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Find (in terms ofLandC) the frequency of electrical oscillations in a series circuit (FigureifR=0orV=0, butI≠0. (When you tune a radio, you are adjustingCand/orLto make this frequency equal to that of the radio station.)

Short Answer

Expert verified

The equation that describes the motion of the current is.I=α1e−iӬt+α2eiӬt

The frequency of the electrical oscillations in such circuit is.Ó¬=1LC

Step by step solution

01

Given information

Given

R=0V=0I≠0

02

Auxiliary equation

Auxiliary equation:

Auxiliary equation is an algebraic equation of degreenupon which depends the solution of a givennth-order differential equation or difference equation. The auxiliary equation can only be formed when the differential or difference equation is linear and homogeneous, and has constant coefficients. Such a differential equation, withyas the dependent variable, superscript () denotingth-derivative, andan,an−1,…,a1,a0as constants,

any(n)+an−1y(n−1)+⋯+a1y'+a0y=0

will have a characteristic equation of the form

anrn+an−1rn−1+⋯+a1r+a0=0

whose solutionsr1,r2,…,rnare the roots from which the general solution can be formed. A linear difference equation of the form

yt+n=b1yt+n−1+⋯+bnyt

has characteristic equation

rn−b1rn−1−⋯−bn=0

03

Order of differential equation

When R=V=0

Ld2Idt2+IC=0d2Idt2+LCI=0

which is a second order differential equation.

04

Solve equation by using auxiliary equation

solve this differential equation starting by write the auxiliary equation

(D2+Ó¬2)I=0

Here Dis the differential operator, and. Ó¬2=1/LCDifferential equation is, start by find the roots of the auxiliary equation

(D+iӬ)(D−iӬ)=0I=0D=±iӬ

05

The general solution in other forms

Then solve the simpler equations (D+iӬ)I=0and (D−iӬ)I=0,

dIdt=−iӬ I=α1e−iӬtdIdt=iӬI=α2eiӬt

and the linear combination of these two solutions is the general solution for our differential equation

I=α1e−iӬt+α2eiӬt

Notice that,write the general solution in other forms. Conclude that the frequency of the electrical oscillations in such circuit isÓ¬=1LC

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Most popular questions from this chapter

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(a) Show that D(eaxy)=eax(D+a)y,D2(eaxy)=eax(D+a)2y, and so on; that is, for any positive integral n, Dn(eaxy)=eax(D+a)ny.

Thus, show that ifis any polynomial in the operator D, then L(D)(eaxy)=eaxL(D+a)y.

This is called the exponential shift.

(b) Use to show that (D-1)3(exy)=exD3y,(D2+D-6)(e-3xy)=e-3x(D2-5D)y..

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This is called the inverse exponential shift.

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(D2-D-6)y=10×e2x; multiplying both sides by e-3xand using (c), we get

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