/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 An \(R L C\) circuit with input ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 17

An \(R L C\) circuit with input voltage \(E(t)\) is described. Find the current \(I(t)\) using the given initial current (in amperes) and charge on the capacitor (in coulombs). $$ \begin{aligned} &R=16 \Omega, L=2 \mathrm{H}, C=0.02 \mathrm{~F}; \\ &E(t)=100 \mathrm{~V} ; \quad I(0)=0, Q(0)=5 \end{aligned} $$

Short Answer

Expert verified
Find the current \(I(t)\) by solving the system of differential equations with the given initial conditions.

Step by step solution

01

Set Up the Differential Equation for the RLC Circuit

For an RLC circuit, the relation between voltage, current, and charge can be expressed as:\[ L \frac{dI}{dt} + RI + \frac{Q}{C} = E(t) \]Given that \(Q = C \cdot V\), and that the voltage across the capacitor \(V_c\) is \(\frac{Q}{C}\), the equation becomes\[ L \frac{dI}{dt} + RI + \frac{Q}{C} = E(t) \]Substitute the given values: \(R = 16\), \(L = 2\), \(C = 0.02\), and \(E(t) = 100\). The equation transforms into:\[ 2 \frac{dI}{dt} + 16I + 50Q = 100 \]
02

Relate Charge and Current

We know that the charge \(Q\) is related to current \(I\) by the derivative:\[ \frac{dQ}{dt} = I \]This is because the current is the rate of change of charge over time.
03

Formulate and Solve the System of Differential Equations

We now have two equations:1. \(\frac{dI}{dt} = \frac{1}{2}(100 - 16I - 50Q)\) from our modified RLC differential equation.2. \(\frac{dQ}{dt} = I\) from the rate of change of charge.These two equations form a system of differential equations. We need initial conditions to solve these: \(I(0) = 0\) and \(Q(0) = 5\).
04

Solve the Differential Equations Using Initial Conditions

Apply the initial conditions \(I(0) = 0\) and \(Q(0) = 5\) to solve the system of equations. After implementing the initial conditions, use methods such as substitution, elimination, or matrix exponentiation (for linear systems) to transform and solve these equations. For instance, convert these equations into matrix form and solve using eigenvectors/eigenvalues for the system.
05

Derive General Solution for Current I(t)

After applying methods like Laplace transforms or matrix diagonalization, we can determine the current \(I(t)\). Using known methods, derive:\[ I(t) = Ae^{\lambda_1 t} + Be^{\lambda_2 t} + \frac{50}{16} \]Where \(A\) and \(B\) are constants found from initial conditions, and \(\lambda_1, \lambda_2\) are the roots of the characteristic equation formed from the differential equations.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Differential Equations
In the realm of electrical engineering, differential equations are vital for modeling the behavior of various circuit components over time. These equations express how a circuit's current, voltage, or charge changes as time progresses. For RLC circuits, which contain a resistor (R), an inductor (L), and a capacitor (C), we use differential equations to understand the relationship between these elements. This is crucial because, in real-world applications, we often deal with circuits that have time-varying inputs, like changing voltages. By setting up differential equations for RLC circuits, engineers can predict how the circuit will respond to different signals.
Electrical Engineering
Electrical engineering is a broad field that involves the study and application of electricity, electronics, and electromagnetism. One key area within this field is circuit analysis, which involves understanding how electrical circuits behave. RLC circuit analysis is a fundamental part of this as it helps engineers design and optimize circuits for various applications鈥攆rom simple power supply networks to complex communication systems. By assessing an RLC circuit, engineers can optimize components to ensure stability, efficiency, and reliability, which are crucial for both consumer electronics and industrial systems.
Initial Conditions
When solving differential equations for circuits, initial conditions are essential. These are the values of the circuit's variables at the start of the analysis鈥攗sually at time zero. They are critical for uniquely determining the behavior of the system over time. In our example, the initial conditions are given as the initial current \( I(0) = 0 \) and the initial charge on the capacitor \( Q(0) = 5 \). These values tell us the state of the system at the beginning, allowing us to find specific solutions that fit the circuit's real-life behavior, rather than just a general pattern.
Laplace Transforms
Laplace Transforms are a powerful mathematical tool used to simplify the analysis of linear time-invariant systems, including RLC circuits. By transforming differential equations into algebraic ones, Laplace Transforms make it manageable to solve complex circuits analytically. In our exercise, after setting up the differential equations, the Laplace Transform could be used to easily solve for \( I(t) \). It enables the direct application of initial conditions into the solutions, finding unique solutions for the system's future behavior effectively. This transforms a complex, time-domain problem into a simpler, frequency-domain analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find a function \(y(x)\) such that \(y^{(4)}(x)=y^{(3)}(x)\) for all \(x\) and \(y(0)=18, y^{\prime}(0)=12, y^{\prime \prime}(0)=13\), and \(y^{(3)}(0)=7\)

Problems pertain to the solution of differential equations with complex coefficients. Use the quadratic formula to solve the following equations. Note in each case that the roots are not complex conjugates. (a) \(x^{2}+i x+2=0\) (b) \(x^{2}-2 i x+3=0\)

Consider an \(L C\) circuit - that is, an \(R L C\) circuit with \(R=\) 0 -with input voltage \(E(t)=E_{0} \sin \omega t .\) Show that unbounded oscillations of current occur for a certain resonance frequency; express this frequency in terms of \(L\) and \(C\).

First verify by substitution that \(y_{1}(x)=x^{-1 / 2} \cos x\) is one solution (for \(x>0\) ) of Bessel's equation of order \(\frac{1}{2}\), $$ x^{2} y^{v}+x y^{\prime}+\left(x^{2}-\frac{1}{4}\right) y=0 $$ Then derive by reduction of order the second solution \(y_{2}(x)=x^{-1 / 2} \sin x\)

Use the method of variation of parameters to find a particular solution of the given differential equation. $$ y^{\prime \prime}+4 y=\sin ^{2} x $$
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.