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#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Impedances at different Frequencies  \\ <fs medium>(written test, approx. 18 % of a 60-minute written test, WS2022)</fs> #@TaskText_HTML@#    

Calculate the **resistor values** which have to be used in the following circuits.

1. A resistor $R_1$ shall have the same absolute value of the impedance as a capacitor $C_1=40 ~{\rm nF}$ at $f_1=4 ~{\rm MHz}$.

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\begin{align*}
R_1 &= |\underline{X}_{C1}| \\
    &= {{1}\over{2\pi \cdot f \cdot C_1}} \\
    &= {{1}\over{2\pi \cdot 4 ~{\rm MHz} \cdot 40 ~\rm nF}} \\    
\end{align*}
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#@ResultBegin_HTML~1~@#
\begin{align*}
 R_1  &= 1.00 ~\Omega \\
\end{align*}
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2. A $RL$ series circuit with $L_2=4.7 ~\rm µH$, where an AC voltage source of $U_2=1.0 ~\rm V$ with $f_2=450 ~\rm kHz$ generates a current $I_2=60 ~\rm mA$.

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A series circuit means that the current is constant on every component. \\
The equivalent impedance for $R$ and $L$ combined is given by
\begin{align*}
{{\underline{U}}\over{\underline{I}}} &= R_2 + \underline{X}_{L2} \\
                                      &= R_2 + {\rm j} \cdot \omega L 
\end{align*}
Since ${\rm j} \cdot \omega L $ is perpendicular to $R_2$ this can be simplified to:
\begin{align*}
\left| {{\underline{U}}\over{\underline{I}}} \right|^2 &= |R_2|^2 + |\underline{X}_{L2}|^2 \\
\left( {{U}\over{I}} \right)^2 &= {R_2}^2 + {X_{L2}}^2 \\
\end{align*}

This can be rearranged to get $R_2$:
\begin{align*}
R_2 &= \sqrt{ \left( {{U }\over{I   }} \right)^2 - X_{L2}^2 } \\
    &= \sqrt{ \left( {{1~{\rm V}}\over{60~\rm mA}} \right)^2 - (2\pi \cdot 450~{\rm kHz} \cdot 4.7 ~ {\rm µH})^2 } \\
\end{align*}

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#@ResultBegin_HTML~2~@#

\begin{align*}
 R_2  &= 10.0 ~\Omega \\
\end{align*}

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3. A $RC$ parallel circuit with $C_3=4.7 ~\rm nF$ on an AC current source ($I_{3S}=1.3 ~\rm A$,$f_3=200 ~\rm kHz$), which generates a current of $I_{3R}=1.0 ~\rm A$ through $R_3$.

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Parallel circuit means that the voltage is the same on $R_3$ and $C_3$: \\
\begin{align*}
 \underline{U}_3 = R_3 \cdot \underline{I}_{3R} = -{\rm j}\cdot {X}_{3C} \cdot \underline{I}_{3C} 
\end{align*}
So it gets clear, that $\underline{I}_{3R}$ is perpendicular to $\underline{I}_{3C}$ (It has to, since $R_3$ is perpendicular to  $-{\rm j}\cdot {X}_{3C}$, too). \\
Therefore, the resulting current of the parallel circuit is given as:
\begin{align*}
 \underline{I}_{3}     &=  \underline{I}_{3R}    +  \underline{I}_{3C} \\
 |\underline{I}_{3}|^2 &= |\underline{I}_{3R}|^2 + |\underline{I}_{3C}|^2 \\
  {I}_{3C}    &= \sqrt{|{I}_{3}|^2 - |{I}_{3R}|^2}
\end{align*}

Back to the first formula: 
\begin{align*}
R_3 \cdot {I}_{3R} &= {X}_{3C} \cdot {I}_{3C} \\
R_3  &=  {X}_{3C} \cdot {{{I}_{3C}}\over{{I}_{3R}}} \\
     &=  {{1}\over{2\pi \cdot f \cdot C_3}} \cdot {{\sqrt{|{I}_{3}|^2 - |{I}_{3R}|^2}}\over{{I}_{3R}}} \\
\end{align*}

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#@ResultBegin_HTML~3~@#

\begin{align*}
 R_3  &= 70.0 ~\Omega \\
\end{align*}

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