1 Analyzing complex Impedances
(written test, approx. 14 % of a 60-minute written test, WS2022)
A circuit with an ideal voltage source ($U=50 ~\rm V$, $f=330 ~\rm Hz$) and two components ($R$ and $\underline{X}_1$) shall be given.
After analysis, the following formula for the impedance was extracted:
\begin{align*}
\underline{Z} = \left({{2}\over{3+4{\rm j}}}+5{\rm j} \right) \Omega
\end{align*}
1. Calculate the physical values of the two components.
Solution steps
\begin{align*}
\underline{Z} &= \left({{2}\over{3+4{\rm j}}} + 5{\rm j} \right) ~\Omega \\
&= \left({{2}\over{3+4{\rm j}}} \cdot {{3-4{\rm j}}\over{3-4j}} + 5{\rm j} \right) ~\Omega \\
&= \left({{2}\over{9+16 }} \cdot (3-4{\rm j} ) + 5{\rm j} \right) ~\Omega \\
&= \left(0.24 - 0.32{\rm j} + 5{\rm j} \right) ~\Omega \\
&= 0.24 ~\Omega + {\rm j} \cdot 4.68 ~\Omega \\
&= R + {\rm j} X_L \\
\end{align*}
With the complex part comes the physical value:
\begin{align*}
X_L &= \omega L \\
L &= {{X_L}\over{2\pi \cdot f}} \\
&= {{4.68 ~\Omega}\over{2\pi \cdot 300 ~\rm{Hz}}} \\
\end{align*}
Result
\begin{align*}
R &= 0.24 ~\Omega \\
L &= 2.26 ~\rm{mH}
\end{align*}
2. Calculate the phase and absolute value of complex current $\underline{I}$ through the circuit.
Solution steps
\begin{align*}
\underline{I} &= {{\underline{U}}\over{\underline{Z}}} \\
&= {{50 ~\rm{V}}\over{ 0.24 ~\Omega + {\rm j} \cdot 4.68 ~\Omega }} \\
&= {{50 ~\rm{V}}\over{ 0.24 ~\Omega + {\rm j} \cdot 4.68 ~\Omega }} \cdot {{ 0.24 ~\Omega - {\rm j} \cdot 4.68 ~\Omega }\over{ 0.24 ~\Omega - {\rm j} \cdot 4.68 ~\Omega }} \\
&= {{50 ~\rm{V}}\over{(0.24 ~\Omega)^2 + (4.68 ~\Omega)^2 }} \cdot ( 0.24 ~\Omega - {\rm j} \cdot 4.68 ~\Omega ) \\
\end{align*}
The absolute value $|\underline{I}|$ can be calculated as:
\begin{align*}
|\underline{I}| &= {|{\underline{U}|}\over{|\underline{Z}|}} \\
&= {{50 ~\rm{V}}\over{| 0.24 ~\Omega + {\rm j} \cdot 4.68 ~\Omega |}} \\
&= {{50 ~\rm{V}}\over{\sqrt{ (0.24 ~\Omega)^2 + (4.68 ~\Omega)^2 }}}
\end{align*}
The phase $\varphi_i$ can be calculated as
\begin{align*}
\varphi_i &= \arctan \left( {{\Im()}\over{\Re()}} \right) \\
&= \arctan \left( {{-4.68 ~\Omega}\over{0.24 ~\Omega}} \right) \\
\end{align*}
Result
\begin{align*}
|\underline{I}| &= 10.67 ~\rm{A} \\
\varphi_i &= -87.06°
\end{align*}
3. Now an additional component $\underline{X}_2$ shall be added in series to the two components.
This component shall be dimensioned in such a way that the current and voltage are in phase. Calculate these component value!
Solution steps
The current and voltage are in phase once there is only a pure ohmic (= pure real) resulting impedance $\underline{Z} + \underline{X}_2$.
Therefore, the component mus be a capacitor with the same absolute value of impedance: $|\underline{X}_C| = |\underline{X}_L| $
\begin{align*}
X_C &= {{1}\over{\omega \cdot C}} = X_L \\
C &= {{1}\over{\omega \cdot X_L}} \\
&= {{1}\over{2\pi \cdot 300 ~\rm{Hz} \cdot 4.68 ~\Omega}} \\
\end{align*}
Result
\begin{align*}
C = 103 ~\rm{µF}
\end{align*}