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#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Complex voltage dividers        \\ <fs medium>(written test, approx. 16 % of a 60-minute written test, SS2023)</fs>
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The circuit below shall be given with the following values:

  * $R = 1.1   ~\rm k\Omega$
  * $L = 3.5   ~\rm mH$
  * $\underline{U}_\rm I = 5 ~\rm V$
  * $f_0 = 150 ~\rm kHz$

{{drawio>electrical_engineering_1:c9fj1si7l797equsCircuit.svg}}

1. Calculate the impedance $\underline{Z}_L$. 

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\begin{align*}
\underline{Z}_L &= {\rm j} \cdot \omega \cdot L \\
                &= {\rm j} \cdot 2\pi \cdot 150 ~\rm kHz \cdot 3.5 ~\rm mH 
\end{align*}
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\begin{align*}
\underline{Z}_L &= {\rm j}\cdot 3.3 ~\rm k\Omega
\end{align*}
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2. **Draw the two impedance phasors** and the **resulting phasor** for the overall impedance in a diagram.
**Choose an appropriate scaling** factor and write it down.

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{{drawio>electrical_engineering_1:c9fj1si7l797equsPhasors.svg}}

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3. Calculate the output voltage $|\underline{U}_\rm O|$ and the phase shift between $\underline{U}_\rm O$ and $\underline{U}_\rm I$. 

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The formula to start with is the (complex) voltage divider:
\begin{align*}
\underline{U}_{\rm O} &= \underline{U}_{\rm I} \cdot {{R} \over {R+{\rm j}\cdot Z_L }} \\
                      &= \underline{U}_{\rm I} \cdot {{R} \over {R+{\rm j}\cdot Z_L }} \cdot {{R-{\rm j}\cdot Z_L} \over {R-{\rm j}\cdot Z_L }}\\
\end{align*}
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\begin{align*}
\underline{U}_{\rm O} &= 0.5 ~\rm V - j \cdot 1.5 ~V
\end{align*}
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4. Calculate the cut-off frequency of this setup. 

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At cut off frequency the absolute values of impedances $\underline{Z}_L$ is equal to $\underline{Z}_R=R$. This leads to:
\begin{align*}
f_{\rm cutoff} &= {{R} \over {2\pi \cdot L} } \\
                 &= {{1.1 \cdot 10^3 {\rm {V} \over{A} } } \over {2\pi \cdot 3.5  \cdot 10^{-3} {\rm {As} \over{V} } }} 
\end{align*}
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#@ResultBegin_HTML~4~@# 
\begin{align*}
f_{\rm cutoff} &= 50 ~\rm kHz
\end{align*}

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