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Chapter 21: Problem 520

At time \(t=0\) a unit charge is placed on the capacitor of a time-varying RC circuit. From experiments with the circuit it is observed that the resistance \(\mathrm{R}\) varies such that the coefficient \(1 / \mathrm{RC}\) in $$ y^{\prime}+(1 / R C) y=0 $$ oscillates about an average value, which we take to be unity. The capacitance \(\mathrm{C}\) is constant. We further assume that the oscillation of \(1 / \mathrm{RC}\) about unity allows us to write $$ (1 / R C) \equiv q(t)=1+b \cos t $$ where \(b\) is a real number such that \(|b|<1\).

Short Answer

Expert verified
The solution to the given time-varying RC circuit is \(y(t) = e^{-t - b\sin t}\), where \(q(t) = 1+b \cos t\), with |b|<1. This solution is obtained by finding the integrating factor and solving the first-order linear differential equation.

Step by step solution

01

Rewrite the given equation

First, rewrite the given equation as follows: \(y'(t) + q(t)y(t) = 0\) where \(q(t) = 1+b \cos t\).
02

Find an integrating factor

To solve this first-order linear differential equation, we will find an integrating factor. An integrating factor (IF) can be found by taking the exponential of the integral of the function multiplied by dt: IF = \(e^{\int q(t) dt} = e^{\int (1+b \cos t) dt}\)
03

Evaluate the integral for the integrating factor

Evaluate the integral to find the integrating factor: IF = \(e^{\int (1+b \cos t) dt} = e^{t + b \sin t}\)
04

Multiply the given equation by the integrating factor

Multiply the given equation on both sides by the integrating factor to get: \(e^{t + b \sin t}y'(t) + (1+b \cos t)e^{t + b \sin t}y(t) = 0\)
05

Observe that the left side is the derivative of a product

Notice that the left side of the equation is the derivative of a product: \(\frac{d}{dt}(e^{t + b\sin t}y(t)) = 0\)
06

Integrate both sides with respect to t

Integrate both sides of the equation with respect to t: \(\int \frac{d}{dt}(e^{t+b\sin t} y(t)) dt = \int 0 dt\) On integrating, we get: \(e^{t+b\sin t}y(t) = C\), where C is the integration constant.
07

Solve for y(t)

Finally, solve for \(y(t)\): \(y(t) = Ce^{-t - b\sin t}\) At time t=0, a unit charge is placed on the capacitor, which means \(y(0) = 1\). Let's use this information to find the value of the integration constant C: \(y(0) = Ce^{-0 - b\sin 0} = Ce^0 = C\) So, C = 1. Therefore, the solution for this problem is: \(y(t) = e^{-t - b\sin t}\)

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Key Concepts

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

First-Order Linear Differential Equation
Understanding the behavior of a time-varying RC circuit requires us to delve into the principles of differential equations, specifically first-order linear differential equations. These types of equations are foundational to several areas of engineering and physical sciences.

In our exercise, we're given an equation of the form:
\[ y'(t) + q(t)y(t) = 0 \]
This represents a first-order linear differential equation where the derivative of the variable \( y(t) \) is related to itself via a function \( q(t) \).

These equations are \

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Most popular questions from this chapter

An RLC-circuit has an inductance of 1 henry, a resistance of 40 ohms, and a capacitance of \(1 / 20000\) farad. The impressed emf is \(E=120\) cos \(150 \mathrm{t}\). Find the current flawing in the circuit.

Solve the RLC-circuit for the current \(\mathrm{i}=\mathrm{i}(\mathrm{t})\) given an arbitrary applied voltage \(E=E(t)\) using the superposition principle.

Find the current in the simple RL-circuit: \(\mathrm{R}\) is the constant resistance, \(\mathrm{L}\) is the constant inductance and \(E=E_{0} \sin \omega t\) is the impressed voltage.

Solve the initial value problem $$ y^{\prime}-k y=A, y(0)=1 $$ where \(\mathrm{k}\) and \(\mathrm{A}\) are given constants. Determine the transient and steady state responses.

An RLC-circuit has resistance 10 ohms, capacitance \(10^{-2}\) farads and inductance (1/2) henry. The impressed voltage at \(\mathrm{t}=0\) is 12 volts. Assuming no initial current or charge on the capacitor, find the charge \(q=q(t)\) on the capacitor.
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