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.
In single phase energy meter, Current coil is made up from thick wire having few no. of turns. Pressure coil is made up from thin wire having more no. of turns.
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
For lagging power factor = (IR*cosφ + IXsinφ)/V_r For leading power factor = (IR*cosφ - IXsinφ)/V_r Zero voltage regulation occurs on leading power factor only.
EXAMIANS