Chapter 6: Problem 1
Find the current in the \(R L C\) circuit, assuming that \(E(t)=0\) for \(t>0\). \(R=3\) ohms; \(L=.1\) henrys; \(C=.01\) farads; \(Q_{0}=0\) coulombs; \(I_{0}=2\) amperes.
Short Answer
Expert verified
Based on the given RLC circuit and initial conditions, the current flowing through the circuit for t > 0 is:
\(I(t) = e^{-15t}\left(-3\sin{(5t)} + 2\cos{(5t)}\right)\)
Step by step solution
01
Write down the equation of the RLC circuit
Based on Kirchhoff's Voltage Law, we can write the governing equation for the RLC circuit as:
\(\frac{d^2Q}{dt^2} + \frac{R}{L}\frac{dQ}{dt} + \frac{1}{LC} Q = \frac{d^2Q}{dt^2} + \frac{3}{0.1}\frac{dQ}{dt} + \frac{1}{0.1*0.01} Q = E(t)\)
We are given that E(t) = 0 for t > 0, so our equation becomes:
\(\frac{d^2Q}{dt^2} + 30\frac{dQ}{dt} + 1000Q = 0\)
02
Solve the homogeneous equation
The above differential equation is a second order linear homogeneous ordinary differential equation (ODE). To solve it, we will start by finding the characteristic equation, which is:
\(m^2 + 30m + 1000 = 0\)
Solve this quadratic equation to obtain the values of m:
\(m_1 = -15 + 5i\)
\(m_2 = -15 - 5i\)
03
Generate the solutions of the differential equation
Using the complex roots method for homogeneous ODEs, we can write the general solution for the given differential equation as:
\(Q(t) = e^{-15t}(A\cos{(5t)} + B\sin{(5t)})\)
Next, we need to find the values of A and B using the initial conditions provided.
04
Apply the initial conditions
We have been given the initial conditions: Q(0) = 0 and I(0) = 2 A. Since I(t) is the derivative of Q(t), we can write:
\(I(t) = \frac{dQ(t)}{dt} = e^{-15t}(-15A\cos{(5t)} - 15B\sin{(5t)} + 5B\cos{(5t)} - 5A\sin{(5t)})\)
Now, let's apply the initial condition Q(0) = 0 to the general solution
\(Q(0) = e^{-15*0}(A\cos{(5*0)} + B\sin{(5*0)}) = A\)
Therefore, \(A = 0\).
Now, apply the initial condition I(0) = 2 A:
\(I(0) = e^{-15*0}(-15A\cos{(5*0)} - 15B\sin{(5*0)} + 5B\cos{(5*0)} - 5A\sin{(5*0)}) = 2\)
\(2 = 5B \Rightarrow B = \frac{2}{5}\)
05
Write down the final expression for the current
Now that we have the values of A and B, we can write the final expression for the current flowing through the RLC circuit as:
\(I(t) = \frac{dQ(t)}{dt} = e^{-15t}\left(\frac{2}{5}\sin{(5t)} + 5\frac{2}{5}\cos{(5t)} - 15\frac{2}{5}\sin{(5t)}\right)\)
So, the current in the RLC circuit for t > 0 is given by:
\(I(t) = e^{-15t}\left(-3\sin{(5t)} + 2\cos{(5t)}\right)\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kirchhoff's Voltage Law
Kirchhoff's Voltage Law (KVL) is a fundamental principle in electrical engineering. It is essential in analyzing circuits like RLC circuits. KVL states that the sum of the electrical potential differences (voltage) around any closed network is zero.
In an RLC circuit, this means that the voltage drop across the resistor (R), inductor (L), and capacitor (C) must sum to zero when there is no external voltage source affecting the circuit. This allows us to create a differential equation that characterizes the behavior of the circuit's current.
By applying KVL to an RLC circuit, you can derive an equation that helps calculate how current and charge change over time within the circuit. This is crucial in predicting circuit behavior, especially as the circuit's response over time to different inputs and initial conditions.
Ordinary Differential Equations
Ordinary Differential Equations (ODEs) play a vital role in modeling the behavior of RLC circuits. These equations are based on derivatives, representing the relationship between functions and their rates of change. In the context of an RLC circuit, ODEs describe how current and charge evolve over time. For the given problem, we derive an ODE from KVL's application. The equation is a second-order linear homogeneous ODE: \[ \frac{d^2Q}{dt^2} + 30\frac{dQ}{dt} + 1000Q = 0 \] This equation describes how the charge, \( Q(t) \), changes in the circuit when there is no external voltage source. Solving this ODE provides critical insights into the transient response of the circuit, such as how quickly currents decay or how oscillations behave. ODEs are essential tools in electrical engineering, leading to precise control and analysis of circuit systems.
Complex Roots Method
The Complex Roots Method is a powerful mathematical technique used to solve differential equations, especially those found in electrical circuit analysis. This approach is particularly useful when dealing with characteristic equations derived from second-order linear homogeneous ODEs, like the one in our problem. Upon solving the characteristic equation \( m^2 + 30m + 1000 = 0 \), we find complex roots:
- \( m_1 = -15 + 5i \)
- \( m_2 = -15 - 5i \)
Electrical Engineering Education
In electrical engineering education, understanding concepts like Kirchhoff's Voltage Law, ODEs, and methods for solving them is foundational. These principles are essential for students to analyze and design complex circuit systems, including RLC circuits.
RLC circuit analysis provides a practical context in which students can apply mathematical tools to real-world problems. Graduating engineers need to be adept in such analyses since modern technology heavily relies on well-designed circuits. Educational programs often emphasize practical applications and theoretical knowledge, preparing students to address challenges in the field.
By mastering these topics, students become proficient in predicting and controlling circuit behavior, crucial skills in everything from creating consumer electronics to designing large-scale electrical systems. A strong foundational knowledge in these areas is vital for success in electrical engineering and related disciplines.