### 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

$\begin{array}{l}{\stackrel{~}{I}}_{C,5}={\stackrel{~}{I}}_{C,5}^{\mathrm{due}NL}+{\stackrel{~}{I}}_{C,5}^{\mathrm{due}sys},\\ {\stackrel{~}{I}}_{C,5}^{\mathrm{due}NL}=\frac{{Z}_{sys,5}}{{Z}_{sys,5}+{Z}_{C,5}+{R}_{vir}}{\stackrel{~}{I}}_{NL,5},\\ =\frac{0.06+j0.2074}{0.06+j0.2074-j0.408+0.5}{\stackrel{~}{I}}_{NL,5},\\ =\frac{0.06+j0.2074}{0.56-j0.2}{\stackrel{~}{I}}_{NL,5}=\frac{0.2159\angle 73.87\xb0}{0.595\angle -19.65\xb0}{\stackrel{~}{I}}_{NL,5},\\ =\left(0.363\angle 93.52\xb0\right){\stackrel{~}{I}}_{NL,5},\\ =\left(0.363\angle 93.52\xb0\right)\left(\sqrt{2}\left(0.3\right)\angle 0\xb0\right)\mathrm{A}\\ =\sqrt{2}\left(0.109\right)\angle 93.52\xb0\mathrm{A}.\\ {\stackrel{~}{I}}_{C,5}^{\mathrm{due}sys}=\frac{1}{{Z}_{sys,5}+{Z}_{C,5}+{R}_{vir}}{\stackrel{~}{V}}_{sys,5},\\ =\frac{1}{0.595\angle -19.65\xb0}{\stackrel{~}{V}}_{sys,5}=\left(1.68\angle 19.65\xb0\right){\stackrel{~}{V}}_{sys,5},\\ =\left(1.68\angle 19.65\xb0\right)\left(\sqrt{2}\left(3\right)\angle 0\xb0\right)\mathrm{A},\\ =\sqrt{2}\left(5.4\right)\angle 19.65\xb0\mathrm{A}.\\ \Rightarrow \Rightarrow {\stackrel{~}{\mathrm{I}}}_{C,5}={\stackrel{~}{I}}_{C,5}^{\mathrm{due}NL}+{\stackrel{~}{I}}_{C,5}^{\mathrm{due}sys},\\ =\left(\sqrt{2}\left(0.109\right)\angle 93.52\xb0\right)+\left(\sqrt{2}\left(5.4\right)\angle 19.65\xb0\right)\mathrm{A},\\ =\sqrt{2}\left(5.09+j1.93\right)=\sqrt{2}\left(5.44\right)\angle 20.77\xb0\mathrm{A}\text{.}\end{array}$

(E5.6-1)

Note that the magnitude ...

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