Power Quality in Power Systems and Electrical Machines, 2nd Edition

Solution to Application Example 5.6

i) The virtual resistor is only activated at harmonic frequencies; therefore, the 5th harmonic component of the injected current in the capacitor is

$\begin{array}{l}{\tilde{I}}_{C,5}={\tilde{I}}_{C,5}^{\mathrm{due}NL}+{\tilde{I}}_{C,5}^{\mathrm{due}sys},\\ {\tilde{I}}_{C,5}^{\mathrm{due}NL}=\frac{Z_{sys,5}}{Z_{sys,5}+Z_{C,5}+R_{vir}}{\tilde{I}}_{NL,5},\\ =\frac{0.06+j0.2074}{0.06+j0.2074-j0.408+0.5}{\tilde{I}}_{NL,5},\\ =\frac{0.06+j0.2074}{0.56-j0.2}{\tilde{I}}_{NL,5}=\frac{0.2159\angle 73.87°}{0.595\angle -19.65°}{\tilde{I}}_{NL,5},\\ =\left(0.363\angle 93.52°\right){\tilde{I}}_{NL,5},\\ =\left(0.363\angle 93.52°\right)\left(\sqrt{2}\left(0.3\right)\angle 0°\right)\mathrm{A}\\ =\sqrt{2}\left(0.109\right)\angle 93.52°\mathrm{A}.\\ {\tilde{I}}_{C,5}^{\mathrm{due}sys}=\frac{1}{Z_{sys,5}+Z_{C,5}+R_{vir}}{\tilde{V}}_{sys,5},\\ =\frac{1}{0.595\angle -19.65°}{\tilde{V}}_{sys,5}=\left(1.68\angle 19.65°\right){\tilde{V}}_{sys,5},\\ =\left(1.68\angle 19.65°\right)\left(\sqrt{2}\left(3\right)\angle 0°\right)\mathrm{A},\\ =\sqrt{2}\left(5.4\right)\angle 19.65°\mathrm{A}.\\ \Rightarrow \Rightarrow {\tilde{\mathrm{I}}}_{C,5}={\tilde{I}}_{C,5}^{\mathrm{due}NL}+{\tilde{I}}_{C,5}^{\mathrm{due}sys},\\ =\left(\sqrt{2}\left(0.109\right)\angle 93.52°\right)+\left(\sqrt{2}\left(5.4\right)\angle 19.65°\right)\mathrm{A},\\ =\sqrt{2}\left(5.09+j1.93\right)=\sqrt{2}\left(5.44\right)\angle 20.77°\mathrm{A}\text{.}\end{array}$

  (E5.6-1)

Note that the magnitude ...