Answer:
The total system active power P = P_1 + P_2 + P_3 = 34.91 KW
Explanation:
Load 1: Active power P_1 = 20 HP = 14.91kW;
Reactive power
Load 2: Active power P_2 = 20 kW;
Reactive power Q2 = 0 since the load is purely resistive.
Load 3: Active power = 0 due to purely capacitiveload
Reactive power = -20 Var
a) since all three loads are connected in parallel therefore
The total system active power P = P_1 + P_2 + P_3 = 34.91 KW
Total system reactive power Q = Q_1 + Q_2 + Q_3 = 11.18 + 0 -20 = -8.82 kVar
Since Q = 0, the power factor is unity.
Supply current per phase is given by
