{{tag>complex_impedance exam_ee1_WS2022}} {{include_n>8000}} #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Complex Impedance Circuit \\ (written test, approx. 15 % of a 60-minute written test, WS2022) #@TaskText_HTML@# A circuit designed to filter the noise from a signal shall be analyzed.\\ The input is given by a voltage source $u(t) = 3.0 ~{\rm V} \cdot \sin⁡(2\pi \cdot 15 ~{\rm kHz} \cdot t)$ with an internal resistance of $10 ~\Omega$.\\ This linear source is connected with an inductor of $330 ~ {\rm µH}$ and a capacitor of $0.22 ~{\rm µF}$, all in series. 1. Draw the circuit diagram of the given circuit.\\ Label all components, voltages, and currents. #@ResultBegin_HTML~1~@# {{drawio>electrical_engineering_1:kricv9fh7haauo6qCircuit.svg}} #@ResultEnd_HTML@# 2. Calculate the single impedance $|\underline{Z}_C|$, $|\underline{Z}_L|$ such as $|\underline{Z}|$ of the overall circuit. #@PathBegin_HTML~2~@# \begin{align*} Z_C &= {{1}\over{2\pi \cdot f \cdot C}}\\ &= {{1}\over{2\pi \cdot 15 ~{\rm kHz} \cdot 0.22 ~{\rm µF}}}\\ \end{align*}\\ \begin{align*} Z_L &= 2\pi \cdot f \cdot L\\ &= 2\pi \cdot 15 ~{\rm kHz} \cdot 0.22 ~{\rm µF}\\ \end{align*}\\ \begin{align*} Z_C &= {{1}\over{2\pi \cdot f \cdot C}}\\ &= {{1}\over{2\pi \cdot 15 ~{\rm kHz} \cdot 330 ~{\rm µH}}}\\ \end{align*}\\ \begin{align*} \underline{Z} &= R + \underline{Z}_L + \underline{Z}_C \\ &= R + j \cdot {Z}_L - j \cdot {Z}_C \\ &= R + j \cdot ({Z}_L - {Z}_C) \\ |\underline{Z}| &= \sqrt{R^2 + (\underline{Z}_L - \underline{Z}_C)^2 }\\ \end{align*} #@PathEnd_HTML@# #@ResultBegin_HTML~2~@# \begin{align*} Z_L &= 31.1 ~\Omega \\ Z_C &= 48.2 ~\Omega \\ Z &= 19.8 ~\Omega \end{align*} #@ResultEnd_HTML@# 3. Draw the three impedance phasors $|\underline{Z}_C|$, $|\underline{Z}_L|$ and $|\underline{Z}_R|$ in a diagram.\\ Choose an appropriate scaling factor and write it down. #@ResultBegin_HTML~3~@# {{drawio>electrical_engineering_1:kricv9fh7haauo6qCircuit3.svg}} #@ResultEnd_HTML~3~@# 4. Calculate the current $|\underline{I}|$. #@PathBegin_HTML~4~@# \begin{align*} Z &= {{\hat{U}}\over{\hat{I}}} \\ \hat{I} &= {{\hat{U}}\over{Z}} \\ \end{align*} With $I = {{1}\over{\sqrt{2}}}\cdot \hat{I}$: \begin{align*} I &= {{1}\over{\sqrt{2}}}\cdot {{\hat{U}}\over{Z}} \\ &= {{1}\over{\sqrt{2}}}\cdot {{3.0 ~{\rm V}}\over{19.28 ~\Omega}} \\ \end{align*} #@PathEnd_HTML@# #@ResultBegin_HTML~4~@# \begin{align*} I = 107 ~{\rm mA} \end{align*} #@ResultEnd_HTML@# #@TaskEnd_HTML@#