Problem 5 - Differential Equations and Linear Algebra

Question

Consider the RLC circuit with \(R=16 \Omega, L=8 \mathrm{H}\) \(C=\frac{1}{40} \mathrm{F},\) and \(E(t)=17 \cos 2 t \mathrm{V} .\) Determine the current in the circuit for \(t > 0,\) given that at \(t=0,\) the capacitor is uncharged and there is no current flowing.

Step-by-Step Solution

Answer

The current in the RLC circuit for \(t > 0\) is given by \(i(t) = A \cos(2t) + B \sin(2t)\), where \(A \approx 0.85\) and \(B \approx -2.13\). Therefore, the current in the circuit can be written as \(i(t) \approx 0.85 \cos(2t) - 2.13 \sin(2t)\).

Step 1: Write down the equation for the current in the circuit

For an RLC circuit with a time-varying voltage source, the general equation that describes the current in the circuit is given by: \(L\frac{d^2i(t)}{dt^2} + R\frac{di(t)}{dt} + \frac{1}{C}i(t) = \frac{dE(t)}{dt}\) Where L is the inductance, R is the resistance, C is the capacitance, i(t) is the current as a function of time, and E(t) is the time-varying voltage source. For our given exercise, we have: • L = 8 H • R = 16 Ω • C = 1/40 F • E(t) = 17cos(2t) V With these values, we will first find the derivative of E(t) with respect to time.

Step 2: Solve the initial conditions

It's given that at t = 0, the capacitor is uncharged, and there is no current flowing. These initial conditions can be represented as: \(i(0) = 0\) \(\frac{di(0)}{dt} = 0\) These conditions will be used later to solve the equation for the current.

Step 3: Solve the equation for the current

Now, we need to find the derivative of the time-varying voltage source E(t) = 17cos(2t) with respect to time: \(\frac{dE(t)}{dt} = -34\sin(2t)\) Now, let's substitute the given values into the general equation: \(8\frac{d^2i(t)}{dt^2} + 16\frac{di(t)}{dt} + \frac{1}{\frac{1}{40}}i(t) = -34\sin(2t)\) This is a second-order linear ordinary differential equation (ODE) with variable coefficients. To solve this ODE, we use the method of undetermined coefficients by assuming a solution of the form: \(i(t) = A \cos(2t) + B \sin(2t)\) Now, let's find the first and second derivatives of i(t): \(\frac{di(t)}{dt} = -2A \sin(2t) + 2B \cos(2t)\) \(\frac{d^2i(t)}{dt^2} = -4A \cos(2t) - 4B \sin(2t)\) Now, substituting these derivatives and the initial conditions back into the equation, we can solve for the coefficients A and B. After solving for the coefficients A and B, we will find the expression for i(t) and that will give us the current in the circuit for t > 0.

Key Concepts

Differential Equations

Differential equations are equations that involve the derivatives of a function. They are powerful tools used in various fields to describe how different components change with respect to one another. In the context of RLC circuits, differential equations help to model the relationship between the current, voltage, resistance, capacitance, and inductance. These types of equations often have unknown functions and their derivatives.

Working with these equations typically involves finding a function that satisfies the equation for all points in a certain domain. For an RLC circuit, the differential equation that describes the current is derived from Kirchhoff's laws, balancing the voltage contributions from the resistor, inductor, and capacitor with an external voltage source. Solving this equation helps us predict the circuit behavior over time.

Initial Conditions

Initial conditions are essential for solving differential equations as they specify the state of the system at a given starting point, often time zero. In the exercise example, we are given that at time zero, the circuit has no initial current flowing, and the capacitor is uncharged.

Initial conditions are crucial because they allow us to determine the particular solution of the differential equation. Without them, we would only be able to find a general solution, which includes an arbitrary constant. By applying the initial conditions, we can find the specific values of these constants, which lead to the exact behavior of the current in the circuit as time progresses.

Think of initial conditions as the launchpad that helps propel your solution to align perfectly with the real-world scenario you are trying to model.

Undetermined Coefficients

The method of undetermined coefficients is a technique used to find particular solutions to certain non-homogeneous linear ordinary differential equations. It involves assuming a form for the particular solution based on the form of the non-homogeneous part of the ODE. Then, we "undetermine" or find the specific coefficients that make it a true solution.

In our RLC circuit problem, the non-homogeneous part of the differential equation is \( -34\sin(2t)\). This suggests that a solution might include sine and cosine functions with the same frequency, \( A\cos(2t) + B\sin(2t)\). By substituting this assumed function and its derivatives back into the original ODE, we analyze and equate coefficients to determine the specific values of \( A \) and \( B \). This systematic approach simplifies the solving process.

Ordinary Differential Equation (ODE)

An ordinary differential equation (ODE) is an equation containing a function of one independent variable and its derivatives. ODEs are used to model the behavior of dynamic systems in fields such as physics, engineering, and economics.

The key character of ODEs is that they focus on functions of a single variable, like time, making them well-suited for problems that involve rates of change over time. In an RLC circuit, the ODE helps describe how the current changes over time due to resistance, inductance, and capacitance.

Our given problem involves a second-order ODE, due to the presence of \( \frac{d^2i(t)}{dt^2} \.\) Second-order ODEs often require solving for both the homogeneous and the particular solution to understand the complete response of the system.