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electrical_engineering_2:inductances_in_circuits [2023/03/24 14:52]
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mexleadmin [Exercises]
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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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 ==== 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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 ===== 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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 <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 $Q = {{1}\over{R}}\sqrt{{{L}\over{C}}}$. 
  
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