Question 5.8: Using Power Triangles Consider the situation shown in Figure...

By the units given in the figure, we see that load A has an apparent power of 10 kVA. On the other hand, the power for load B is specified as 5 kW.
Furthermore, load A has a power factor of 0.5 leading, which means that the current leads the voltage in load A. Another way to say this is that load A is capacitive. Similarly, load B has a power factor of 0.7 lagging (or inductive).
Our approach is to find the power and reactive power for each load. Then, we add these values to find the power and reactive power for the source. Finally, we compute the power factor for the source and then find the current.
Because load A has a leading (capacitive) power factor, we know that the reactive power Q_A and power angle θ_A are negative. The power triangle for load A is shown in Figure 5.28(a). The power factor is

\cos{\left(\theta_A\right) }=0.5

The power is

P_A=V_\mathrm{rms}I_{A\mathrm{rms}}\cos{(\theta_A)}=10^4(0.5)=5\mathrm{~kW}

Solving Equation 5.64 for reactive power, we have

P^2+Q^2=\left(V_{\mathrm{rms}}I_{\mathrm{rms}}\right)^2     (5.64)

Q_A=\sqrt{\left(V_{\mathrm{rms}}I_{A\mathrm{rms}}\right)^2-P_A^2 } =\sqrt{\left(10^4\right)^2-\left(5000\right)^2  }=-8.660\mathrm{~kVAR}

Notice that we have selected the negative value for Q_A, because we know that reactive power is negative for a capacitive (leading) load.
The power triangle for load B is shown in Figure 5.28(b). Since load B has a lagging (inductive) power factor, we know that the reactive power Q_B and power angle θ_B are positive. Thus,

\theta_B=\mathrm{arccos}\left(0.7\right)=45.57^\circ

Applying trigonometry, we can write

Q_B=P_B\mathrm{tan}(\theta_B) = 5000\mathrm{tan}(45.57^\circ)

Q_B= 5.101\mathrm{~kVAR}

At this point, as shown here we can find the power and reactive power delivered by the source:

P=P_A+P_B=5+5=10\mathrm{~kW}

Q=Q_A+Q_B=-8.660+5.101=-3.559\mathrm{~kVAR}

Because Q is negative, we know that the power angle is negative. Thus, we have

\theta=\mathrm{arctan}\left(\frac{Q}{P} \right)= \mathrm{arctan}\left(\frac{-3.559}{10} \right) =-19.59^\circ

The power factor is

\cos{(\theta)}=0.9421

Power-system engineers frequently express power factors as percentages and would state this power factor as 94.21 percent leading.
The complex power delivered by the source is

\mathrm{S}=P+jQ=10-j3.559=10.61\angle -19.59^\circ \mathrm{~kVA}

Thus, we have

\mathrm{S}=\frac{1}{2}\mathrm{V}_s\mathrm{I}^*=\frac{1}{2}\left(1414\angle 30^\circ \right)\mathrm{I}^*=10.61\times 10^3\angle-19.59^\circ \mathrm{~kVA}

Solving for the phasor current, we obtain:

\mathrm{I}=15.0\angle 49.59^\circ \mathrm{~A}

The phasor diagram for the current and voltage is shown in Figure 5.29. Notice that the current is leading the voltage.