53 lines
1.3 KiB
Markdown
53 lines
1.3 KiB
Markdown
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![[TD_FGE_2025.pdf#page=10]]
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![[b5364d1697.pdf#page=5]]
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# Exercise 1.
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Loi des mailles $\rightarrow E=U_R+UC$
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$E=R_i+U_C$
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$E=RC\frac{dU_C}{dt}+U_C$
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La solution est une courbe de la forme $U_C(t)=A_e^{-t/RC}+B$
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Conditions aux limites:
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À $t=0$, $U_C(t)=0$ par lecture du circuit.
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$U_C(t)=A_e^{-0/RC}+B=A+B$
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$\implies 0=A+B\implies A=-B$
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À $t\rightarrow ∞$, $U_C(t)=E_0$ par lecture du circuit.
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$U_C(∞)=B$ par la forme de la solution.
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Finalement, $U_C(t)=-E_0e^{-t/RC}+E_0=E_0(1-e^{-t/RC})$
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$V=E-U_C(t)$
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$V(t)=E_0e^{-t/RC}$
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# Exercise 2.
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![[Pasted image 20251003152305.png]]
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$\tau=RC=10E^3\times100E^{-6}=1s$
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![[IMG_7370.jpeg]]
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# Exercise 3: Second order DC circuits
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![[Pasted image 20251003160147.png]]
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$\implies I=CL\frac{d^2i_1}{dt^2}+i_1$
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Puisque $0=LC\frac{d^2U}{dt^2}+U+°\times\frac{dU}{dt}$
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$\implies$ Solution: $U(t)=A\cos{(\omega_0t)}+B\sin{(\omega_0t)}$
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À $t=0$, $U(t)=0$ donc $A=0$.
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Donc $U(t)=B\sin{(\omega_0t)}$
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d'où: $\frac{du}{dt}=B\omega_0\cos{(\omega_0t)}$
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Or, $\frac{dU}{dt}=\frac{i_C}{C}$ et $i=i_C+i_L$
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Donc $\frac{dU}{dt}=\frac{i-i_L}{C}$
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À $t=0$:
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$\frac{dU}{dt}=B\omega_0=\frac{I}{C}$
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Parceque $I_L$ est continu:
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$I_L(0^+)=I_L(0^-)=0$
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$\implies B=\frac{I}{i\omega_0}=\frac{I}{\frac{C}{\sqrt{LC}}}=\frac{I\sqrt{LC}}{C}$
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Finalement, $U(t)=\frac{I\sqrt{LC}}{C}\sin{(\frac{1}{\sqrt{LC}}t)}$
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