{{tag>complex_voltage_divider exam_ee2_SS2021}}{{include_n>1240}} #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Complex series circuit \\ (written test, approx. 8 % of a 120-minute written test, SS2021) #@TaskText_HTML@# A series circuit of $C = 4.95~\rm nF$, $R = 200 ~\rm \Omega$ at $f = 40 ~\rm kHz$ shall be given. a) Determine the complex impedance $\underline{Z}_C$. #@HiddenBegin_HTML~9Xy69AxG3GI3NR26_11,Path~@# The complex impedance $\underline{Z}_C$ is given as \begin{align*} \underline{Z}_C &= {{1}\over{{\rm j} \cdot 2\pi \cdot f \cdot C }} \\ &= {{-{\rm j}}\over{2\pi \cdot 40 \cdot 10^3 ~\rm Hz \cdot 4.95 \cdot 10^{-9} ~\rm F }} \\ &= -{\rm j} \cdot 803.81... ~\rm \Omega \\ \end{align*} #@HiddenEnd_HTML~9Xy69AxG3GI3NR26_11,Path~@# #@HiddenBegin_HTML~9Xy69AxG3GI3NR26_12,Result~@# $\underline{Z}_C = -{\rm j} \cdot 804 ~\rm \Omega $ #@HiddenEnd_HTML~9Xy69AxG3GI3NR26_12,Result~@# b) Determine the absolute value of the resulting impedance of the series circuit using an impedance vector diagram. Pay attention to the correct dimensioning. #@HiddenBegin_HTML~9Xy69AxG3GI3NR26_22,Result~@# {{drawio>ee2:9Xy69AxG3GI3NR26_solution1.svg}} Based on the diagram: $|\underline{Z}|= 828 ~\Omega$ #@HiddenEnd_HTML~9Xy69AxG3GI3NR26_22,Result~@# #@TaskEnd_HTML@#