### For the following circuit, determine the complex power as well as P, Q, S and the power factor:

(note: The value of the voltage source is in terms of RMS.)

## Determine the complex power:

We will start by calculating the equivalent impedance of the circuit: $$ \mathbb{Z} = 2 + \frac{1}{\frac{1}{10-j5} + \frac{1}{8+j6}} $$ $$ \mathbb{Z} = 8.19 \angle 5.389^{\circ} $$ Now determine the phasor current flowing out of the voltage source: $$ \mathbb{I}_{rms} = \frac{\mathbb{V}_{rms}}{\mathbb{Z}} = \frac{50\angle 50^{\circ}}{8.19 \angle 5.389^{\circ}} $$ $$ \mathbb{I}_{rms} = 6.105 \angle 44.61^{\circ} $$ Recall the third expression for complex power. (Derived on the previous page): $$ \mathbb{S} = \mathbb{V}_{rms} \mathbb{I}_{rms}* $$ ... and use it to calculate the complex power for our circuit (don't forget we need to take the complex conjugate of the current.): $$ \mathbb{S} = [50\angle 50^{\circ}] \; [6.105 \angle (-44.61^{\circ})] $$

$$ \mathbb{S} = 305.3 \angle 5.39^{\circ} = (304+j28.7) \; VA $$

## Determine the real power (P):

Recall the definition of real power in terms of complex power. (Also derived on the previous page ): $$ P = R_e \{ \mathbb{S} \} = R_e \{ 304+j28.7 \} $$

$$ P = 304 \; W $$

## Determine the reactive power (Q):

Recall the definition of reactive power in terms of complex power. (Also derived on the previous page ): $$ Q = I_m \{ \mathbb{S} \} = I_m \{ 304+j28.7 \} $$

$$ Q = 28.7 \; VAR $$

## Determine the apparent power (S):

Recall the expression for apparent power: $$ S = V_{rms} I_{rms} $$ ...and applied to our circuit gives us: $$ S = (50)(6.105) $$

$$ S = 305.3 \; VA $$

## Determine the power factor:

Recall the following expression for the power factor: $$ P_f = \frac{P}{S} $$ Using our values for P and S gives us: $$ P_f = \frac{304}{305.3} $$

$$ P_f = 0.9957 $$

Let's look at another example:

Continue on to Complex Power example problem #2...