Unterschiede

Hier werden die Unterschiede zwischen zwei Versionen angezeigt.

--- electrical_engineering_2:inductances_in_circuits [2024/06/05 02:13] – [Exercises] mexleadmin
+++ electrical_engineering_2:inductances_in_circuits [2024/06/06 11:08] (aktuell) – [Exercises] mexleadmin
@@ Zeile 232: / Zeile 232: @@
 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.
+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:
   - $\omega_0 = {{1}\over{\sqrt{LC}}}$
-  - $X_C = X_L$
+  - $X_{C0} = - X_{L0}$
-  - $Z = sqrt{R^2 + (X_L - X_C)^2}$, based on the sum of the impedances $\underline{Z_\rm eq} = \underline{X_R} + \underline{X_C} + \underline{X_L}$ and the Pythagorean theorem$
+  - $Z = \sqrt{R^2 + (X_L + X_C)^2}$, based on the sum of the impedances $\underline{Z}_{\rm eq} = \underline{X}_R + \underline{X}_C + \underline{X}_L$ and the Pythagorean theorem$
+\\
-Based on the task, the following is also known
+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 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 ~@#