## Given the following circuit:

## A) Determine the power factor of the parallel load combination.

## B) Calculate the value of the parallel capacitor required to correct the power factor to unity.

### A) Start by determining an expression for the complex power of the parallel load combination:

## Obtain an expression for the complex power of load "A":

Real Power for load "A": $$ P_A = 24 \; kW $$ Apparent power for load "A": $$ S_A = \frac{P_A}{P_{f_A}} = \frac{24 \; kW}{0.8} = 30 \; kVA $$ Power factor angle for load "A": $$ \cos(\theta_A) = 0.8 $$ $$ \theta_A = \cos^{-1}(0.8) = 36.87^{\circ} $$ Reactive power for load "A": $$ Q_A = S_A \sin(\theta_A) $$ $$ = 30,000 \sin(36.87^{\circ}) $$ $$ Q_A = 18 \; kVAR $$ The complex power for load "A": can now be expressed as: $$\mathbb{S}_A = P_A + jQ_A $$ $$ \mathbb{S}_A = 24,000 + j 18,000 \qquad (Eqn \; 1) $$

## Obtain an expression for the complex power of load "B":

Real Power for load "B": $$ P_B = 40 \; kW $$ Apparent power for load "B": $$ S_B = \frac{P_B}{P_{f_B}} = \frac{40 \; kW}{0.95} = 42,105 \; VA $$ Power factor angle for load "B": $$ \cos(\theta_B) = 0.95 $$ $$ \theta_B = \cos^{-1}(0.95) = 18.19^{\circ} $$ Reactive power for load "B": $$ Q_B = S_B \sin(\theta_B) $$ $$ = 42,105 \sin(18.19^{\circ}) $$ $$ Q_B = 13,144 \; VAR $$ The complex power for load "B": can now be expressed as: $$\mathbb{S}_B = P_B + jQ_B $$ $$ \mathbb{S}_B = 40,000 + j 13,144 \qquad (Eqn \; 2) $$

## Apply the law of conservation of AC power to get the complex power of the circuit:

$$ \mathbb{S} = \mathbb{S}_A + \mathbb{S}_B $$ Substituting equations 1 and to into the above expression gives us: $$ \mathbb{S} = (24,000 + j 18,000) + (40,000 + j 13,144) $$ $$ \mathbb{S} = 64,000 + j31,144 = 71,175 \angle 25.95^{\circ} \qquad (Eqn \; 3) $$ The real power of the circuit will be: $$ P = 64,000 \; W $$ From equation #3 we see that : $$ 25.95^{\circ} = (\theta_v - \theta_i) = power \; factor \; angle $$ We can now determine the power factor of the original circuit: $$ P_f = \cos(25.95^{\circ}) $$

$$ P_f = 0.899 $$

### B) Determine the capacitor value required to correct power factor to unity:

In order to get a power factor of "1" we need the new power factor angle to be 0 degrees (cos[0]=1). Recall the formula for the capacitance value required to achieve a unity power factor for an inductive load (found on the previous page) $$ C = \frac{P[\tan(\theta_1) - \tan(\theta_2)]}{\omega V^2_{rms}} $$ ...where: $$ \theta_2 = new \; power \; factor \; angle = 0^{\circ} $$ Recall that : $$ \omega = 2 \pi f $$ and the frequency (f) of our circuit was given to us as 60 Hz. Therefore: $$ \omega = 120 \pi $$ We now have all of the necessary information to determine the required capacitor value: $$ C = \frac{64,000[\tan(25.95^{\circ}) - \tan(0^{\circ})]}{120 \pi (120)^2} $$

$$ C = 5.74 \; \mu F $$