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--- electrical_engineering_2:inductances_in_circuits [2023/09/19 23:51] – mexleadmin
+++ electrical_engineering_2:inductances_in_circuits [2024/06/06 11:08] (aktuell) – [Exercises] mexleadmin
@@ Zeile 7: / Zeile 7: @@
 ==== 6.1.1 Series Circuits ====
-Based on $L = {{ \Psi(t)}\over{i}}$ and Kirchhoff's mesh law ($i=const$) the series circuit of inductions can be interpreted as a single current $i$ which generates multiple linked fluxes $\Psi$. Since the current must stay constant in the series circuit, the following applies for the equivalent inductor of a series connection of single ones:
+Based on $L = {{ \Psi(t)}\over{i}}$ and Kirchhoff's mesh law ($i=\rm const$) the series circuit of inductions can be interpreted as a single current $i$ which generates multiple linked fluxes $\Psi$. Since the current must stay constant in the series circuit, the following applies for the equivalent inductor of a series connection of single ones:
-\begin{align*} L_{eq} &= {{\sum_i \Psi_i}\over{I}} = \sum_i L_i \end{align*}
+\begin{align*} L_{\rm eq} &= {{\sum_i \Psi_i}\over{I}} = \sum_i L_i \end{align*}
 A similar result can be derived from the induced voltage $u_{ind}= L {{{\rm d}i}\over{{\rm d}t}}$, when taking the situation of a series circuit (i.e. $i_1 = i_2 = i_1 = ... = i_{\rm eq}$ and $u_{\rm eq}= u_1 + u_2 + ...$):
@@ Zeile 22: / Zeile 22: @@
 ==== 6.1.2 Parallel Circuits ====
-For parallel circuits one can also start with the principles based on Kirchhoff's mesh law:
+For parallel circuits, one can also start with the principles based on Kirchhoff's mesh law:
 \begin{align*} u_{\rm eq}= u_1 = u_2 = ... \\ \end{align*}
@@ Zeile 51: / Zeile 51: @@
 ==== 6.1.3 in AC Circuits ====
-For AC circuits (i.e. with sinusoidal signals) the impedance $Z$ based on the real part $R$ and imaginary part $X$ has to be considered. In order to do so, one has to solve:
+For AC circuits (i.e. with sinusoidal signals) the impedance $Z$ based on the real part $R$ and imaginary part $X$ has to be considered. To do so, one has to solve:
 \begin{align*} \underline{Z} = {{\underline{u}}\over{\underline{i}}} = {{1}\over{\underline{i}}} \cdot \underline{u}  \end{align*}
@@ Zeile 82: / Zeile 82: @@
 ===== 6.3 Resonance Phenomena =====
-Similar to the approach last semester we now focus on circuits with inductors $L$. For preparation, please recap the chapter [[:electrical_engineering_1:circuits_under_different_frequencies|Circuits under different Frequencies]] from last semester.
+Similar to last semester's approach, we now focus on circuits with inductors $L$. For preparation, please recap the chapter [[:electrical_engineering_1:circuits_under_different_frequencies|Circuits under different Frequencies]] from last semester.
 ==== 6.3.1 RLC - Series Resonant Circuit ====
@@ Zeile 216: / Zeile 216: @@
 <panel type="info" title="Exercise 6.3.1 Series Resonant Circuit I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-A $R$-$L$-$C$ series circuit uses a capacity of $C=1 ~\rm µF$. The circuit is fed by a voltage source with $U_I$ at $f_1 = 50~\rm Hz$.
+A $R$-$L$-$C$ series circuit uses a capacity of $C=100 ~\rm µF$. A voltage source with $U_I$ feeds the circuit at $f_1 = 50~\rm Hz$.
   - Which values does $R$ and $L$ need to have, when the resonance voltage $|\underline{U}_L|$ and $|\underline{U}_C|$ at $f_1$ shall show the double value of the input voltage $U_I$?
@@ Zeile 225: / Zeile 225: @@
 <panel type="info" title="Exercise 6.3.2 Series Resonant Circuit II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-A given $R$-$L$-$C$ series circuit is fed with a frequency, which is $20~\%$ larger than the resonance frequency keeping the amplitude of the input voltage constant. In this situation, the circuit shows a current that is $30~\%$ lower than the maximum current value.
+A given $R$-$L$-$C$ series circuit is fed with a frequency, $20~\%$ larger than the resonance frequency keeping the amplitude of the input voltage constant. In this situation, the circuit shows a current $30~\%$ lower than the maximum current value.
+Calculate the Quality $Q = {{1}\over{R}}\sqrt{{{L}\over{C}}}$.
+#@HiddenBegin_HTML~63211,Solution~@#
+The solution looks hard at first since no insights for the values of $R$, $C$, and $L$ are given.
+However, it is possible and there are multiple ways to solve it. \\ \\
+<fs large>__What we know__</fs>
+But first, add some more info, which is always true from resonant circuits at the resonant frequency:
+  - $\omega_0 = {{1}\over{\sqrt{LC}}}$
+  - $X_{C0} = - X_{L0}$
+  - $Z = \sqrt{R^2 + (X_L + X_C)^2}$, based on the sum of the impedances $\underline{Z}_{\rm eq} = \underline{X}_R + \underline{X}_C + \underline{X}_L$ and the Pythagorean theorem$
+\\
+From the task, the following is also known.
+  - Using "a frequency, $20~\%$ larger than the resonance frequency":
+    - $f = 1.2 \cdot f_0 $ and
+    - $\omega = 1.2 \cdot \omega_0 $
+  - The circuit shows a current $30~\%$ lower than the maximum current value:
+    - The maximum current for the series resonant circuit is given for the minimum impedance $Z$. \\ The minimum impedance $Z$ is given at resonance frequency, and is $Z_{\rm min} = R$
+    - Therefore: $Z = {{1}\over{0.7}} \cdot R$
+\\
+<fs large>__Solution 2: The fast path__</fs>
+We start with $Z = \sqrt{R^2 + (X_L + X_C)^2}$ for the cases: (1) at the resonant frequency $f_0$ and (2) at the given frequency $f = 1.2 \cdot f_0 $
+\begin{align*}
+(1): && Z_0 &= R \\
+(2): && Z   &= \sqrt{R^2 + (X_L + X_C)^2} \\
+\end{align*}
+In formula $(2)$ the impedance $X_L$ and $X_C$ are:
+  * $X_L= \omega \cdot L$ and therefore also $X_L = 1.2 \cdot \omega_0 \cdot L = 1.2 \cdot X_{L0}$
+  * $X_C= - {{1}\over {\omega \cdot C}}$ and therefore also $X_C = - {{1}\over {1.2 \cdot \omega \cdot C}} = - {{1}\over {1.2}} \cdot X_{C0}$
+With $X_{C0} = X_{L0}$ we get for $(1)$:
+\begin{align*}
+Z &= \sqrt{R^2 + \left(1.2\cdot X_{L0} - {{1}\over{1.2}} X_{L0} \right)^2} \\
+  &= \sqrt{R^2 + X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2} \\
+\end{align*}
+Since we know that $Z = {{1}\over{0.7}} \cdot R$ and $Z_0 = R$, we can start by dividing $(2)$ by $(1)$:
+\begin{align*}
+{{(2)}\over{(1)}} : &&  {{Z}\over{Z_0}}                 &= {{\sqrt{R^2 +                     (X_L + X_C)^2}                    }\over{R}}   & &| \text{put in the info from before}\\
+                    &&  {{1}\over{0.7}}                 &= {{\sqrt{R^2 + X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2} }\over{R}}   & &| (...)^2 \\
+                    &&  {{1}\over{0.7^2}}               &= {{     {R^2 + X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2} }\over{R^2}} & &| \cdot R^2 \\
+                    &&  {{1}\over{0.7^2}}     \cdot R^2 &=        {R^2 + X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2}              & &| -R^2 \\
+         && \left({{1}\over{0.7^2}} -1\right) \cdot R^2 &=        {      X_{L0}^2 \cdot \left(1.2 - {{1}\over{1.2}} \right)^2}              & &| : R^2 \quad | : \left(1.2 - {{1}\over{1.2}} \right)^2 \\
+                && {{X_{L0}^2}\over{R^2}}           &={ { {{1}\over{0.7^2}} -1 } \over { \left(1.2 - {{1}\over{1.2}} \right)^2 } }          & &| \sqrt{...} \\
+                    && {{X_{L0}}\over{R}}           &={ \sqrt{ {{1}\over{0.7^2}} -1 } \over  { 1.2 - {{1}\over{1.2}} } }                    & &| \text{with } X_{L0} = \omega_0 \cdot L = {{1}\over{\sqrt{LC}}} \cdot L = \sqrt{ {L} \over {C} }\\
+                    && {{1}\over{R}}\cdot \sqrt{ {L} \over {C} }           &={ \sqrt{ {{1}\over{0.7^2}} -1 } \over  { 1.2 - {{1}\over{1.2}} } }
+\end{align*}
+#@HiddenEnd_HTML~63211,Solution ~@#
+#@HiddenBegin_HTML~63212,Result~@#
+\begin{align*}
+Q             &= {{ \sqrt{{{1}\over{0.7^2}} - 1} }\over{ 1.2 - {{1}\over{1.2}} }} \\
+              &= 2.782... \\
+\rightarrow Q &= 2.78
+\end{align*}
+#@HiddenEnd_HTML~63212,Result~@#
-  - Calculate the Quality $Q = {{1}\over{R}}\sqrt{{{L}\over{C}}}$.
 </WRAP></WRAP></panel>