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I need to know how to solve the discharge of a capacitor whose resistances are 10kilo ohms, 330 ohm, with a voltage source of 7volts, the capacitor has 470uF

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Answer to a math question I need to know how to solve the discharge of a capacitor whose resistances are 10kilo ohms, 330 ohm, with a voltage source of 7volts, the capacitor has 470uF

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Ali
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92 Answers
Para calcular la descarga de un condensador a través de resistencias, podemos usar la ley de descarga de un condensador, que se expresa como:

V_c = V_f \cdot e^{-\frac{t}{RC}}

Donde:
- V_c es el voltaje en el condensador en el tiempo t .
- V_f es el voltaje inicial en el condensador.
- R es la resistencia total en el circuito.
- C es la capacitancia del condensador.
- t es el tiempo transcurrido.

Primero, necesitamos calcular la resistencia total en el circuito. Dado que las resistencias están en paralelo, podemos calcular la resistencia total usando la fórmula:

\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}

Donde R_1 = 10k\Omega = 10,000\Omega y R_2 = 330\Omega . Calculando R_{total} :

\frac{1}{R_{total}} = \frac{1}{10,000\Omega} + \frac{1}{330\Omega}
\frac{1}{R_{total}} = 0.0001 + 0.00303
R_{total} = \frac{1}{0.00313}
R_{total} \approx 318.47\Omega

Ahora, con R_{total} y la capacitancia C = 470\mu F = 0.00047 F , y el voltaje inicial V_f = 7V , podemos usar la fórmula de descarga del condensador para resolver el problema. Vamos a calcular el tiempo t cuando el voltaje en el condensador V_c = 0V .

0V = 7V \cdot e^{-\frac{t}{318.47\Omega \cdot 0.00047 F}}
e^{-\frac{t}{150.0151}} = 0
-\frac{t}{150.0151} = \ln(0)
\text{No hay solución real}

El condensador nunca se descargará completamente a 0V en este circuito, ya que requeriría un tiempo infinito. La descarga será exponencial pero nunca alcanzará completamente 0V.

**Respuesta:** El condensador no se descargará completamente a 0V en este circuito.

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