Answer:
I = 27.65A < 40.59°
PowerFactor = 0.76
Explanation:
Current on the heating load is:
I1 = 30KW / 4KV = 7.5A < 0°
Current on the inductive load:
I2 = (150KVA*0.6) /4KV = 22.5A with an angle of acos(0.6)=53.1°
The sum of both currents is:
It = I1 + I2 = 7.5<0° + 22.5<53.1° = 27.65<40.59°
Now, the power factor will be:
pf = cos (40.59°) = 0.76