# Complex power calculation across parallel loads

I have the following problem which require the calculation of Load L3 apparent power, real power and reactive power. I have done the following as the described in the figure trying to get 2 equations of 2 unknowns but got to a dead end how can i proceed with such problem with small information the global power factor is also giving but with no details on whether it is lagging or leading. i assumed it is lagging since L1 and L2 are inductive loads based on the description

Intuitively, if all the load power factors are lagging, the overall power factor is lagging too.

% outer vertical array of arrays \begin{array}{c} % inner horizontal array of arrays \begin{array}{c|c} % inner array of minimum values \begin{align} S_1&=20\; \;\text{kVA}\\ V&=600 \angle{0}^{\circ} \\ I_1&=\frac{S}{V}=\frac{20k}{600}=33.3 \;\text{A}\\ \therefore {\overline{I}_{1}}&=33.3\angle{-36.87^{\circ}} \end{align} & % inner array of maximum values \overline{I}_{2}=\frac{V}{Z}=\frac{600}{15+30j}=8\sqrt{5}\angle{-63.43^{\circ}}\\ \end{array} \\ % inner array of delta values \hline \begin{matrix} &\overline{I}_{source}=\overline{I}=I\angle{-\arccos(0.707)}=I\angle{-45^{\circ}} \Longleftarrow\\ &\overline{I}_3=I_3\angle{-\arccos(0.6)}=I_3\angle{-53.13^{\circ}}\\ &{\overline{I}}={\overline{I}_{1}}+{\overline{I}_{2}}+{\overline{I}_{3}}\\ &\Im\{{\overline{I}_{1}}+{\overline{I}_{2}}\}=-35.98j\\ &\therefore I_3\sin(-53.13^{\circ})=-45^{\circ}-(-35.98^{\circ})=-4.02^{\circ}\\ &\text{So}\quad \overline{I}_{3}=5.03\angle{-53.13^{\circ}}\Longleftarrow\\ & \therefore I=54.94\angle{-45^{\circ}}\Longleftarrow\\ \end{matrix} \end{array}

$$\begin{array} &S_{L_3}={V}\times{I}_3=600\times5.03=3018\; \text{VA}\\ P_{L_3}=3018\times 0.6=1810.8 \;\text{W}\\ Q_{L_3}=\sqrt{3018^2-1810.8^2}=2414.4 \;\text{VAr}\\ P_{loss}=I^2R=54.94^2\times 0.3=905.5 \;\text{W}\\ V_i=\overline{V}+\overline{I}*\overline{Z}_{\text{line}}=600+(54.94\angle{-45^{\circ}})\cdot(0.3+j0.8)=643.0\angle1.73^{\circ} \end{array}$$

You start by finding the impedance for each of the loads.

$$I_{1}=\frac{20\times 10^{3}}{600}\angle -36.8 = 33.33A \angle -36.8$$ $$Z_{1}=\frac{V}{I_{1}}= \frac{600}{33.33\angle-36.8}=18\angle36.8$$

$$Z_{2}=R_{2}\left | \right |X_{2}$$

$$Z_{2}=((15)^{-1}+(j30)^{-1})^{-1}=13.42\angle 26.565$$

$$\cos(\tan^{-1}(\frac{X_{3}}{R_{3}}))=0.6$$ $$X_{3}=3.33R_{3}$$