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Chapter 5: Problem 9

Ein Kondensator der Kapazit?t \(C\) wird zun?chst auf die Spannung \(u_{0}\) aufgeladen und dann über einen ohmschen Widerstand \(R\) entladen. Die Differentialgleichung für diesen zur Zeit \(t=0\) einsetzenden Ausschaltvorgang lautet: $$ R C \frac{d u_{C}}{d t}+u_{C}=0 $$ Berechnen Sie den zeitlichen Verlauf der Kondensatorspannung \(u_{C}=u_{C}(t)\) durch Trennung der Variablen.

Short Answer

Expert verified
The voltage across the capacitor over time is \( u_{C}(t) = u_{0} e^{-\frac{t}{RC}} \).

Step by step solution

01

- Understand the Differential Equation

The given differential equation is \[ R C \frac{d u_{C}}{d t}+u_{C}=0 \]This equation describes the discharge of the capacitor over time.
02

- Rearrange the Equation

Rearrange the equation to isolate the differential part. Divide both sides by \(RC\):\[ \frac{d u_{C}}{d t} + \frac{u_{C}}{RC} = 0 \]
03

- Separate the Variables

Separate the variables to facilitate integration. Rearrange to get:\[ \frac{d u_{C}}{u_{C}} = -\frac{1}{RC} dt \]
04

- Integrate Both Sides

Integrate both sides of the separated equation:\[ \int \frac{1}{u_{C}} \, du_{C} = -\int \frac{1}{RC} \, dt \]This will give:\[ \ln|u_{C}| = -\frac{t}{RC} + C_{1} \]where \(C_{1}\) is the integration constant.
05

- Solve for \(u_{C}\)

Exponentiate both sides to solve for \(u_{C}\):\[ u_{C} = e^{ -\frac{t}{RC} + C_{1}} \]This simplifies to\[ u_{C} = e^{C_{1}} e^{ -\frac{t}{RC}} \]Let \(e^{C_{1}} = u_{0}\), where \(u_{0}\) is the initial voltage at \(t = 0\). So,\[ u_{C} = u_{0} e^{-\frac{t}{RC}} \]
06

- Finalize the Solution

The final expression for the capacitor voltage as a function of time is:\[ u_{C}(t) = u_{0} e^{-\frac{t}{RC}} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
In physics and engineering, we often use differential equations to model the behavior of systems over time. A differential equation relates a function with its derivatives, representing how a quantity changes over time based on current conditions.

In our capacitor discharge problem, the differential equation is: \[ R C \frac{d u_{C}}{d t} + u_{C} = 0 \] This equation describes how the voltage across the capacitor, denoted as \( u_{C} \), decreases over time. Understanding the differential equation allows us to predict how the system will evolve from a given starting condition.
RC Circuit
An RC circuit is composed of a resistor (R) and a capacitor (C) connected either in series or parallel. In the context of capacitor discharge, we are examining a series RC circuit.

When the capacitor discharges, it releases its stored electrical energy through the resistor. The rate at which this process occurs is governed by both the resistor and capacitor values.

The key parameters are:
  • R (Resistance) - governs how easily current flows through the circuit.
  • C (Capacitance) - determines how much electrical energy the capacitor can store.

The product \(RC\) is known as the time constant \(\tau \), and it represents the characteristic time over which the voltage across the capacitor decreases.
Separation of Variables
Separation of variables is a mathematical technique used to solve differential equations. It involves rearranging the equation so that each variable and its differentials are isolated on opposite sides of the equation.

For our RC circuit, we start with: \[ \frac{d u_{C}}{d t} + \frac{u_{C}}{RC} = 0 \] To separate variables, we divide both sides by \( u_{C} \): \[ \frac{d u_{C}}{u_{C}} = -\frac{1}{RC} dt \]
With each side now containing only one variable, we can integrate separately. This process translates the problem into a simpler form that is more straightforward to solve.
Exponential Decay
Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value. In our context, the voltage across the capacitor decreases exponentially over time.

After using separation of variables and integrating, we find the voltage function: \[ u_{C}(t) = u_{0} e^{-\frac{t}{RC}} \]
Here, \( u_{0} \) is the initial voltage at time \( t = 0 \). The term \( e^{-\frac{t}{RC}} \) shows an exponential decay from the initial value, influenced by the time constant \( \tau = RC \).
  • At \( t = 0 \), the voltage is \( u_{C}(0) = u_{0} \).
  • As \( t \) increases, \( u_{C}(t) \) exponentially approaches zero.

Exponential decay models are common and capture many real-world processes where things gradually diminish over time.

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Most popular questions from this chapter

Gegeben ist die nichtlineare Differentialgleichung 1. Ordnung $$ y^{\prime}=\sqrt{x+y} $$ und der Anfangswert \(y(1)=1\). Bestimmen Sie näherungsweise den Ordinatenwert der L?sungskurve an der Stelle \(x_{1}=1,2\) a) nach dem Eulerschen Srreckenzugverfahren, b) nach dem Runge-Kutta-Verfahren 4. Ordning. W?hlen Sie als Schrittweite \(h=0,05\). F?hren Sie ferner eine Zweitrechnung (Grobrechnung) mit doppelter Schrittweite durch und geben Sie eine Abschätzung des Fehlers.

Lösen Sie die folgenden Schwingungsgleichungen (freie ungedämpfte Schwingungen): a) \(\ddot{x}+4 x=0, \quad x(0)=2, \quad \dot{x}(0)=1\) b) \(\ddot{x}+x=0, \quad x(0)=1, \quad \dot{x}(0)=-2\) c) \(\quad \ddot{x}+a^{2} x=0, \quad x(0)=0, \quad \dot{x}(0)=v_{0} \quad(a \neq 0)\)

Die Aufladung eines Kondensators der Kapazit?t \(C\) über einen ohmschen Widerstand \(R\) auf die Endspannung \(u_{0}\) erfolgt nach dem Exponentialgesetz $$ u_{C}(t)=u_{0}\left(1-\mathrm{e}^{-\frac{t}{R C}}\right) \quad(t \geqslant 0) $$ Zeigen Sie, daB diese Funktion eine (partikul?re) Lôsung der Differentialgleichung 1. Ordnung $$ R C \frac{d u_{C}}{d t}+u_{C}=u_{0} $$ ist, die diesen Einschaltworgang beschreibt (sog. RC-Glied, Bild V-63).

Zeigen Sie: Die Funktionen $$ y_{1}(x)=\mathrm{e}^{2 x} \quad \text { und } \quad y_{2}(x)=x \cdot \mathrm{e}^{2 x} $$ bilden eine Fundamentalbasis der homogenen Differentialgleichung 2 . Ordnung \(y^{\prime \prime}-4 y^{\prime}+4 y=0\)

Lösen Sie die folgenden Differentialgleichungen 1. Ordnung durch Variation der Konstanten: a) \(y^{\prime}+x y=4 x\) b) \(y^{\prime}+\frac{y}{1+x}=\mathrm{e}^{2 x}\) c) \(x y^{\prime}+y=x \cdot \sin x\) d) \(y^{\prime} \cdot \cos x-y \cdot \sin x=1\) e) \(y^{\prime}-(2-\cos x) \cdot y=\cos x\) f) \(x y^{\prime}-y=x^{2}+4\)
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