The value of the current, i(t) shown in the given circuit with the assumptions i(0)=0,R=1kΩ and L=1H and using the Laplace transforms.

The value of the current, i(t) is i(t)={0, 0<t<π−40106+1cost+40⋅103106+1sint−40106+1e−1000(t−π), t>π}_.

Definition used:

Laplace transform of derivatives:

“The first and second derivatives of f(t) satisfy

ℒ(f′)=sℒ(f)−f(0) ⋯(1)ℒ(f″)=s2ℒ(f)−sf(0)−f′(0) ⋯(2)

Formula (1) holds if f(t) is continuous for all t≥0 and satisfies the growth restriction (2) and f′(t) is piecewise continuous on every finite interval on the semi axis t≥0. Similarly (2) holds if f(t) and f′(t) are continuous for all t≥0 and satisfy the growth restriction and f″(t) is piecewise continuous on every finite interval on the semi axis t≥0.”

Unit step function or Heaviside Function:

The unit step function or the Heaviside function denoted by u(t−a) is defined as follows.

u(t−a)={0, if t<a1, if t>a}

Second shifting theorem

If f(t) has the transform F(s), then the shifted function

f˜(t)=f(t−a)u(t−a)={0 if t<af(t−a) if t>a}

has the transform e−asF(s). That is, if ℒ(f(t))=F(s) then,

ℒ(f(t−a)u(t−a))=e−asF(s) ⋯(1)

Calculation:

Consider the given conditions as follows.

R=1kΩ and L=1H

Also, the value of voltage is given as shown below.

v={0 if 0<t<π40sint if t>π}

Use the above definition of unit step function and rewrite the given voltage function in terms of unit step function as shown below.

v(t)=40sintu(t−π)

It is known that the differential equation of a voltage of an RL circuit is LdIdt+RI=V(t).

Substitute the values R=1kΩ and L=1H in the equation LdIdt+RI=V(t) as shown below.

LdIdt+RI=V(t)dIdt+1000I=V(t)

Take Laplace transform on the both sides of the above equation; apply definition of Laplace transform of derivatives as shown below.

ℒ(dIdt+1000I)=ℒ(V(t))sℒ(I)+1000ℒ(I)=ℒ(40sintu(t−π))(s+1000)ℒ(I)=−40e−πsℒ(sin(t+π))(s+1000)ℒ(I)=−40e−πss2+1ℒ(I)=−40e−πs(s2+1)(s+1000)

Now, apply partial fractions on the right hand side of the above equation as follows.

1(s2+1)(s+1000)=As+1000+Bs+Cs2+11(s2+1)(s+1000)=A(s2+1)+(Bs+C)(s+1000)(s2+1)(s+1000)1=A(s2+1)+(Bs+C)(s+1000)1=As2+A+Bs2+1000Bs+Cs+1000C

Equate the coefficient of s2 on both sides of the above equation as follows.

0=A+BA=−B

Equate constant terms on both sides as follows.

1=A+1000C1=−B+1000C ⋯(3)

Equate the coefficient of s on both sides of the above equation as follows.

0=1000B+C ⋯(4)

Now, multiply equation (3) by 1000 and add the equation (4) with the resulting equation as shown below.

1000=−1000B+106C0=1000B+C_1000=(106+1)C

On further simplifications of the above equation the following is obtained.

103=(106+1)C (Divide both sides by 106+1)103106+1=(106+1)(106+1)C103106+1=C

Substitute the value of C in the equation (4) and solve for B as shown below.

0=103B+103106+1−103106+1=103B−1106+1=B

Therefore, the value of A is 1106+1.

Substitute the values of A,B and C in the equation 1(s2+1)(s+1000)=As+1000+Bs+Cs2+1 as follows.

1(s2+1)(s+1000)=As+1000+Bs+Cs2+1=1106+1⋅1s+1000−1106+1⋅ss2+1+103106+1⋅1s2+1

Thus, ℒ(i)=−40e−πs(1106+1⋅1s+1000−1106+1⋅ss2+1+103106+1⋅1s2+1).

Now, take inverse Laplace transform on both sides as follows.

i(t)=−40106+1(ℒ−1(e−πss+1000)−L−1(se−πss2+1)+103L−1(e−πss2+1))=(−40106+1e−1000(t−π)u(t−π)+40106+1cos(t−π)u(t−π)+−40106+1sin(t−π)u(t−π))

Therefore, applying the definition of the Heaviside function the value of i(t) at the intervals 0<t<π and t>π is obtained as follows.

i(t)={0, 0<t<π−40106+1cost+40⋅103106+1sint−40106+1e−1000(t−π), t>π}

Thus, the value of i(t) is i(t)={0, 0<t<π−40106+1cost+40⋅103106+1sint−40106+1e−1000(t−π), t>π}_.

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Chapter 6 Solutions