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 = 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 W1 = -10 kW and W2 = 40 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)
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| 3 and 5 3 and 6 5 and 3 6 and 3 3 and 5 3 and 6 5 and 3 6 and 3 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) Transistors were used in ____ of Computers. Fourth Generation First Generation Second Generation Third Generation Fourth Generation First Generation Second Generation Third Generation ANSWER DOWNLOAD EXAMIANS APP
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. 1135 mH 1270 mH 730 mH 932 mH 1135 mH 1270 mH 730 mH 932 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) પાણી : પાણીગ્રહણ :: હસ્ત : ??? છુટાછેડા પ્રેમાલાપ હસ્તમેળાપ વિવાહ છુટાછેડા પ્રેમાલાપ હસ્તમેળાપ વિવાહ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A circular loop has its radius increasing at a rate of 2 m/s. The loop is placed perpendicular to a constant magnetic field of 0.1 Wb/m². When the radius of the loop is 2 m, the emf induced in the loop will be 0.4π V 0.2π V zero 0.8π V 0.4π V 0.2π V zero 0.8π V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP emf = dΨ/dt = change of flux linkageemf = dф/dtemf = d(B*Area)/dtemf = B*d(area)/dtemf = B*d(πr²)/dtemf = Bπ*2r*dr/dtemf = Bπ*2r*change of radius per unit timeemf = 0.1*π*2*2emf = 4π