I had this in a practice test and would like some help solving.

Q. An industrial plant has an inductive load of 400kVA operating at a power factor of 0.6 lag and a resistive load of 40 kWatts. A synchronous motor is used to correct the power factor to 0.9 lag.

Assuming the synchronous motor produces no watts the corrected plant kVARS will be ?

Now i was wondering if i have to find the kVARS or Q of the 0.6 PF first, then the kVARS of the 0.9 PF, and take the difference as the corrected plant kVARS? The answer given is 135.5 kVARS, but i cannot seem to arrive at this number.


Any assistance is much appreciated.

 

I believe you have to determine the resistive component of the inductive load at 0.6 lag and then calculate from there. But it's not clear to me whether the 0.6 lag includes the 40kW resistive load.