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.
The capacity of reactor = 50 MVAr when it is operated at 415 kV. Impedance of reactor = V2/MVAr = (415*415)2/50 = 3444.5 Ω For 400 kV, Reactive power observed = (400*400)/3444.5 = 46.45 MVAR
Or
VAr capacity of capacitor or reactor is proportional to V²
Equivalent resistance of the circuit, Req = (12||18) + 8 Req = 7.2 + 8 Req = 15.2 ohm Source current = current through 8 ohm resistor, I = Vr/R8 I = 20/8 I = 2.5 A Supply votage V = I*Req V = 2.5*15.2 V = 38 V
EXAMIANS