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12.3: Analógica eléctrica

Última actualización

30 oct 2022
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- 12.2: Movimiento Oscilatorio Forzado
- 13: Mecánica Lagrangiana

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Supongamos que $E=\hat{E}\sin\omega t$ se aplica una diferencia de potencial alterno a través de un circuito LCR. Nos referimos a la Ecuación 11.6.3, y vemos que la ecuación que gobierna la carga en el condensador es

$L\ddot{Q}+R\dot{Q}+\frac{Q}{C}=\hat{E}\sin\omega t. \label{12.3.1}$

Podemos diferenciar ambas partes con respecto al tiempo, y dividirlas por $L$ , y de ahí ver que la corriente viene dada por

$\ddot{I}+\frac{R}{L}\dot{I}+\frac{1}{LC}I=\frac{\hat{E}\omega}{L}\cos\omega t. \label{12.3.2}$

Podemos comparar esto directamente con la Ecuación 12.2.2, para que tengamos

$\gamma = \frac{R}{L},\quad \omega_{0}^{2}=\frac{1}{\sqrt{LC}},\quad \hat{f}=\frac{\hat{E}\omega}{L}. \label{12.3.3}$

Entonces, en comparación con la Ecuación 12.2.5, vemos que voy a quedar atrás $E$ por $\alpha$ , donde

$\tan\alpha =\frac{\frac{R\omega}{L}}{\frac{1}{LC}-\omega^{2}}=\frac{R}{\frac{1}{C\omega}-L\omega}. \label{12.3.4}$

Esto es justo lo que obtenemos del enfoque numérico complejo más familiar para los circuitos de corriente alterna.

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