Problem 34 - Advanced Engineering Mathematics

Question

Recall that the differential equation for the instantaneous charge \(q(t)\) on the capacitor in an \(L R C\) -series circuit is $$ L \frac{d^{2} q}{d t^{2}}+R \frac{d q}{d t}+\frac{1}{C} q=E(t) $$ See Section 3.8. Use the Laplace transform to find \(q(t)\) when \(L=1 \mathrm{~h}, R=20 \Omega, C=0.005 \mathrm{f}, E(t)=150 \mathrm{~V}, t>0, q(0)=0\) and \(i(0)=0\). What is the current \(i(t) ?\)

Step-by-Step Solution

Answer

The current is \( i(t) = 0.75 \, e^{-10t} (20 \sin(10t) - 20 \cos(10t)) \).

Step 1: Write the Differential Equation in Standard Form

From the given problem, the differential equation is: \[ \frac{d^{2} q}{d t^{2}} + 20 \frac{d q}{d t} + 200 q = 150. \] This is because \(L = 1\), \(R = 20\), and \(\frac{1}{C} = 200\) since \(C = 0.005\).

Step 2: Take the Laplace Transform

Apply the Laplace transform to the entire differential equation. Using the initial conditions \(q(0) = 0\) and \(i(0) = 0\), the Laplace transform of the equation becomes: \[ s^2 Q(s) + 20s Q(s) + 200Q(s) = \frac{150}{s}. \] Here, \(Q(s)\) is the Laplace transform of \(q(t)\).

Step 3: Solve for Q(s)

Reorganize the equation to solve for \(Q(s)\): \[ Q(s)(s^2 + 20s + 200) = \frac{150}{s}. \] Therefore, \[ Q(s) = \frac{150}{s(s^2 + 20s + 200)}. \]

Step 4: Partial Fraction Decomposition

Perform partial fraction decomposition on \[ Q(s) = \frac{150}{s(s^2 + 20s + 200)}. \] This can be written as: \[ \frac{A}{s} + \frac{Bs + C}{s^2 + 20s + 200}. \] Solve for \(A, B,\) and \(C\) to begin inverse transformation.

Step 5: Calculate Coefficients

Setting up the equation: \[ 150 = A(s^2 + 20s + 200) + (Bs + C)s. \] Equating coefficients of powers of \(s\), we get: - For \(s^2\): \(A + B = 0\) - For \(s^1\): \(20A + C = 0\) - For constant: \(200A = 150\) Solving these gives: \(A = 0.75\), \(B = -0.75\), \(C = -15\).

Step 6: Inverse Laplace Transform

Insert the coefficients back into the partial fraction decomposition and take the inverse Laplace Transform:\[ q(t) = 0.75(1 - e^{-10t} \cos(10t) - e^{-10t} \sin(10t)). \]

Step 7: Current i(t)

Since current \(i(t)\) is the derivative of charge \(q(t)\), find \(i(t)\) by differentiating \(q(t)\):\[ i(t) = \frac{dq}{dt} = 0.75 \left(10e^{-10t} \sin(10t) - 10e^{-10t} \cos(10t) - 10e^{-10t} \cos(10t) \right). \] Simplifying, \[ i(t) = 0.75 \, e^{-10t} (20 \sin(10t) - 20 \cos(10t)). \]

Key Concepts

Differential Equations

Differential equations are fundamental in describing how physical systems change over time. They provide us with equations that involve derivatives which indicate rates of change. In the context of this exercise, we're focusing on an electrical circuit, which is expressed in terms of the charge on a capacitor. Typically in a simple RLC circuit, the relationship can be modeled as

• Second-order differential equation: Captures the dynamics involving resistance (R), inductance (L), and capacitance (C).
• Initial conditions: Essential for describing the complete behavior of the system.

The standard form of the differential equation derived from the RLC circuit includes terms that signify the resistance, inductance, and capacitance, alongside the applied voltage (E(t)). Solving this type of equation helps us predict how the electrical charge changes over time.

RLC Circuit Analysis

RLC circuit analysis involves studying how resistors (R), inductors (L), and capacitors (C) work together. These components are widely used in electronic devices, functioning as filters or oscillators. In this exercise, we're focusing on a series RLC circuit. Here's why each component matters:

• Resistor (R): Opposes the flow of electric current, reducing the overall current in the circuit.
• Inductor (L): Resists changes in current and can store energy in a magnetic field.
• Capacitor (C): Stores electrical energy in an electric field and can oppose changes in voltage.

Understanding these elements together with the external voltage supply allows us to derive the differential equation, providing insights into circuit behavior over time.

Partial Fraction Decomposition

Partial fraction decomposition is a tool used in calculus to simplify complex rational expressions into a sum of simpler fractions. This is particularly useful when solving equations with the Laplace Transform method. In our example:

• We decompose the transform function into parts that are easier to handle mathematically.
• Specifically, terms like \(\frac{A}{s}\) and \(\frac{Bs + C}{s^2 + 20s + 200}\) are used to represent individual components of the solution spectrum.

The goal is to solve for constants (A, B, and C) to make inverse transformation straightforward. Turning a complex fraction into simpler parts is key to tackling inverse Laplace Transforms efficiently.

Inverse Laplace Transform

The inverse Laplace transform is a mathematical operation that takes us from the frequency domain back into the time domain. After transforming an equation into the s-domain using the Laplace Transform, the goal is to revert it back to express the solution regarding time. In our circuit problem:

• We obtained a rational expression in terms of \(Q(s)\).
• The inverse Laplace technique helps us convert this back to \(q(t)\), the time-dependent charge expression.

In simpler terms, it's like translating a complex equation into practical terms (like how charge changes over time). The inverse process often involves standard Laplace transform pairs and properties, translating mathematical expressions we can't easily solve in their original form.