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Magazine Chapter 31: Problem 32
Solve the given differential equations by Laplace transforms. The function is subject to the given conditions. A 20 -mH inductor, a \(40-\Omega\) resistor, a \(50-\mu \mathrm{F}\) capacitor, and a voltage source of \(100 e^{-1000 t}\) are connected in series in an electric circuit. Find the charge on the capacitor as a function of time \(t\), if \(q=0\) and \(i=0\) when \(t=0\)
Short Answer
Expert verified
Find the charge function by solving the RLC Laplace-form equation and using inverse transforms.
Step by step solution
01
Understand the System
The given system is an RLC circuit with a series connection. The circuit parameters are an inductor with inductance \(L = 20\) mH, a resistor with resistance \(R = 40\, \Omega\), a capacitor with capacitance \(C = 50 \mu\, F\), and a voltage source \(V(t) = 100 e^{-1000 t}\). The problem asks for the charge on the capacitor as a function of time, \(q(t)\).
02
Write the Differential Equation
For an RLC circuit, the differential equation for charge \(q(t)\) is given by: \[ L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C} q = V(t). \] Substituting the known values, we get: \[ 0.02 \frac{d^2q}{dt^2} + 40 \frac{dq}{dt} + 20000 q = 100 e^{-1000 t}. \]
03
Apply Laplace Transform
Apply Laplace transform to both sides of the differential equation, considering initial conditions \(q(0) = 0\) and \(i(0) = 0\), where current \(i = \frac{dq}{dt}\). The transform of the equation becomes: \[ 0.02 s^2 Q(s) + 40 s Q(s) + 20000 Q(s) = \frac{100}{s + 1000}. \]
04
Solve for Q(s)
Solve for \(Q(s)\) in the Laplace domain: \[ Q(s) = \frac{100}{0.02 s^2 + 40 s + 20000} \times \frac{1}{s + 1000}. \] Simplifying further gives: \[ Q(s) = \frac{100}{0.02s^2 + 40s + 20000} \frac{1}{s+1000}. \]
05
Inverse Laplace Transform
To find \(q(t)\), perform an inverse Laplace transform of \(Q(s)\). This typically involves partial fraction decomposition and using Laplace transform tables. The inverse transform will yield \(q(t)\) as a function of time.
06
Compute Partial Fractions
Decompose the expression \(\frac{100}{0.02 s^2 + 40 s + 20000} \frac{1}{s + 1000}\) into simpler fractions before applying the inverse Laplace transform. Find the roots of the quadratic equation and solve for partial fractions.
07
Conclude Solution
Find the function \(q(t)\) after applying the inverse Laplace transform to each term from the partial fraction decomposition and verify by plugging the inverse transform terms. This function will represent the charge in the capacitor over time.
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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 mathematical expressions that involve functions and their derivatives. They are used to model and solve real-world problems involving rates of change. In the context of an RLC circuit, such as the one outlined in the exercise, a differential equation describes how the charge on the capacitor changes over time in response to the electrical components and the applied voltage source.
- Second-order differential equations: In the RLC circuit, the relation is defined as a second-order differential equation, owing to the inductor which brings the second derivative component.
- Initial conditions: The conditions given at the start, like the charge (\(q = 0\)) and current (\(i = 0\)), help in solving the differential equation uniquely.
RLC Circuit
An RLC circuit is composed of three fundamental electrical components: a Resistor (\(R\)), an Inductor (\(L\)), and a Capacitor (\(C\)). These components are connected in series or parallel and respond to the time-varying voltage source provided.
- Resistor (\(R\)): Impedes current flow, managing energy dissipation as heat, described by resistance (measured in Ohms).
- Inductor (\(L\)): Stores energy in a magnetic field temporarily when current passes through it, resulting in potential difference changes (measured in Henry).
- Capacitor (\(C\)): Stores energy in an electric field and releases it at certain intervals, primarily affecting voltage (measured in Farads).
Inverse Laplace Transform
The Inverse Laplace Transform is the process used to convert functions in the frequency domain back to the time domain. When you apply a Laplace Transform to a differential equation in a circuit, you convert the problem to algebraic form, which is simpler to handle.
- Frequency to Time: Once solved algebraically, you'll need to use the inverse transform to interpret the function as time to understand physical response.
- Utilizing tables: Common inverse transforms are often found using pre-calculated tables for ease.
Partial Fraction Decomposition
Partial Fraction Decomposition is an algebraic technique used to break complex rational expressions into simpler fractions. This is crucial when you're dealing with inverse Laplace Transforms, as it makes the transformation process much more manageable.
- Decomposing: Start by breaking down the numerator/denominator expressions into simpler parts for easier transformation.
- Solving steps: Use identified roots or factors to express as sums of simpler fractions.
- Integration with Laplace: Integrate these simpler fractions with known Laplace transforms to find inverse solutions more straightforwardly.
