Complex Power

Problem-Solving Strategy for solving the Complex Power

STEP 1. Calculate the amplitude of $\mathbf S_L$ using $S_L=\frac{P}{\cos\theta}=\frac{P}{pf}$ and then determine

$\mathbf S_L=S\angle \theta$

STEP 2. Use following equation to calculate $\mathbf I_L$

$\mathbf I_L=[\frac{\mathbf S_L}{\mathbf V_L}]^*$

STEP 3. Then the complex power losses in the line are

$\mathbf S_{loss}=I_L^2\mathbf Z_{loss}$

STEP 4. Therefore, the complex power at the source is

$\mathbf S_{S}=\mathbf S_L+\mathbf S_{loss}$

Note: This post is based on Irwin's Basic Engineering Circuit Analysis 11th ed.

If there are any errors in solution, please let me know in the comments

Complex power of each load is

$\mathbf S_1=\frac{20000}{0.7}\angle -\cos^{-1}(0.7)=28571.43\angle -45.573^{\circ}VA$

$\mathbf S_2=12000\angle \cos^{-1}(0.9)=12000\angle 25.842^{\circ}VA$

So the total complex power of loads is

$\mathbf S_L=\mathbf S_1+\mathbf S_2=30800-15173.4j=34334.7\angle -26.227^{\circ}VA$

Since $\mathbf S_L=\mathbf V_L \mathbf I_L^*$

$\mathbf I_L=[\frac{34334.7\angle -26.227^{\circ}}{240\angle 0^{\circ}}]^*=128.33+63.223j=143.06\angle 26.227^{\circ}A rms$

Also, the complex power losses in the line is

$\mathbf S_{loss}=(143.06)^2(0.05+0.2j)=1023.3+4093.23j=4219.21\angle 75.964^{\circ}VA$

Therefore, the complex power supplied by the source is

$\mathbf S_S=31823.26-11080.254j=33697.06\angle -19.197^{\circ}VA$

The power factor of the source is

$pf=\cos(-19.197^{\circ})=0.9444$ leading

And the source voltage is

$\mathbf V_S=\frac{33697.06\angle -19.197^{\circ}}{[143.06\angle 26.227^{\circ}]^*}=233.774+28.828j=235.545\angle 7.03^{\circ}Vrms$

$v_s(t)=333.111\cos(377t+7.03^{\circ})V$