Mundra Vindhayachal Talcher 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) A coil having an inductance of 100 mH is magnetically coupled to another coil having an inductance of 900 mH. The coefficient of coupling between the coils is 0.45. Calculate the equivalent inductance if the two coils are connected in series opposing. 1270 mH 932 mH 1135 mH 730 mH 1270 mH 932 mH 1135 mH 730 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP M = k√(L₁L₂)M= 0.45√(100*900)M = 135 mHL_eq for opposite connection of coils = L₁ + L₂ - 2ML_eq = 900 + 100 - 2*135L_eq = 730 mH
MGVCL Exam Paper (30-07-2021 Shift 2) Tree Topology Font style Web layout Page Orientation Paper size Font style Web layout Page Orientation Paper size ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A MOSFET rated for 15 A, carries a periodic current as shown in figure. The ON state resistance of the MOSFET is 0.15 Ω. The average ON state loss in the MOSFET is 33.8 W 7.5 W 15 W 3.8 W 33.8 W 7.5 W 15 W 3.8 W ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Ip = 10 ATon = πT = 2πIavg = Ip*√(Ton/T)Iavg = 10*√(1/2)= 7.07 AThe average ON state loss in MOSFET, P = Iavg²*RP = 7.07²*0.15P = 7.5 W
MGVCL Exam Paper (30-07-2021 Shift 2) પાણી : પાણીગ્રહણ :: હસ્ત : ??? પ્રેમાલાપ છુટાછેડા હસ્તમેળાપ વિવાહ પ્રેમાલાપ છુટાછેડા હસ્તમેળાપ વિવાહ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 4 hours. What is the permissible overload for a duration of four hours. Assume the permissible load kVA as a fraction of rated kVA is 1.24. 930 kVA 1240 kVA 2480 kVA 620 kVA 930 kVA 1240 kVA 2480 kVA 620 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
EXAMIANS