MGVCL Exam Paper (30-07-2021 Shift 2) The Great Barrier Reef is located in which country? Germany Australia Canada India Germany Australia Canada India ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A wire pilot is generally economical for distances up to 10 km to 15 km 75 km to 100 km 20 to 25 km 30 km to 60 km 10 km to 15 km 75 km to 100 km 20 to 25 km 30 km to 60 km ANSWER EXPLANATION DOWNLOAD EXAMIANS APP A wire pilot is generally economical for distances up to 5 or 10 miles, beyond which a carrier-current pilot usually becomes more economical.
MGVCL Exam Paper (30-07-2021 Shift 2) In the context of smart meters, the term 'SMETS' stand for Standard Monitoring Equipment Technical Specifications Standard Metering Equipment Technical Specifications Smart Metering Equipment Technical Specifications Smart Monitoring Equipment Technical Specifications Standard Monitoring Equipment Technical Specifications Standard Metering Equipment Technical Specifications Smart Metering Equipment Technical Specifications Smart Monitoring Equipment Technical Specifications ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) The necessary equation to be solved during the load flow analysis using Fast-Decoupled method is given below:where B' and B'' are formed using the imaginary part of the bus admittance matrix, Ybus. If the system consists of 6 buses of which one of the bus is slack bus and 2 bues are voltage regulated buses, then then the matrix B' is of order ____and the matrix B'' is of order____, respectivity.∆P/|Vi| = - B'∆δ∆Q/|Vi| = - B''∆|V| 6 and 3 5 and 3 3 and 6 3 and 5 6 and 3 5 and 3 3 and 6 3 and 5 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In Fast-decoupled method,Order of B' = No. of PV buses = 5Order of B'' = No. of PQ buses = 3
MGVCL Exam Paper (30-07-2021 Shift 2) A short-shunt dc compound generator supplies 50 A at 300 V. If the shunt field resistance, series field resistance and armature resistance are 30 Ω, 0.03Ω, and 0.05 Ω respectively, determine the e.m.f. generated. 304.5 V 306.5 V 302.5 V 308.5 V 304.5 V 306.5 V 302.5 V 308.5 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,Vt = 300 VRse = 0.03 Ω, Rsh = 30 Ω, Ra = 0.05 ΩI_L = 50 AVsh = V + I_L*Rse = 300 + 50*0.03 = 301.5 VIsh = Vsh/Rsh = 301.5/30 = 10.05 AIa = Ish + I_L = 60.05 AEg = Vt + I_L*Rse + IaRaEg = 300 + 50*0.03 + 60.05*0.05 = 304.5025 V
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√(LC) t A i(t) = (E/R) cos[1/√(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√(LC) t A i(t) = (E/R) cos[1/√(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
EXAMIANS