MGVCL Exam Paper (30-07-2021 Shift 2) In a typical DC micro grid, hybrid storage system is a combination of Lead acid battery and Capacitor Lead acid battery and Lithium ion battery Lead acid battery and ultra capacitor / super capacitor None of these Lead acid battery and Capacitor Lead acid battery and Lithium ion battery Lead acid battery and ultra capacitor / super capacitor None of these ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Hybrid storage system is a combination of microturbines, fuel cells, photovoltaics (PV), Lead acid battery and ultra or super capacitors etc.
MGVCL Exam Paper (30-07-2021 Shift 2) The Great Barrier Reef is located in which country? Germany Canada India Australia Germany Canada India Australia ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) શિક્ષકે વિદ્યાર્થીને વાર્તા કહી'1. વાક્ય અક્મર્ક છે.2. વાક્ય દ્રીકર્મક છે.3. વાક્યમાં વિદ્યાર્થીને 'પ્રધાન કર્મ' છે અને વાર્તા 'ગૌણ કર્મ' છે.4.વાક્યમાં વાર્તાને 'પ્રધાન કર્મ' છે અને વિધાથીને 'ગૌણ કર્મ' છે 1,3 1,3 2,3 2,4 1,3 1,3 2,3 2,4 ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In the circuit shown in figure, the voltage across 8 Ω resistor is 20 V. Calculate the supply voltage. 47 V 18 V 27 V 38 V 47 V 18 V 27 V 38 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equivalent resistance of the circuit,Req = (12||18) + 8Req = 7.2 + 8Req = 15.2 ohmSource current = current through 8 ohm resistor, I = Vr/R8I = 20/8I = 2.5 ASupply votage V = I*ReqV = 2.5*15.2V = 38 V
MGVCL Exam Paper (30-07-2021 Shift 2) A circuit breaker is rated 2500 A, 2000 MVA, 33 kV, 3 sec, 3-phase oil circuit breaker. Determine the breaking current. 25 kA 350 kA 35 kA 2500 A 25 kA 350 kA 35 kA 2500 A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Breaking capacity of circuit breaker = √3*VL*IbVL - line voltage in voltIL - Line current/breaking in ampereVL = 33 kVS = 2000 MVAIb = 30*1000 (in kVA)/(√3*33)Ib = 34.99 kA
MGVCL Exam Paper (30-07-2021 Shift 2) A single phase motor connected to 415 V, 50 Hz supply takes 30 A at a power factor of 0.7 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.9 lagging. 66.32 µF 96.32 µF 86.32 µF 76.32 µF 66.32 µF 96.32 µF 86.32 µF 76.32 µF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = VIcosφ = 8715 wattQ = VAR required to improve PF = P(tanφ₁ - tanφ₂) = 4670 VARCapacitive Reactance, Xc = V²/Q = 36.8790 ohm.C = (2πfXc)⁻¹ = 86.314 μF
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