/*! 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 10 A resistor \((20 \Omega)\) and c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A resistor \((20 \Omega)\) and capacitor \((0.1 \mathrm{~F})\) are joined in series with an electromotive force (emf) \(E=E(t)\), as shown in Figure 4 . If there is no charge on the capacitor at time \(t=0\), find the ensuing charge on the capacitor at time \(E(t)=4 \cos 3 t \mathrm{~V}\)

Short Answer

Expert verified
The charge on the capacitor at time \(t\) is \(q(t) = -\frac{1}{7}e^{-0.5t} + \frac{1}{7} \cos 3t + \frac{3}{35} \sin 3t\).

Step by step solution

01

Understanding the Circuit

The electric circuit consists of a resistor and a capacitor in series with an alternating EMF, given by \(E(t) = 4 \cos 3t\, \text{V}\). At \(t=0\), there is no charge on the capacitor.
02

Recall the Governing Differential Equation

For a series R-C circuit, the governing equation for charge \(q(t)\) on the capacitor is:\[R \frac{dq}{dt} + \frac{q}{C} = E(t)\]Here, \(R = 20 \,\Omega\), \(C = 0.1\,\text{F}\), and \(E(t) = 4 \cos 3t\).
03

Substitute Values

Substitute the given values into the differential equation:\[20 \frac{dq}{dt} + \frac{q}{0.1} = 4 \cos 3t\]
04

Simplify the Equation

The equation becomes:\[20 \frac{dq}{dt} + 10q = 4 \cos 3t\]This can be rewritten as:\[\frac{dq}{dt} + 0.5q = 0.2 \cos 3t\]
05

Solve the Homogeneous Equation

First, solve the associated homogeneous equation:\[\frac{dq_H}{dt} + 0.5q_H = 0\]The solution is:\[q_H(t) = A e^{-0.5t}\]
06

Solve the Particular Solution

Use the method of undetermined coefficients for the particular solution. Assume a solution of the form:\[q_P(t) = B \cos 3t + C \sin 3t\]Substitute \(q_P\) into the left side of the non-homogeneous differential equation and compare coefficients to solve for \(B\) and \(C\).
07

Substitute and Solve Coefficients

Substitute \(q_P(t) = B \cos 3t + C \sin 3t\) and its derivative into:\[\frac{dq_P}{dt} = -3B \sin 3t + 3C \cos 3t\]Then:\[-3B \sin 3t + 3C \cos 3t + 0.5(B \cos 3t + C \sin 3t) = 0.2 \cos 3t\]From this, solve the linear equations for \(B = \frac{1}{7}\) and \(C = \frac{3}{35}\).
08

General Solution

Combine the homogeneous and particular solutions to form the general solution:\[q(t) = A e^{-0.5t} + \left(\frac{1}{7} \cos 3t + \frac{3}{35} \sin 3t\right)\]
09

Apply Initial Condition

Use the initial condition \(q(0) = 0\) to solve for \(A\):\[0 = A + \frac{1}{7}\]Thus, \(A = -\frac{1}{7}\).
10

Final Expression for Charge

Substitute \(A\) back into the general solution:\[q(t) = -\frac{1}{7}e^{-0.5t} + \frac{1}{7}\cos 3t + \frac{3}{35} \sin 3t\]This is the expression for the charge on the capacitor over time.

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 91Ó°ÊÓ!

Key Concepts

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

R-C Circuit
A Resistive-Capacitive (R-C) Circuit is a key foundation in electrical engineering and circuit analysis. It consists of a resistor (R) and a capacitor (C) arranged in a series, across which an external voltage called electromotive force (EMF) is applied. Such circuits are fundamental in filtering applications where they help charge or discharge a capacitor through the resistor controlled by a differential equation over time.
In this specific exercise, we are dealing with an R-C circuit containing a resistor of 20 ohms and a capacitor of 0.1 farad. The EMF is given as a function of time, specifically as a cosine function, which introduces an alternating current component to the circuit.
These kind of circuits are studied not only for their foundational elements, but also their behavior over time, such as the rate at which capacitors charge and discharge and how this behavior can be modeled mathematically.
Electromotive Force
Electromotive Force (EMF), denoted as \(E(t)\), is essentially the voltage supplied to the circuit. It drives the electric current through the circuit components like resistors and capacitors. In our example, the EMF is represented by an alternating cosine function \(E(t) = 4 \cos 3t\).
This EMF determines how the charge on the capacitor changes over time. Since the function varies with time, it signifies that the electric pressure pushing the charge through the circuit is also changing, causing the capacitor to charge and discharge in a periodic manner.
Understanding EMF and its variations is crucial because it directly affects the performance of the circuit. Engineers and physicists often need to calculate the current and voltage drop across each circuit component given the EMF, especially when it is time-dependent as seen here.
Particular Solution
The concept of a particular solution emerges when solving non-homogeneous differential equations. These are equations which have terms independent of the function and its derivatives. In simpler terms, they include an external function (here, \(E(t)\)) contributing to the solution.
In the context of our R-C circuit problem, the particular solution can be thought of as the component of the solution that responds directly to the alternating EMF, given by the cosine function. We assumed a form \(q_P(t) = B \cos 3t + C \sin 3t\) because the EMF is a cosine function. By substituting \(q_P(t)\) into the differential equation and adjusting for coefficients, we isolate the terms causing this non-standard behavior.
The particular solution essentially allows the system's response to the EMF to be articulated in mathematical terms. It's a vital aspect, especially in circuits with time-varying inputs, helping to distinguish between the natural response of the system and the forced response due to the EMF.
Initial Condition
Initial conditions are used to solve differential equations uniquely. They specify the state of the system, often at the start, ensuring that the obtained solution meets certain known values.
For our problem, the initial condition is stated as there being no charge on the capacitor at \(t = 0\), denoted by \(q(0) = 0\). By applying this to the general solution of our differential equation, we can find the arbitrary constants (such as \(A\) in our homogeneous solution).- Initial conditions thus stabilize the solution over time, providing the puzzle piece needed to fit the particular and homogeneous solutions into a comprehensive description of charge over time.- Without them, the solution would remain ambiguous, potentially representing a multitude of possible scenarios.
This concept ensures a tailored answer pertinent to the specific scenario described, reflecting real world settings where devices are usually initialized from an uncharged or zero state.

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

A certain bacterium is known to grow according to the Malthusian model, doubling itself every 4 hours. If a biologist starts with a culture of 10,000 bacteria, at what minimal rate does he need to harvest the culture so that it won't overwhelm the container with bacteria?

Adriana opens a savings account with an initial deposit of \(\$ 1000\). The annual rate is \(6 \%\), compounded continuously. Adriana pledges that each year her annual deposit will exceed that of the previous year by \(\$ 500\). How much will be in the account at the end of the tenth year?

A certain bacterium, given plenty of nutrient and room, is known to grow according to the Malthusian model with reproductive rate \(r\). Suppose that the biologist working with the culture harvests the bacteria at a constant rate of \(h\) bacteria per hour. Use qualitative analysis to discuss the fate of the culture.

A population, left alone, obeys the logistic law with an initial population of 1000 doubling itself in about \(2.3\) hours. It is known that the environment can sustain approximately 10,000 individuals. Harvesting is introduced into this environment, with 1500 individuals removed per hour, but only during the last 4 hours of a 24 -hour day. Suppose that the population numbers 6000 at the beginning of the day. Use a numerical solver to sketch a graph of the population over the course of the next three days. Find approximately the size of the population at the end of each day.

Don and Heidi would like to buy a home. They've examined their budget and determined that they can afford monthly payments of \(\$ 1000\). If the annual interest is \(7.25 \%\) and the term of the loan is 30 years, what amount can they afford to borrow?

See all solutions

Recommended explanations on Math Textbooks

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.