Solved Problems about AC Powers

Problem 52-9 Source: Problem created by the author of the site.

The powers in a circuit are calculated, where the average power equal to P = 1 300 W and a power apparent value of S = 1 900 VA. After power factor correction, the measured apparent power was S_new = 1 410 VA. Find:

a) the value of PF before and after correction.

b) the value of Q (reactive power) before and after correcting the PF.

Solution of the Problem 52-9

Before PF Correction

Before of the PF correction, the values of P and S were provided. It is then possible to calculate the PF_old using the equation:

PF_old = P/S = 1 300/1 900 = 0.68

To calculate the reactive power (Q) before the PF correction, we use the equation:

Q_old = √(S² - P²) = √(1 900)² - (1 300)²

Carrying out the calculation, we find:

Q_old = 1 385.64 VAr

After PF Correction

In this case, we must be aware that when we correct the PF, the average or real power (P) does not change, that is, its value remains the same. Thus, using the new apparent power (S_new) given in the problem statement we can find the new value of PF, or:

FP_new = P/S_new = 1 300/1 410 = 0.92

To calculate the new reactive power (Q_new) we use the same equation as above, or:

Q_new = √(S_new² - P²) = √(1 410)² - (1 300)²

Carrying out the calculation, we find:

Q_new = 546 VAr

Note the big difference in reactive power after the PF correction.

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