MGVCL Exam Paper (30-07-2021 Shift 2) When different types of consumers are charged at different uniform per unit rates, it is called as ____ block rate tariff two-part tariff maximum demand tariff flat rate tariff block rate tariff two-part tariff maximum demand tariff flat rate tariff ANSWER EXPLANATION DOWNLOAD EXAMIANS APP When different types of consumers are charged at different uniform per unit rates, it is said to be Flat Rate Tariff.In this type, the consumers are grouped into different classes.Each class is charged at different uniform rate.
MGVCL Exam Paper (30-07-2021 Shift 2) What is the full form of UN in 'UN Human Rights Prize'? United Nations Uniform Nacional Unity Nation Uniformed Nations United Nations Uniform Nacional Unity Nation Uniformed Nations ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A three-phase, 12 kV, 50 Hz and 300 kW load is operating at 0.8 pf lag. If the pf is to be improved to unity, using star connected capacitor, the per phase value of the reactive VAR required is: 100 kVAR 25 kVAR 75 kVAR 50 kVAR 100 kVAR 25 kVAR 75 kVAR 50 kVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The reactive VAR required = kW*(tanφ1 -tanφ2)Cosφ1 = 0.8φ1 = 36.87 deg.Cosφ2 = 0φ2 = 0 deg.kVAR required = kW*(tan(36.87) - tan(0))= 300*1000(0.75)= 225.00 kVARPer phase value = 225.00/3= 75 kVAR
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 = (546.41+j165.47) AIa₂ = (-146.41-j65.47) A Ia₀ = (100+j50) AIa₁ = (200+ j107.75) AIa₂ = (200-j107.75) A Ia₀ = (100+j50) AIa₁ = (546.41+ j165.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (146.41+ j65.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (546.41+j165.47) AIa₂ = (-146.41-j65.47) A Ia₀ = (100+j50) AIa₁ = (200+ j107.75) AIa₂ = (200-j107.75) A Ia₀ = (100+j50) AIa₁ = (546.41+ j165.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (146.41+ j65.47) 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) 1. આપને મુખ્યમત્રીએ ખાસ કામ માટે નિયુક્ત કર્યા છે.2. તેનાથી આંખો ઝીણી કરાઈ3. પ્રધાનમંત્રીએ સ્કુલનું ઉદ્રઘાટન કર્યું.આરોપીએ બધા ગુના કબુલ કરી લીધા.ઉક્ત વાક્યોમાં કેટલા કર્મણી વાક્યો પ્રયોગના છે? કુલ 1 કુલ 2 બધા જ કર્તરી વાક્ય પ્રયોગના છે બધા જ વાક્યો કર્મણી વાક્ય પ્રયોગના છે કુલ 1 કુલ 2 બધા જ કર્તરી વાક્ય પ્રયોગના છે બધા જ વાક્યો કર્મણી વાક્ય પ્રયોગના છે ANSWER DOWNLOAD EXAMIANS APP
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. 86.32 µF 96.32 µF 76.32 µF 66.32 µF 86.32 µF 96.32 µF 76.32 µF 66.32 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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