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...