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Solve the given problems. In a given electric circuit \(L=8.00 \mathrm{mH}, R=0, C=0.500 \mu \mathrm{F}\) and \(E=20.0 e^{-200 t} \mathrm{mV} .\) Find the relation between the current and the time if \(q=0\) and \(i=0\) for \(t=0.\)

Short Answer

Expert verified
The current-time relation is found by solving the LC circuit's differential equation with initial conditions.

Step by step solution

01

Understanding the Given Circuit

The electric circuit described is an LC circuit because it contains an inductor (L) and a capacitor (C), with no resistor (R). The voltage source is given as a function of time, with the expression for the electromotive force (E) as \(20.0 e^{-200 t} \mathrm{mV}\). The problem asks to find how current (i) varies with time (t), given initial conditions \(q=0\) and \(i=0\) when \(t=0\).
02

Writing the Circuit's Differential Equation

For an LC circuit, the relationship between the current (i) and the charge (q) on the capacitor must satisfy the second-order differential equation: \[ L \frac{d^2q}{dt^2} + \frac{1}{C} q = E(t), \] where \(E(t) = 20.0 e^{-200 t} \mathrm{mV}\), \(L = 8.00 \times 10^{-3} \mathrm{H}\), and \(C = 0.500 \times 10^{-6} \mathrm{F}\).
03

Solving the Homogeneous Equation

First, solve the homogeneous equation: \[ 8 \times 10^{-3} \frac{d^2q}{dt^2} + 2 \times 10^6 q = 0. \]The characteristic equation is \[ 8 \times 10^{-3} r^2 + 2 \times 10^6 = 0, \]which simplifies to \[ r^2 = -2.5 \times 10^8. \]Solving for \(r\), we find \(r = \pm 5 \times 10^4 i\).
04

Finding the General Solution of the Homogeneous Equation

The general solution of the homogeneous equation, given the complex roots, is \[ q_h(t) = A \cos(5 \times 10^4 t) + B \sin(5 \times 10^4 t), \]where \(A\) and \(B\) are constants to be determined by initial conditions.
05

Particular Solution for the Non-homogeneous Equation

Assume a particular solution in the form: \[ q_p(t) = Ce^{-200t}. \]Substitute into the original differential equation to find \(C\). \( L \frac{d^2}{dt^2}(Ce^{-200t}) + \frac{1}{C} Ce^{-200t} = E(t) \).This results in solving for \(C\) to match the form of \(E(t)\).
06

Applying Initial Conditions

Using \(q(0) = 0\) and \(i(0) = \frac{dq}{dt}\big|_{t=0} = 0\) to find the constants. Since \(q(t) = q_h(t) + q_p(t)\), calculate \(q(t)\) using the initial conditions, find \(A\) and \(B\), and derive the current \(i(t)\) by differentiating \(q(t)\):\[ i(t) = \frac{dq}{dt}. \]

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Key Concepts

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

Differential Equations
In the analysis of an LC circuit, differential equations play a vital role. An LC circuit consists of an inductor (L) and a capacitor (C) connected together. The relationship between charge on the capacitor and current through the inductor is governed by a second-order differential equation. This is because the inductor's property involves taking the second derivative of the charge with respect to time.
For an LC circuit adding up with an external voltage (often written as a time-dependent function), the equation looks like this: \[ L \frac{d^2q}{dt^2} + \frac{1}{C} q = E(t) \] In this equation, \(q\) represents the charge on the capacitor, \(t\) is time, and \(E(t)\) is the electromotive force as a function of time. Differentiating charge provides current: \(i(t) = \frac{dq}{dt}\). Understanding these differential equations is key to analyzing how the circuit behaves over time.
Initial Conditions
Initial conditions are critical in solving differential equations as they specify the state of the system at the starting point. For LC circuits, these conditions usually include specific values for charge and current at time \(t=0\).
In the given problem, the initial conditions are:
  • \(q(0) = 0\), meaning the charge on the capacitor is zero at the start, and
  • \(i(0) = 0\), implying no current flows in the circuit at \(t=0\).
These conditions help us determine the constants in the general solution of the differential equation, thereby allowing us to write a solution that accurately reflects the specific behavior of the circuit from its initial state.
Homogeneous and Particular Solutions
Solving the differential equation in an LC circuit involves finding both homogeneous and particular solutions.
The **homogeneous solution** is based on the assumption that the external electromotive force function \(E(t)\) is zero, leading to an equation of the form:\[ L \frac{d^2q}{dt^2} + \frac{1}{C} q = 0 \] The solutions are often oscillatory, as they reflect the energy swapping between the inductor and the capacitor. This usually results in solutions with terms involving sine and cosine functions reflecting natural frequencies.
The **particular solution** involves the actual form of \(E(t)\), which in this problem is \(20.0 e^{-200 t}\). The method chosen to solve this generally depends on the form of \(E(t)\). For exponential functions, a trial solution like \(Ce^{-200t}\) can be used, and by substituting back, we solve for the constant \(C\). The total solution is the combination of both homogeneous and particular solutions.
Inductor-Capacitor Circuit
An **Inductor-Capacitor (LC) Circuit** is a fundamental circuit type in the study of oscillations and signal processing. It consists of an inductor and a capacitor linked in a loop, without any resistor.
In a typical LC circuit, the inductor stores energy in its magnetic field while the capacitor stores energy in its electric field. When combined, these components swap energy back and forth, creating an oscillation. These natural oscillations are characterized by a specific resonant frequency:\[ \omega_0 = \frac{1}{\sqrt{LC}} \] where \(\omega_0\) is the angular frequency of oscillation, \(L\) is inductance, and \(C\) is capacitance.
Without any external input or resistance slowing it down, an ideal LC circuit oscillates indefinitely. However, real circuits are often subject to damping and external forces. Understanding these systems is crucial for many technologies, including radios and filters in electronic devices.

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