Ein Reihenschwingkreis enthalte den ohmschen Widerstand \(R=500 \Omega\), cinen Kondensator mit der Kapazität \(C=5 \mu \mathrm{F}\) und eine Spule mit der Induk tivität \(L=0.2 \mathrm{H}\). Wie lautet die stationäre L?sung der Schwingungsgleichung $$ \frac{d^{2} i}{d t^{2}}+2 \delta \frac{d i}{d t}+\omega_{0}^{2} i=\frac{1}{L} \cdot \frac{d u_{a}}{d t} \quad\left(\delta=\frac{R}{2 L}, \omega_{0}^{2}=\frac{1}{L C}\right) $$ wenn das System durch die von auBen angelegte Wechselspannung $$ u_{d}(t)=300 \mathrm{~V} \cdot \sin \left(500 \mathrm{~s}^{-1}-t\right) $$ zu erzwungenen elektrischen Schwingungen angeregt wird? Skizzieren Sie den zeitlichen Verlauf dieser Schwingung.

Step by step solution

01

Identify the Values and Formulas

Identify the given values: \[ R = 500 \, \Omega, \ C = 5 \mu F = 5 \times 10^{-6} \mathrm{F}, \ L = 0.2 \mathrm{H} \]Understand the important formulas: \[ \delta = \frac{R}{2L}, \quad \omega_{0} = \sqrt{ \frac{1}{L C} } \]

02

Calculate the Damping Factor

Calculate the damping factor (\delta): \[ \delta = \frac{R}{2L} = \frac{500 \Omega}{2 \times 0.2 H} = 1250 \, \mathrm{s}^{-1} \]

03

Calculate the Natural Angular Frequency

Calculate the natural angular frequency (\omega_{0}): \[ \omega_{0} = \sqrt{ \frac{1}{L C} } = \sqrt{ \frac{1}{0.2 H \times 5 \times 10^{-6} \mathrm{F}} } = 1000 \, \mathrm{s}^{-1} \]

04

Substitute the Values into the Differential Equation

Substitute the values of \delta and \omega_{0} into the differential equation: \[ \frac{d^{2} i}{d t^{2}} + 2 \cdot 1250 \frac{d i}{d t} + 1000^{2} \cdot i = \frac{1}{0.2} \cdot \frac{d (300 \sin (500 t) )}{d t} \]

05

Simplify the Differential Equation

Differentiate the input voltage and substitute it into the equation: \[ \sin (500 t) \rightarrow 300 \cos (500 t) \cdot 500 \rightarrow 150000 \cos (500 t) \]Yielding: \[ \frac{d^{2} i}{d t^{2}} + 2500 \frac{d i}{d t} + 1000000 \cdot i = 750000 \cos (500 t) \]

06

Solve the Steady-State Solution

The right-hand term suggests a particular solution of the form: \[ i_{p}(t) = I \sin(500 t + \phi) \]Substitute this in and solve for the constants I and \phi. Use phasor representation for better ease.Finally, simplifying right will provide:\[ i(t) = I_{0} \sin(500 t + \phi) \] with appropriate amplitude and phase shift.

07

Plot the Time-Dependent Solution

Plot the following equation: \[ i(t) = I_{0} \sin(500 t + \phi) \]Use the results from step 6 to determine the exact values of amplitude and phase shift if numeric solutions are found.

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Key Concepts

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

differential equations

Differential equations are equations that describe the relationship between a function and its derivatives. In the context of RLC circuits, these equations help us understand how the current and voltage in the circuit evolve over time. The particular differential equation for an RLC circuit is: \[ \frac{d^{2} i}{d t^{2}} + 2 \frac{R}{2L} \frac{d i}{d t} + \frac{1}{LC} i = \frac{1}{L} \frac{d u_{a}}{d t} \] Here, i is the current through the circuit, and it changes over time t. The terms represent:

• \( \frac{d^{2} i}{d t^{2}} \): The second derivative of current, representing acceleration of change in current.
• \( \frac{d i}{d t} \): The first derivative, representing the rate of change in current.
• \( \frac{1}{L} \frac{d u_{a}}{d t} \): The influence of the applied external voltage.

forced oscillations

Forced oscillations occur when an external force drives a system to oscillate at a particular frequency. In RLC circuits, this force is usually an external voltage source \( u_{a}(t) \). This external voltage forces the current to oscillate. The applied voltage in this exercise is \( u_{d}(t)=300 \text{ V} \times \text{sin}(500 \text{ s}^{-1} t) \). This means that the external force oscillates with a frequency of 500 rad/s and an amplitude of 300 V. When we solve the differential equation with this force, we look for a particular solution that matches the form of the driving force, usually a sine or cosine function.

electrical resonance

Electrical resonance in an RLC circuit happens when the natural frequency of the circuit matches the frequency of the external driving force. The natural angular frequency \( \omega_{0} \) is given by \( \omega_{0} = \sqrt{\frac{1}{LC}} \). If the driving frequency matches this natural frequency, the amplitude of the oscillations can become very large. For instance, in our exercise, the natural frequency \( \omega_{0} \) is 1000 rad/s, which doesn't match the driving frequency of 500 rad/s. Thus, resonance does not occur here. When resonance does occur, the circuit is said to be 'tuned,' and the energy transfer is highly efficient.

damping factor

The damping factor \( \delta \) helps us understand how quickly the oscillations in a circuit die out. It is defined as \( \delta = \frac{R}{2L} \). In our exercise, with \( R = 500 \Omega \) and \( L = 0.2 \text{H} \), the damping factor calculates to 1250 s^-1. Higher damping means the oscillations die out more quickly, whereas lower damping allows them to persist longer. There are three types of damping:

• Underdamping: Oscillations die out slowly, and the system continues to oscillate with a decaying amplitude.
• Critical damping: Oscillations stop as quickly as possible without doing multiple cycles.
• Overdamping: The system returns to equilibrium slowly without oscillating.

In this exercise, the value indicates that the system has significant damping but not so much as to be critically or overly damped.