200 lines
No EOL
9 KiB
Markdown
200 lines
No EOL
9 KiB
Markdown
![[TD_FGE_2025.pdf#page=15]]
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# Exercise 1. Fresnel vector
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![[Pasted image 20251006094239.png]]
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a) polar form (modulus and argument)
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$\overrightarrow{v_1}=(15A,\frac{7\pi}{6})=15e^{j(\phi)}=15e^{j210}$
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b) rectangular form (a+jb)
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$\overrightarrow{v_1}=(\begin{matrix}15A\cos{(\frac{7\pi}{6})} \\ 15A\sin{(\frac{7\pi}{6})}\end{matrix})=(\begin{matrix}-12.99A \\ -7.5A\end{matrix})$
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$\overrightarrow{v_1}=a+jb$ avec $a=15\cos{210}$ et $b=15\sin{210}$
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$\overrightarrow{v_1}=-12.99 - 7.5j$
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# Exercise 2:
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![[Pasted image 20251006100509.png]]
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a)
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$\overrightarrow{V_1}=63+j16=\sqrt{63^2+16^2}e^{j\arctan{(\frac{16}{63})}}=65e^{j0.249}$
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$\overrightarrow{V_2}=-5-j12=\sqrt{(-5)^2+(-12)^2}e^{j\arctan{(\frac{-12}{-5})}}=13e^{j(1.176+\pi)}$
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$\overrightarrow{V_1}\wedge\overrightarrow{V_2}=65\times13\times e^{j(0.249+\pi+1.176)}$
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b)
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$\overrightarrow{V_1}/\overrightarrow{V_2}=\frac{65}{13}e^{j(0.249-\pi-1.176)}$
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# Exercise 3:
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![[Pasted image 20251006101747.png]]
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## Figure 2: ![[Drawing 2025-10-06 10.49.08.excalidraw]]
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a)
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$U=\underline{z}_1i_1\implies i_1=\frac{\underline{U}}{\underline{z_1}}=\frac{\underline{U}}{R}$
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$U=\underline{z}_2i_2\implies i_2=\frac{\underline{U}}{\underline{z_2}}=\frac{\underline{U}}{jL\omega}$
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$||\underline{i_1}||=||\frac{\underline{U}}{R}||=\frac{||\underline{U}||}{||R||}=5A$
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$||\underline{i_2}||=||\frac{\underline{U}}{\underline{Z_2}}||=\frac{||\underline{U}||}{||\underline{Z_2}||}=2A$
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b)
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$\underline{i}=\underline{i_1}+\underline{i_2}$
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$i_2=\frac{\underline{U}}{jL\omega}=\frac{-j\underline{U}}{L\omega}$
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![[Drawing 2025-10-06 10.55.18.excalidraw]]
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$||\underline{i}||=\sqrt{||\underline{i_1}||^2+||\underline{i_2}||^2}=\sqrt{29}\approx5.3851648071A$
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$\phi=\arctan{(\frac{||\underline{i_2}||}{||\underline{i_1}||})}=\arctan{(\frac{-2}{5})}\approx-0.3805063771 rad$
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c)
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$\frac{1}{Z_{eq}}=\frac{1}{Z_R}+\frac{1}{Z_L}\implies Z_{eq}=\frac{\underline{Z_L}\underline{Z_R}}{\underline{Z_L}+\underline{Z_R}}$
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$\frac{1}{Z_{eq}}=\frac{1}{R}+\frac{1}{jL\omega}=\frac{1}{R}-j\frac{1}{L\omega}$
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ou
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$Z_{eq}=\frac{jRL\omega}{R+jL\omega}=\frac{(R-jL\omega)(jRL\omega)}{(R+jL\omega)(R-jL\omega)}=\frac{RL^2\omega^2+jR^2L\omega^2}{R^2+L^2\omega^2}=\frac{RL^2\omega^2}{R^2+L^2\omega^2}+j\frac{R^2L\omega}{R^2+L^2\omega^2}$
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$||Z_{eq}||=\frac{||R||||Z_L||}{||R+Z_L||}=\frac{1000}{\sqrt{20^2+50^2}}\approx\frac{1000}{53.8516480713}=18.5695338177\Omega$
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d)
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$\underline{U}=\underline{Z_{eq}}\underline{I}$
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$||\underline{U}||=||\underline{Z_{eq}}||||\underline{I}||\approx18.5695338177\times5.3851648071=99.9999999993V$
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## Figure 3: ![[Drawing 2025-10-06 11.17.18.excalidraw|1000]]
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a)
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$U=\underline{z}_1i_1\implies i_1=\frac{\underline{U}}{\underline{z_1}}=\frac{\underline{U}}{R}$
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$U=\underline{z_2}i_2\implies i_2=\frac{\underline{U}}{\underline{z_2}}=\frac{\underline{U}}{\frac{1}{jC\omega}}$
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$||\underline{i_1}||=||\frac{\underline{U}}{R}||=\frac{||\underline{U}||}{||R||}=\frac{120}{20}=6A$
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$||\underline{i_2}||=||\frac{\underline{U}}{\underline{Z_2}}||=\frac{||\underline{U}||}{||\underline{Z_2}||}=\frac{120}{30}=4A$
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b)
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$\underline{i}=\underline{i_1}+\underline{i_2}$
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$i_2=\frac{\underline{U}}{\frac{1}{jC\omega}}=\underline{U}jC\omega$
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![[Drawing 2025-10-06 11.25.11.excalidraw]]
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$||\underline{i}||=\sqrt{||\underline{i_1}||^2+||\underline{i_2}||^2}=\sqrt{52}\approx7.2111025509A$
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$\phi=\arctan{(\frac{||\underline{i_2}||}{||\underline{i_1}||})}=\arctan{(\frac{4}{6})}\approx0.5880026035 rad$
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c)
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$\frac{1}{Z_{eq}}=\frac{1}{Z_R}+\frac{1}{Z_C}\implies Z_{eq}=\frac{\underline{Z_C}\underline{Z_R}}{\underline{Z_C}+\underline{Z_R}}$
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$\frac{1}{Z_{eq}}=\frac{1}{R}+\frac{1}{\frac{1}{jC\omega}}=\frac{1}{R}+jC\omega$
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ou
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$||Z_{eq}||=\frac{||R||||Z_C||}{||R+Z_C||}=\frac{20\times30}{\sqrt{20^2+30^2}}\approx\frac{600}{36.0555127546}=16.6410058868\Omega$
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d)
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$\underline{U}=\underline{Z_{eq}}\underline{I}$
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$||\underline{U}||=||\underline{Z_{eq}}||||\underline{I}||\approx16.6410058868\times7.2111025509=119.9999999998V$
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## Figure 4: ![[Drawing 2025-10-06 11.38.13.excalidraw|1000]]
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a)
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$U=\underline{z}_1i_1\implies i_1=\frac{\underline{U}}{\underline{z_1}}=\frac{\underline{U}}{\frac{1}{jC\omega}}$
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$U=\underline{z}_2i_2\implies i_2=\frac{\underline{U}}{\underline{z_2}}=\frac{\underline{U}}{jL\omega}$
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$||\underline{i_1}||=||\frac{\underline{U}}{\underline{Z_1}}||=\frac{||\underline{U}||}{||\underline{Z_1}||}=\frac{100}{20}=5A$
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$||\underline{i_2}||=||\frac{\underline{U}}{\underline{Z_2}}||=\frac{||\underline{U}||}{||\underline{Z_2}||}=\frac{100}{30}=3\frac{1}{3}A$
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b)
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$\underline{i}=\underline{i_1}+\underline{i_2}$
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$i_1=\frac{\underline{U}}{\frac{1}{jC\omega}}=\underline{U}jC\omega$
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$i_2=\frac{\underline{U}}{jL\omega}=\frac{-j\underline{U}}{L\omega}$
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![[Drawing 2025-10-06 11.43.56.excalidraw]]
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$||\underline{i}||=||\underline{i_1}||-||\underline{i_2}||=5-3\frac{1}{3}=1\frac{2}{3}A$
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$\phi=180°$ ou $\pi rad$
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c)
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$\frac{1}{Z_{eq}}=\frac{1}{Z_L}+\frac{1}{Z_C}\implies Z_{eq}=\frac{\underline{Z_C}\underline{Z_L}}{\underline{Z_C}+\underline{Z_L}}$
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$\frac{1}{Z_{eq}}=\frac{1}{jL\omega}+\frac{1}{\frac{1}{jC\omega}}=-j\frac{1}{L\omega}+jC\omega$
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ou
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$||Z_{eq}||=\frac{||Z_L||||Z_C||}{||Z_L+Z_C||}=\frac{20\times30}{||20-30||}\approx\frac{600}{10}=60\Omega$
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d)
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$\underline{U}=\underline{Z_{eq}}\underline{I}$
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$||\underline{U}||=||\underline{Z_{eq}}||||\underline{I}||\approx60\times1\frac{2}{3}=100V$
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## Figure 5:![[Drawing 2025-10-06 12.02.59.excalidraw|1000]]
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a)
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$U=\underline{z}_1i_1\implies i_1=\frac{\underline{U}}{\underline{z_1}}=\frac{\underline{U}}{\frac{1}{jC\omega}}$
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$U=\underline{z}_2i_2\implies i_2=\frac{\underline{U}}{\underline{z_2}}=\frac{\underline{U}}{jL\omega}$
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$U=\underline{z_3}i_3\implies i_3=\frac{\underline{U}}{\underline{z_3}}=\frac{\underline{U}}{R}$
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$||\underline{i_1}||=||\frac{\underline{U}}{\underline{Z_1}}||=\frac{||\underline{U}||}{||\underline{Z_1}||}=\frac{100}{30}=3\frac{1}{3}A$
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$||\underline{i_2}||=||\frac{\underline{U}}{\underline{Z_2}}||=\frac{||\underline{U}||}{||\underline{Z_2}||}=\frac{100}{60}=1\frac{2}{3}A$
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$||\underline{i_3}||=||\frac{\underline{U}}{R}||=\frac{||\underline{U}||}{||R||}=\frac{100}{20}=5A$
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b)
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$\underline{i}=\underline{i_1}+\underline{i_2}+i_3$
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$i_1=\frac{\underline{U}}{\frac{1}{jC\omega}}=\underline{U}jC\omega$
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$i_2=\frac{\underline{U}}{jL\omega}=\frac{-j\underline{U}}{L\omega}$
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![[Drawing 2025-10-06 12.09.17.excalidraw]]
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$||\underline{i}||=\sqrt{(||\underline{i_1}+\underline{i_2}||)^2+||i_3||^2}=\sqrt{(3\frac{1}{3}-1\frac{2}{3})^2+5^2}=\sqrt{27.7777777778}\approx5.2704627669A$
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$\phi=\arctan{(\frac{||\underline{i_{1+2}}||}{||\underline{i_3}||})}=\arctan{(\frac{1\frac{2}{3}}{5})}\approx0.3217505544 rad$
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c)
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$\frac{1}{Z_{eq}}=\frac{1}{Z_L}+\frac{1}{Z_C}+\frac{1}{Z_R}=\frac{Z_RZ_C+Z_CZ_L+Z_RZ_L}{Z_RZ_LZ_C}$
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$Z_{eq}=\frac{Z_RZ_LZ_C}{Z_RZ_C+Z_CZ_L+Z_RZ_L}$
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ou
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$||Z_{eq}||=\frac{||Z_R||||Z_L||||Z_C||}{||Z_RZ_C||+||Z_CZ_L||+||Z_RZ_L||}$
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# Exercise 4: Application
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![[Pasted image 20251006122351.png]]
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a)
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![[Drawing 2025-10-06 12.24.01.excalidraw]]
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$Z_{eq}=30+\frac{80}{j}=30-j80$
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$Z_C=\frac{1}{jC\omega}=\frac{80}{j}$
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b) ![[Drawing 2025-10-06 12.28.41.excalidraw]]
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$Z_{eq}=(r,\theta)=(\sqrt{30^2+80^2}, \arctan{(\frac{-80}{30})})=(85.4400374532, -1.2120256565 rad)$
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$V=85.44e^{-1.21j}$
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# Exercise 5: Application, Study of a real inductor
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The series model of a real inductor is defined by its series resistance $R_S$ and its series inductance $L_S$. It is possible to define a parallel model equivalent to the series model as shown in Figure 6.
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![[Pasted image 20251008135058.png]]
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a)
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Modèle en série $\implies Z_{eq_1}=L_S+R_S=R_S+jL_S\omega$
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Modèle parallèle $\implies \frac{1}{Z_{eq_2}}=\frac{1}{L_P}+\frac{1}{R_P}$ $\implies Z_{eq_2}=\frac{R_PjL_P\omega}{R_P+jL_P\omega}$
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Multiplication par le conjugué
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$\frac{jR_PL_P\omega(R_P-jL_P\omega)}{(R_P+jL_P\omega)(R_P-jL_P\omega)}=\frac{R_PL_P^2\omega^2+jR_P^2L_P\omega}{R_P^2+L_P^2\omega^2}$
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Or, $Z_{eq_1}=Z_{eq_2}\implies\begin{cases}R_S=\frac{R_PL_P^2\omega^2}{R_P^2+L_P^2\omega^2} \\ L_S\omega=\frac{R_P^2L_P\omega}{R_P^2+L_P^2\omega^2}\end{cases}$
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$R_S+jL_S\omega=\frac{R_P^2L_P\omega}{R_P^2+L_P^2\omega^2}+j\frac{R_PL_P^2\omega^2}{R_P^2+L_P^2\omega^2}$
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b)
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$U(t)=U_2\cos{(\omega t)}$
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$\underline{U}=U_2e^{j\omega t}$
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$\underline{I}=\frac{\underline{U}}{\underline{Z_{eq}}}=\frac{U_2e^{j\omega t}}{\underline{Z_{eq}}}$
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c)
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![[Drawing 2025-10-08 14.39.38.excalidraw|1000]]
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$V_{total}^2=V_{L_S}^2+(V_0+V_{R_S})^2$
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$V_{total}^2=V_{L_S}^2+V_0^2+2V_0V_{R_S}+V_{R_S}^2$
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$V_{bobine}^2=V_{L_S}^2+V_{R_S}^2$
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D'après Pythagore:
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$$
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\begin{cases}
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V_{bobine}^2=V_{L_S}^2+V_{R_S}^2 \\
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V_{total}^2=V_{L_S}^2+(V_0+V_{R_S})^2 \\
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V_0=R_0I
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\end{cases}
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\implies
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\begin{cases}
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V_{bobine}^2=L_S^2\omega^2I^2+R_S^2I^2 \\
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V_{total}^2=L_S^2\omega^2I^2+(R_S+R_0)^2I^2 \\
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V_0=R_0I
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\end{cases}
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$$
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![[IMG_7432.jpeg|400]]
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$$
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\begin{cases}
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L_S^2=(V_{bobine}^2-\frac{R_S^2V_0^2}{R_0^2})\times\frac{R_0^2}{V_0^2\omega^2} \\
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R_S=\frac{V_{tot}^2-V_{bobine}^2-V_0^2}{2V_0}\times R_0
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\end{cases}
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$$
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d)
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Application Numérique: $\begin{matrix}R_S=4.93\Omega \\ L_S=32mH\end{matrix}$
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e)
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Au lieu de mettre $R_0$ en série pour mesurer les tensions, on peut le mettre en parallèle pour mesurer les courants. |