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[LELEC1370] Exam circmes Juin 2022 #889
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YEAR=2022 | ||
include ../../../exam.mk |
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MONTH=Juin | ||
include ../../2022.mk |
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MINMAJ=Majeure | ||
include ../Juin.mk |
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\documentclass[fr]{../../../../../../eplexam} | ||||||||
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\hypertitle{circmes}{4}{ELEC}{1370}{2022}{Juin}{Majeure} | ||||||||
{Quentin De Laet} | ||||||||
{Christophe Craeye, Bruno Dehez and Claude Oestges} | ||||||||
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\usepackage[oldvoltagedirection]{circuitikz} | ||||||||
\usepackage{amsmath} | ||||||||
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\begin{document} | ||||||||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Le \begin{document} est déjà inclus dans le document de base donc pas besoin de le rajouter
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\section{Question Oestges : Bode et quadripôles} | ||||||||
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On considère le montage suivant : | ||||||||
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\begin{center} | ||||||||
\begin{circuitikz} | ||||||||
\draw (0,0) node[above]{$v_i$} to[short, o-*] ++(1,0) coordinate(SPL1) -- ++(0,-2) to[R=$R_H$] ++(2,0) to[C=$C_H$] ++(2,0) | ||||||||
node[op amp, anchor=-](OAH){} to[short,*-] ++(0,1.25) coordinate(OAHm) to[R=$R_H$] (OAHm -| OAH.out) to[short,-*] (OAH.out) | ||||||||
to[R=$R_i$] ++(2,0) coordinate(EndH) | ||||||||
(OAH.+) node[ground]{} | ||||||||
(SPL1) -- ++(0,1.5) to[R=$R_L$] ++(4,0) node[op amp, anchor=-](OAL){} to[short,*-] ++(0,1.25) coordinate(OALm) | ||||||||
to[R=$R_L$, *-*] (OALm -| OAL.out) | ||||||||
(OALm) -- ++(0,1.25) coordinate(OALm2) to[C=$C_L$] (OALm2 -| OAL.out) -- (OAL.out) to[R=$R_i$, *-] ++(2,0) | ||||||||
(OAL.+) node[ground]{} | ||||||||
(EndH) -- ++(0,2) coordinate(OAf_in) -- (OAf_in |- OAL.out) | ||||||||
(OAf_in) to[short, *-*] ++(1,0) node[op amp, anchor=-](OAf){} -- ++(0,1.25) coordinate(OAfm) to[R=$R_f$] (OAfm -| OAf.out) -- (OAf.out) | ||||||||
to[short,*-o] ++(1,0) node[above]{$v_o$} | ||||||||
(OAf.+) node[ground]{} | ||||||||
; | ||||||||
\end{circuitikz} | ||||||||
\end{center} | ||||||||
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On demande : | ||||||||
\begin{enumerate} | ||||||||
\item le gain $g_f(\omega) = \frac{v_o}{v_i}$ du montage | ||||||||
\item le diagramme de Bode de ce montage (en dB et en phase) avec les valeurs suivantes (unités omises) : | ||||||||
\begin{itemize} | ||||||||
\item $R_f = 2R_i = 2000$ | ||||||||
\item $R_L C_L = 10^{-5}$ | ||||||||
\item $R_H C_H = 3\cdot10^{-6}$ | ||||||||
\end{itemize} | ||||||||
\item l'utilité de ce montage | ||||||||
\item les valeurs $Z_{in}$ et $Z_{out}$ du quadripôle lorsqu'une charge de valeur $Z_L$ est placée à sa sortie, | ||||||||
sachant que la source de tension connectée en entrée possède une résistance interne de valeur $R_S$. | ||||||||
\end{enumerate} | ||||||||
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\nosolution | ||||||||
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\newpage | ||||||||
\section{Question Dehez : circuits magnétiques couplés et puissance} | ||||||||
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Soit le montage suivant, fonctionnant à une fréquence de $50Hz$ : | ||||||||
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\begin{center} | ||||||||
\begin{circuitikz}[american] | ||||||||
\draw (0,0) to[V, l=$24V\angle 0$] ++(0,2) -- node[flowarrow]{$\Bar{I_a}$} ++(1,0) -- ++(2,0) to[R=$1\Omega$] ++(2,0) coordinate(SPLIT) | ||||||||
to[american inductor=$j2\Omega$] node[below](a){$\bullet$} ++(2,0) to[C=$-j2\Omega$] ++(2,0) node[below](b){$\bullet$} | ||||||||
to[american inductor=$j2\Omega$] ++(2,0) to[R=$1\Omega$] ++(0,-2) -- (0,0); | ||||||||
\draw (SPLIT) to[C=$-j1\Omega$,*-*] ++(0,-2); | ||||||||
\draw (a) to[open] ++(0,1.1) [ultra thick] [stealth-stealth, scale=1.02] to[bend left] node[above,pos=0.1]{$j1\Omega$} ++(2,0); | ||||||||
\end{circuitikz} | ||||||||
\end{center} | ||||||||
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On demande : | ||||||||
\begin{enumerate} | ||||||||
\item le facteur de dispersion entre les inductances couplées | ||||||||
\item la valeur du courant $\bar{I_a}$ (amplitude et phase) | ||||||||
\item la puissance active, réactive et apparente fournie par la source de tension | ||||||||
\item la valeur de l'impédance $Z$ à placer (voir circuit ci-dessous) afin d'annuler la puissance réactive fournie par la source de tension | ||||||||
\end{enumerate} | ||||||||
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\begin{center} | ||||||||
\begin{circuitikz}[american] | ||||||||
\draw (0,0) to[V, l=$24V\angle 0$] ++(0,2) -- node[flowarrow]{$\Bar{I_a}$} ++(1,0) to[generic=$Z$] ++(2,0) to[R=$1\Omega$] ++(2,0) | ||||||||
coordinate(SPLIT) to[american inductor=$j2\Omega$] node[below](a){$\bullet$} ++(2,0) to[C=$-j2\Omega$] ++(2,0) node[below](b){$\bullet$} | ||||||||
to[american inductor=$j2\Omega$] ++(2,0) to[R=$1\Omega$] ++(0,-2) -- (0,0); | ||||||||
\draw (SPLIT) to[C=$-j1\Omega$,*-*] ++(0,-2); | ||||||||
\draw (a) to[open] ++(0,1.1) [ultra thick] [stealth-stealth, scale=1.02] to[bend left] node[above,pos=0.1]{$j1\Omega$} ++(2,0); | ||||||||
\end{circuitikz} | ||||||||
\end{center} | ||||||||
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\nosolution | ||||||||
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\newpage | ||||||||
\section{Question Craeye : transitoire} | ||||||||
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Soit le circuit suivant, avec l'interrupteur passant de A à B en $t=0$. \\ | ||||||||
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Suggested change
Il est préférable d'utiliser 2 retours à la ligne plutôt que des |
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Sachant que $\omega = 2\cdot10^6$ $rad/s$, $R = 1$ $k\Omega$, $L = 1$ $mH$ et $C = 1/6$ $nF$, calculer la valeur de $V_o(t)$ pour $t > 0$.\newline | ||||||||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. C'est préférable d'utiliser la commande \SI{Nombre}{Unité}
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Exprimez le terme harmonique de la réponse sous forme paramétrique.\newline | ||||||||
\textbf{Point bonus :} trouvez la valeur des coefficients du terme harmonique. | ||||||||
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\begin{center} | ||||||||
\begin{circuitikz}[american] | ||||||||
\draw (0,0) to[sV, v^<=$\cos{(\omega t)}$] ++(0,2) to[R=$R$] ++(0,2.5) -- ++(2,0) node[above](B){\textbf{B}} | ||||||||
node[spdt, scale=-1, anchor=out 2](Sw){}; | ||||||||
\draw (0,0) -- (0,0 -| Sw.out 1) coordinate(a) to[V=$2V$] ++(0,2) to[R=$R$] (Sw.out 1) node[left]{\textbf{A}}; | ||||||||
\draw (Sw.in) -- ++(1,0) coordinate(b) to[C, a_=$C$] (b |- a) coordinate(c) -- (a); | ||||||||
\draw (b) to[L=$L$] ++(2,0) coordinate(d) to[R, a_=$R$,v^=$V_o$] (d |- c) -- (c); | ||||||||
\end{circuitikz} | ||||||||
\end{center} | ||||||||
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\nosolution | ||||||||
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\end{document} |
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Pour avoir accès aux unités (voir plus bas)