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electrical_engineering_2:inductances_in_circuits [2023/03/17 12:52] – mexleadminelectrical_engineering_2:inductances_in_circuits [2024/06/06 11:08] (aktuell) – [Exercises] mexleadmin
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-====== 6. Inductances in Circuits ======+====== 6 Inductances in Circuits ======
  
 ===== 6.1 Basic Circuits ===== ===== 6.1 Basic Circuits =====
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 ==== 6.1.1 Series Circuits ==== ==== 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 $u_{\rm eq}= u_1 + u_2 + ...$): 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 $u_{\rm eq}= u_1 + u_2 + ...$):
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 ==== 6.1.2 Parallel Circuits ==== ==== 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*} \begin{align*} u_{\rm eq}= u_1 = u_2 = ... \\ \end{align*}
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 \begin{align*}  \begin{align*} 
      u_{\rm ind}          &= L {{{\rm d}i}\over{{\rm d}t}} \quad \quad \quad \quad \bigg| \int(){\rm d}t \\       u_{\rm ind}          &= L {{{\rm d}i}\over{{\rm d}t}} \quad \quad \quad \quad \bigg| \int(){\rm d}t \\ 
-\int u_{\rm ind} {\rm d}t &= L \cdot I \\ +\int u_{\rm ind} {\rm d}t &= L \cdot i \\ 
                         i &= {{1}\over{L}} \cdot \int u_{\rm ind} {\rm d}t \\                          i &= {{1}\over{L}} \cdot \int u_{\rm ind} {\rm d}t \\ 
 \end{align*} \end{align*}
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 ==== 6.1.3 in AC Circuits ==== ==== 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*} \begin{align*} \underline{Z} = {{\underline{u}}\over{\underline{i}}} = {{1}\over{\underline{i}}} \cdot \underline{u}  \end{align*}
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 \end{align*} \end{align*}
  
-Without limiting the generality, one can assume the current $i$ to be: $i = I \cdot \sqrt{2} \cdot e^{j \cdot \omega t + \varphi_0}$.\\+Without limiting the generality, one can assume the current $i$ to be: $i = I \cdot \sqrt{2} \cdot {\rm e}^{{\rm j} \cdot \omega t + \varphi_0}$.\\
 Once inserted, the formula gets: Once inserted, the formula gets:
  
 \begin{align*}  \begin{align*} 
-\underline{Z} &= {{1} \over {I \cdot \sqrt{2} \cdot e^{j \cdot \omega t + \varphi_0}}} \cdot L {{ {\rm d}} \over {{\rm d}t} } \left( I \cdot \sqrt{2} \cdot e^{j \cdot \omega t + \varphi_0} \right)  \\  +\underline{Z} &= {{1} \over {I \cdot \sqrt{2} \cdot {\rm e}^{{\rm j} \cdot \omega t + \varphi_0}}} \cdot L {{ {\rm d}} \over {{\rm d}t} } \left( I \cdot \sqrt{2} \cdot {\rm e}^{{\rm j} \cdot \omega t + \varphi_0} \right)  \\  
-              &= {{1} \over {I \cdot \sqrt{2} \cdot e^{j \cdot \omega t + \varphi_0}}} \cdot L                                       I \cdot \sqrt{2} \cdot {{ {\rm d}} \over {{\rm d}t} } \left( e^{j \cdot \omega t + \varphi_0} \right)  \\  +              &= {{1} \over {I \cdot \sqrt{2} \cdot {\rm e}^{{\rm j} \cdot \omega t + \varphi_0}}} \cdot L                                       I \cdot \sqrt{2} \cdot {{ {\rm d}} \over {{\rm d}t} } \left( {\rm e}^{{\rm j} \cdot \omega t + \varphi_0} \right)  \\  
-              &= {{1} \over {\qquad\quad\; e^{j \cdot \omega t + \varphi_0}}} \cdot L                                       \qquad\; \cdot {{ {\rm d}} \over {{\rm d}t} } \left( e^{j \cdot \omega t + \varphi_0} \right)  \\  +              &= {{1} \over {\qquad\quad\; {\rm e}^{{\rm j} \cdot \omega t + \varphi_0}}} \cdot L                                       \qquad\; \cdot {{ {\rm d}} \over {{\rm d}t} } \left( {\rm e}^{{\rm j} \cdot \omega t + \varphi_0} \right)  \\  
-              &= {{1} \over {\qquad\quad\; e^{j \cdot \omega t + \varphi_0}}} \cdot L                                       \qquad\; \cdot j\omega \cdot e^{j \cdot \omega t + \varphi_0}   \\ \\ +              &= {{1} \over {\qquad\quad\; {\rm e}^{{\rm j} \cdot \omega t + \varphi_0}}} \cdot L                                       \qquad\; \cdot j\omega \cdot {\rm e}^{{\rm j} \cdot \omega t + \varphi_0}   \\ \\ 
-\underline{Z} &= L \cdot j\omega  \\ +\underline{Z} &= L \cdot {\rm j}\omega  \\ 
 \end{align*} \end{align*}
  
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 ===== 6.3 Resonance Phenomena ===== ===== 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 ==== ==== 6.3.1 RLC - Series Resonant Circuit ====
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 \begin{align*}  \begin{align*} 
-\underline{U}_I        &=        R \cdot \underline{I} + j \omega L \cdot \underline{I} +         \frac {1}{j\omega C }        \cdot \underline{I} \\  +\underline{U}_I        &=        R \cdot \underline{I} + {\rm j} \omega L \cdot \underline{I} +               \frac {1}{j\omega C }         \cdot \underline{I} \\  
-\underline{U}_I        &= \left( R                     + j \omega L                     - j \cdot \frac {1}{ \omega C } \right) \cdot \underline{I} \\  +\underline{U}_I        &= \left( R                     + {\rm j} \omega L                     - {\rm j} \cdot \frac {1}{ \omega C } \right) \cdot \underline{I} \\  
-\underline{Z}_{\rm eq} &=        R                     + j \omega L                     - j \cdot \frac {1}{ \omega C } +\underline{Z}_{\rm eq} &=        R                     + {\rm j} \omega L                     - {\rm j} \cdot \frac {1}{ \omega C } 
 \end{align*} \end{align*}
  
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 \begin{align*}  \begin{align*} 
 \varphi_u = \varphi_Z  \varphi_u = \varphi_Z 
-         &= arctan \frac{\omega L - \frac{1}{\omega C}}{R} +         &= \arctan \frac{\omega L - \frac{1}{\omega C}}{R} 
 \end{align*} \end{align*}
  
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 |voltage $U_C$ \\ at the capacitor |  |  $\boldsymbol{\LARGE{U}}$ \\ because $\frac{1}{\omega C}$ becomes very large  |  |  $\boldsymbol{\frac{1}{\omega_0 C} \cdot I = \frac{1}{\omega_0 C} \cdot \frac{U}{R} = \color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}\cdot U}}$  |  |  $\boldsymbol{\small{0}}$ \\ because $\frac{1}{\omega C}$ becomes very small  | |voltage $U_C$ \\ at the capacitor |  |  $\boldsymbol{\LARGE{U}}$ \\ because $\frac{1}{\omega C}$ becomes very large  |  |  $\boldsymbol{\frac{1}{\omega_0 C} \cdot I = \frac{1}{\omega_0 C} \cdot \frac{U}{R} = \color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}\cdot U}}$  |  |  $\boldsymbol{\small{0}}$ \\ because $\frac{1}{\omega C}$ becomes very small  |
  
-The calculation in the table shows that in the resonance case, the voltage across the capacitor or inductor deviates from the input voltage by a factor $\color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}}$. This quantity is called **quality or Q-factor**  $Q_S$: +The calculation in the table shows that in the resonance case, the voltage across the capacitor or inductor deviates from the input voltage by a factor $\color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}}$. This quantity is called **quality or Q-factor**  $Q_{\rm S}$: 
  
 \begin{align*}  \begin{align*} 
-\boxed{ Q_S = \left.\frac{U_C}{U} \right\vert_{\omega = \omega_0}  +\boxed{ Q_{\rm S} = \left.\frac{U_C}{U} \right\vert_{\omega = \omega_0}  
-            = \left.\frac{U_L}{U} \right\vert_{\omega = \omega_0}  +                  = \left.\frac{U_L}{U} \right\vert_{\omega = \omega_0}  
-            = \color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}} } +                  = \color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}} } 
 \end{align*} \end{align*}
  
-The quality can be greater than, less than, or equal to 1. The quality $Q$ does not have a unit and should not be confused with the charge $Q$.+The quality can be greater than, less than, or equal to 1. The quality $Q_{\rm S}$ does not have a unit and should not be confused with the charge $Q$.
  
   * If the quality is very high, the overshoot of the voltages at the impedances becomes very large in the resonance case. This is useful and necessary in various applications, e.g. in an $RLC$ element as an antenna.   * If the quality is very high, the overshoot of the voltages at the impedances becomes very large in the resonance case. This is useful and necessary in various applications, e.g. in an $RLC$ element as an antenna.
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 The reciprocal of the $Q$ is called **attenuation**  $d_{\rm S}$. This is specified when using the circuit as a non-overshooting filter. The reciprocal of the $Q$ is called **attenuation**  $d_{\rm S}$. This is specified when using the circuit as a non-overshooting filter.
  
-\begin{align*} \boxed{ d_{\rm S} = \frac{1}{Q_S} = R \sqrt{\frac{C}{L}} } \end{align*}+\begin{align*} \boxed{ d_{\rm S} = \frac{1}{Q_{\rm S}} = R \sqrt{\frac{C}{L}} } \end{align*}
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
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 <panel type="info" title="Exercise 6.3.1 Series Resonant Circuit I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <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 $U_I$?   - 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 $U_I$?
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 <panel type="info" title="Exercise 6.3.2 Series Resonant Circuit II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <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> </WRAP></WRAP></panel>