Question: The circuit shown in Figure Q3 contains resistor, inductor and capacitor components. The ODE of the circuit is LI”+RI’+1cI=E'(t). Assuming

The circuit shown in Figure Q3

contains resistor, inductor and capacitor components. The ODE of

the circuit is LI”+RI’+1cI=E'(t).

Assuming R=4Ω,L=0.1H,C=0.025F,E=10sin10tV,

the initial current and charge are zero.

risuitus(

a)

Find the general solution for the homogeneous part of the ODE.(

b)

Solve the ODE by considering the initial conditions.

Sure, here are the solutions for parts (a) and (b) of the problem:

(a) Homogeneous solution

The homogeneous part of the ODE is:

LI′′+RI′+1cI=0

The characteristic equation for this ODE is:

Lr2+Rr+1c=0

Substituting the given values of R, L, and C, we get:

0.1r2+4r+40=0

Using the quadratic formula, we can solve for the roots of the characteristic equation:

r=2(0.1)−4±42−4(0.1)(40)

​​

r=−20±2199

Therefore, the general solution for the homogeneous part of the ODE is:

Ih​(t)=A1​e−20t+A2​e(−20+2199​)t

(b) Particular solution

The particular solution for the ODE can be found using the method of undetermined coefficients. Since the forcing term E(t) is of the form sin(10t), we assume a particular solution of the form:

Ip​(t)=Bsin(10t)+Ccos(10t)

Substituting this particular solution into the ODE, we get:

−100BLsin(10t)−100CLcos(10t)+4Bsin(10t)+4Ccos(10t)+40Bsin(10t)+40Ccos(10t)=100cos(10t)

Matching coefficients, we get the following system of equations:

−100BL+4B+40B=0 −100CL+4C+40C=100

Solving for B and C, we get:

B=1001​ C=401​

Therefore, the particular solution is:

Ip​(t)=1001​sin(10t)+401​cos(10t)

General solution

The general solution for the ODE is the sum of the homogeneous solution and the particular solution:

I(t)=Ih​(t)+Ip​(t)

I(t)=A1​e−20t+A2​e(−20+2199​)t+1001​sin(10t)+401​cos(10t)

Initial conditions

The initial conditions are I(0) = 0 and I'(0) = 0. Substituting these conditions into the general solution, we get the following system of equations:

A1​+A2​+401​=0 −20A1​+2(−20+2199​)A2​+101​=0

Solving for A_1 and A_2, we get:

A1​=201​ A2​=80−4199​−1​

Therefore, the complete solution for the ODE is:

$$I(t)=\frac{1}{20}e^{-20t}+\frac{-1}{80-4

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