MGVCL Exam Paper (30-07-2021 Shift 2) Magnetic tape is not practical for applications where data must be quickly recalled because tape is a/an____. None of these Sequential access medium expensive storage medium Random access medium None of these Sequential access medium expensive storage medium Random access medium 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 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -5 kW and W2 = 35 kW W1 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 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) A delta-connected balanced three-phase load is supplied from a three-phase, 400 V supply. The line current is 20 A and the power taken by the load is 10 kW. Find the power factor. 0.72 0.52 0.82 0.62 0.72 0.52 0.82 0.62 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = 10 kWVL = 400 VIL = 20 A3- phse power, P = √3*VL*IL*cosφcosφ = 10*1000/(√3*400*20)= 0.7216
MGVCL Exam Paper (30-07-2021 Shift 2) Choose the word which expresses nearly the opposite meaning of the given word "WHIMSICAL" Curious Fantastic Regular Funny Curious Fantastic Regular Funny ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) The line currents in amperes in phases a, b and c respectively are 500 +j150, 100 – j600 and – 300 + j600 referred to the same reference vector with phase sequence of abc. Find the symmetrical component of currents. Ia₀ = (100+j50) AIa₁ = (200+ j107.75) AIa₂ = (200-j107.75) A Ia₀ = (100+j50) AIa₁ = (546.41+j165.47) AIa₂ = (-146.41-j65.47) A Ia₀ = (100+j50) AIa₁ = (146.41+ j65.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (546.41+ j165.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (200+ j107.75) AIa₂ = (200-j107.75) A Ia₀ = (100+j50) AIa₁ = (546.41+j165.47) AIa₂ = (-146.41-j65.47) A Ia₀ = (100+j50) AIa₁ = (146.41+ j65.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (546.41+ j165.47) AIa₂ = (546.41-j165.47) A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Calculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 500 + j150 + (- 0.5 + j*0.866)*(100 – j6000) + (-0.5 - j*0.866)*(– 300 + j600)= (546.41 + j*156.47) AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= 500 + j150 + (- 0.5 - j*0.866)*(100 – j6000) + (-0.5 + j*0.866)*(– 300 + j600)= (-146.41 - j*56.47) AIao = 1/3*(Ia + Ib + Ic)= 1/3*(500 + j150 + 100 – j6000 – 300 + j600)= (100 + j*50) A
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) cos[1/√(LC)]t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) sin√(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[1/√(LC)]t A i(t) = (E/R) sin√(LC) t A i(t) = (E/R) cos√(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
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