Vibra: probs c4

Main cap.4

4.1 Show that the relaxation time for very heavily damped LCR circuit is RC.

R= Nuestra formula para oscilaciones es: ${\frac {d^{2}\psi }{dt^{2}}}+\gamma {\frac {d\psi }{dt}}+\omega _{0}^{2}\psi =0$ ... (1)

con: $\gamma ={\frac {b}{m}}$

$\omega _{0}^{2}={\frac {s}{m}}$

En este caso de oscilaciones en un circuito LCR óLRC tenemos la formula (por Kirchhoff)

$L\left({\frac {d^{2}\psi }{dt^{2}}}\right)+R\left({\frac {d\psi }{dt}}\right)+\left({\frac {1}{C}}\right)\psi =0$

dividiendo todo por L.

${\frac {d^{2}\psi }{dt^{2}}}+{\frac {R}{L}}\left({\frac {d\psi }{dt}}\right)+{\frac {1}{LC}}\psi =0$ ... (2)

comparando (1) y(2) observamos que

$\gamma ={\frac {R}{L}}$

$\omega _{0}^{2}={\frac {1}{LC}}$

por lo tanto, el “relaxation time” esta dado por:

$\tau _{r}={\frac {\gamma }{\omega _{0}^{2}}}={\frac {\frac {R}{L}}{\frac {1}{LC}}}={\frac {RLC}{L}}=RC$

$\Rightarrow \tau _{r}=RC$

--Leticia González Zamora (discusión) 15:00 25 may 2013 (CDT)

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4.2 De las Ecuaciones (4.1) y (3.3)

Solucion para amortiguamiento critico

$\psi (t)=(C_{1}+C_{2}w_{0}t)$

Condiciones iniciales

1) $\psi (0)=V_{1}C$ Entonces $V_{1}C=C_{1}\Longrightarrow C_{1}=V_{1}C$

2) $i(0)={\frac {d\psi }{dt}}(0)=0$

$i(t)={\frac {d}{dt}}\psi (t)$

$i(t)={\frac {d}{dt}}\left[(C_{1}+C_{2}w_{0}t)\exp ^{-w_{0}t}\right]$

$i(t)=C_{2}w_{0}\exp ^{-w_{0}t}+(C_{1}+C_{2}w_{0}t)(-w_{0})\exp ^{-w_{0}t}$

$i(t)=(C_{2}-C_{1})w_{0}\exp ^{-w_{0}t}-C_{2}w_{0}^{2}\exp ^{-w_{0}t}$

Entonces $0=i(0)=C_{2}w_{0}-w_{0}C_{1}$

Asi que $C_{1}=C_{2}\Longrightarrow C_{2}=V_{1}C$

Asi $\psi (t)=V_{1}C(1+w_{0}t)\exp ^{-w_{0}t}$

$i(t)=V_{1}Cw_{0}^{2}t\exp ^{-w_{0}t}$

Con $i(t)=\left|{\frac {d\psi }{dt}}\right|$

${\frac {di}{dt}}=V_{1}Cw_{0}^{2}\left[\exp ^{-w_{0}t}-w_{0}t\exp ^{-w_{0}t}\right]$

Si ${\frac {di}{dt}}=0$ Entonces $\exp ^{-w_{0}t}-w_{0}t\exp ^{-w_{0}t}=0$

$\exp ^{-w_{0}t}(1-w_{0}t)=0$

Asi que $(1-w_{0}t)=0$

${\dot {t}}={\frac {1}{w_{0}}}$

Luego $i_{max}({\dot {t}})=V_{1}Cw_{0}^{2}{\dot {t}}\exp ^{-w_{0}{\dot {t}}}$

$i_{max}({\dot {t}})=V_{1}Cw_{0}^{2}{\frac {1}{w_{0}}}\exp ^{-w_{0}{\frac {1}{w_{0}}}}$

Con $w_{0}^{2}={\frac {1}{LC}}$

$i_{max}({\dot {t}})=V_{1}Cw_{0}\exp ^{-1}$

$i_{max}({\dot {t}})=V_{1}C{\sqrt {\frac {1}{LC}}}\exp ^{-1}$

$i_{max}({\dot {t}})=v_{1}{\sqrt {\frac {c}{L}}}\exp ^{-1}$

Puesto que la resistecia critica de amortiguamiento es

$R=2w_{0}L$

$R=2{\sqrt {\frac {1}{LC}}}L$

$R=2{\sqrt {\frac {L}{C}}}\Longrightarrow {\sqrt {\frac {C}{L}}}={\frac {2}{R}}$

Asi entonces $i_{max}={\frac {2V_{1}}{\exp R}}$

--Mario Moranchel (discusión) 12:58 19 jun 2013 (CDT)

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--mfg-wiki (discusión) 12:00 9 may 2013 (CDT)