/*! 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 28 Assume \(L d i / d t+R i=E(t), E... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 28

Assume \(L d i / d t+R i=E(t), E(t)=E_{0} \sin \omega t,\) and \(i(0)=i_{0}\) so that $$i=\frac{E_{0} R}{L^{2} \omega^{2}+R^{2}} \sin \omega t-\frac{E_{0} L \omega}{L^{2} \omega^{2}+R^{2}} \cos \omega t+c e^{-R t / L}$$ since \(i(0)=i_{0}\) we obtain \(c=i_{0}+\frac{E_{0} L \omega}{L^{2} \omega^{2}+R^{2}}\)

Short Answer

Expert verified
The constant \( c \) is \( c = i_0 + \frac{E_0 L \omega}{L^2 \omega^2 + R^2} \).

Step by step solution

01

Understand the Given Differential Equation

We start with the differential equation \( L \frac{di}{dt} + R i = E(t) \), where \( E(t) = E_0 \sin(\omega t) \). Here, \( L \) is the inductance, \( R \) is the resistance, and \( i(t) \) is the current through the circuit as a function of time \( t \).
02

Analyze the Specific Solution

The given solution to the differential equation is \[ i(t) = \frac{E_0 R}{L^2 \omega^2 + R^2} \sin(\omega t) - \frac{E_0 L \omega}{L^2 \omega^2 + R^2} \cos(\omega t) + c e^{-\frac{R t}{L}} \]. This solution includes a particular solution involving sine and cosine functions, and a homogeneous solution involving an exponential decay term.
03

Apply Initial Condition

We are given the initial condition \( i(0) = i_0 \). Substitute \( t = 0 \) into the solution: \[ i(0) = \frac{E_0 R}{L^2 \omega^2 + R^2} \sin(0) - \frac{E_0 L \omega}{L^2 \omega^2 + R^2} \cos(0) + c e^{0} = i_0 \]. Since \( \sin(0) = 0 \) and \( \cos(0) = 1 \), this simplifies to \[ -\frac{E_0 L \omega}{L^2 \omega^2 + R^2} + c = i_0 \].
04

Solve for Constant \( c \)

Rearrange the equation from Step 3 to solve for \( c \): \[ c = i_0 + \frac{E_0 L \omega}{L^2 \omega^2 + R^2} \]. This is the value of the constant necessary for the solution to satisfy the initial condition \( i(0) = i_0 \).

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Key Concepts

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

Inductance
In the realm of differential equations, especially when dealing with circuits, the concept of inductance is crucial. Inductance, denoted by the letter 'L', refers to the ability of a coil or an inductor to store energy in a magnetic field when current flows through it. This property is central to the behavior of inductors in electrical circuits, influencing how they react to changes in current.

Inductance is measured in units called henries (H). The larger the inductance, the more opposition there is to changes in current, which significantly affects the circuit's behavior over time. In this exercise, the inductance determines the rate at which the current changes, as represented in the differential equation by the term \( L \frac{di}{dt} \).

Key points about inductance:
  • It's a property of inductors in a circuit.
  • Stores energy in a magnetic field.
  • Affects how quickly a circuit reaches a steady state.
  • Opposes changes in current flow.
Resistance
Resistance is another fundamental concept when solving differential equations for electrical circuits. Denoted as 'R', resistance measures how strongly an object opposes the flow of electric current. It's a vital factor that influences how currents change in response to applied voltages.

Resistance is measured in ohms (惟). It plays a role in determining the current in the circuit and affects both the homogeneous and particular solutions to the differential equation. In circuits, higher resistance implies more energy lost as heat, and it affects how the transient and steady-state behaviors manifest in the circuit's response.

Important aspects of resistance include:
  • Measured in ohms.
  • Determines how easy or difficult it is for the current to flow through a component.
  • Affects energy dissipation in the circuit.
  • Integral to calculating heat generated by electrical components.
Initial Condition
Initial conditions are vital when solving differential equations because they help determine particular solutions from a family of solutions. In this exercise, the initial condition is represented by \( i(0) = i_0 \). It means the current at time \( t = 0 \) is a specific value \( i_0 \).

This condition allows us to solve for any constant terms in the solution, such as the constant 'c' seen in the exponential term of the solution. Applying the initial condition ensures the particular solution appropriately reflects the physical situation at the starting point of the observation.

Reasons initial conditions are important:
  • Help specify a unique solution among many possible ones.
  • Reflect the physical setup at the starting point.
  • Essential for accurately modeling real-world dynamical systems.
Particular Solution
A particular solution in differential equations provides a solution that satisfies not just the differential equation, but also the non-homogeneous part, which in this case arises from the source \( E(t) = E_0 \sin(\omega t) \). Particular solutions are not necessarily the most general solutions, but they are specific enough to account for external forces or inputs.

The particular solution in this exercise includes sine and cosine terms, reflecting the sinusoidal input voltage \( E(t) \). Each component of the particular solution links directly to this driving force, accounting for its frequency and amplitude.

Characteristics of a particular solution:
  • Addresses the driving function or input.
  • Usually found through methods like undetermined coefficients or variation of parameters.
  • Ensures the solution fits the specific form of external influences.
Homogeneous Solution
Homogeneous solutions to differential equations arise when we set the differential equation equal to zero, focusing on the natural behavior of the system without external forces. In our exercise, the term \( c e^{-\frac{R t}{L}} \) represents the homogeneous solution.

This exponential term characterizes how the system evolves over time by itself, primarily dictated by resistance \( R \) and inductance \( L \). The exponential decay means the effects diminish over time, often representing how a system returns to equilibrium.Main aspects of homogeneous solutions:
  • Solutions to the equation when set to zero.
  • Reflect the intrinsic dynamics of the system.
  • Often involve exponential terms.
  • Describe how a system naturally evolves without external inputs.

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Most popular questions from this chapter

From \(\frac{1}{N} d N=\left(t e^{t+2}-1\right) d t\) we obtain \(\ln |N|=t e^{t+2}-e^{t+2}-t+c\) or \(N=c_{1} e^{t e^{t+2}-e^{t+2}-t}\).

Write the differential equation as $$\frac{d v}{d x}+\frac{1}{x} v=32 v^{-1},$$ and let \(u=v^{2}\) or \(v=u^{1 / 2} .\) Then $$\frac{d v}{d x}=\frac{1}{2} u^{-1 / 2} \frac{d u}{d x},$$ and substituting into the differential equation, we have $$\frac{1}{2} u^{-1 / 2} \frac{d u}{d x}+\frac{1}{x} u^{1 / 2}=32 u^{-1 / 2} \quad \text { or } \quad \frac{d u}{d x}+\frac{2}{x} u=64.$$ The latter differential equation is linear with integrating factor \(e^{\int(2 / x) d x}=x^{2},\) so $$\frac{d}{d x}\left[x^{2} u\right]=64 x^{2}$$ and $$x^{2} u=\frac{64}{3} x^{3}+c \quad \text { or } \quad v^{2}=\frac{64}{3} x+\frac{c}{x^{2}}.$$

Let \(I=I(t)\) be the intensity, \(t\) the thickness, and \(I(0)=I_{0} .\) If \(d I / d t=k I\) and \(I(3)=0.25 I_{0},\) then \(I=I_{0} e^{k t}\) \(k=\frac{1}{3} \ln 0.25,\) and \(I(15)=0.00098 I_{0}\).

Separating variables, we have \\[\frac{d y}{y-y^{3}}=\frac{d y}{y(1-y)(1+y)}=\left(\frac{1}{y}+\frac{1 / 2}{1-y}-\frac{1 / 2}{1+y}\right) d y=d x\\]. Integrating, we get \\[\ln |y|-\frac{1}{2} \ln |1-y|-\frac{1}{2} \ln |1+y|=x+c\\]. When \(y>1,\) this becomes \\[\ln y-\frac{1}{2} \ln (y-1)-\frac{1}{2} \ln (y+1)=\ln \frac{y}{\sqrt{y^{2}-1}}=x+c.\\] Letting \(x=0\) and \(y=2\) we find \(c=\ln (2 / \sqrt{3}) .\) Solving for \(y\) we get \(y_{1}(x)=2 e^{x} / \sqrt{4 e^{2 x}-3},\) where \(x>\ln (\sqrt{3} / 2)\) When \(0\ln (\sqrt{3} / 2)\).

From \(d A / d t=4-A / 50\) we obtain \(A=200+c e^{-t / 50} .\) If \(A(0)=30\) then \(c=-170\) and \(A=200-170 e^{-t / 50}\).
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