The solution of one loop circuit with L = 1 H , R = 5 3 Ω , C = 3 2 F , Q = 10 − 6 C, I ( 0 ) = 0 A and E ( t ) = 0 V .

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Answer 1

Question

Chapter 7, Problem 1RE

(a)

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=0 V.

(a)

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=1500000(−32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

Explanation of Solution

Result used:

From Kirchhoff’s voltage law, RI+LdIdt+1CQ−E(t)=0.

From the above equation, the system of differential equations {dQdt=IdIdt=−1LCQ−RLI+1LE(t).

The initial condition Q(0)=Q0 and I(0)=I0 on charge and current, respectively.

Here, I(t)=current(dQ(t)dt), Q(t)=charge, R is resistance, C is capacitance, E is voltage and L is inductance.

Calculation:

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=0 V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11(0)

Simplify the system of equations as follows.

{dQdt=IdIdt=−23Q−53I

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(00)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further as,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Note that,

X(0)=(Q(0)I(0))=(10−60)

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(00)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+0}=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)

Simplify further as,

X(t)=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)=((3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)=(1500000(32e−23t−e−t)1500000(−e−23t+e−t))

The solution of one-loop L-R-C circuit is, Q(t)=1500000(32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

b.

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=e−t V.

b.

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

Explanation of Solution

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=e−t V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11e−t

On further simplification.

{dQdt=IdIdt=−23Q−53I+e−t

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(0e−t)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Here, the matrix X(0)=(10−60).

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(0e−u)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−53u3)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)}

Simplify further as,

X(t)=(−3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)=((−3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)+(−95e−23t(e−53t−1)−3te−t−15e−23t(6e−53t−6)−3te−t)=((−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t)

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

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Answer 2

Question

Chapter 7, Problem 1RE

(a)

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=0 V.

(a)

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=1500000(−32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

Explanation of Solution

Result used:

From Kirchhoff’s voltage law, RI+LdIdt+1CQ−E(t)=0.

From the above equation, the system of differential equations {dQdt=IdIdt=−1LCQ−RLI+1LE(t).

The initial condition Q(0)=Q0 and I(0)=I0 on charge and current, respectively.

Here, I(t)=current(dQ(t)dt), Q(t)=charge, R is resistance, C is capacitance, E is voltage and L is inductance.

Calculation:

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=0 V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11(0)

Simplify the system of equations as follows.

{dQdt=IdIdt=−23Q−53I

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(00)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further as,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Note that,

X(0)=(Q(0)I(0))=(10−60)

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(00)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+0}=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)

Simplify further as,

X(t)=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)=((3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)=(1500000(32e−23t−e−t)1500000(−e−23t+e−t))

The solution of one-loop L-R-C circuit is, Q(t)=1500000(32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

b.

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=e−t V.

b.

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

Explanation of Solution

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=e−t V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11e−t

On further simplification.

{dQdt=IdIdt=−23Q−53I+e−t

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(0e−t)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Here, the matrix X(0)=(10−60).

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(0e−u)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−53u3)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)}

Simplify further as,

X(t)=(−3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)=((−3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)+(−95e−23t(e−53t−1)−3te−t−15e−23t(6e−53t−6)−3te−t)=((−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t)

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

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Answer 3

Question

Chapter 7, Problem 1RE

(a)

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=0 V.

(a)

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=1500000(−32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

Explanation of Solution

Result used:

From Kirchhoff’s voltage law, RI+LdIdt+1CQ−E(t)=0.

From the above equation, the system of differential equations {dQdt=IdIdt=−1LCQ−RLI+1LE(t).

The initial condition Q(0)=Q0 and I(0)=I0 on charge and current, respectively.

Here, I(t)=current(dQ(t)dt), Q(t)=charge, R is resistance, C is capacitance, E is voltage and L is inductance.

Calculation:

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=0 V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11(0)

Simplify the system of equations as follows.

{dQdt=IdIdt=−23Q−53I

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(00)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further as,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Note that,

X(0)=(Q(0)I(0))=(10−60)

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(00)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+0}=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)

Simplify further as,

X(t)=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)=((3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)=(1500000(32e−23t−e−t)1500000(−e−23t+e−t))

The solution of one-loop L-R-C circuit is, Q(t)=1500000(32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

b.

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=e−t V.

b.

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

Explanation of Solution

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=e−t V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11e−t

On further simplification.

{dQdt=IdIdt=−23Q−53I+e−t

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(0e−t)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Here, the matrix X(0)=(10−60).

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(0e−u)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−53u3)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)}

Simplify further as,

X(t)=(−3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)=((−3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)+(−95e−23t(e−53t−1)−3te−t−15e−23t(6e−53t−6)−3te−t)=((−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t)

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

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Answer 4

Question

Chapter 7, Problem 1RE

(a)

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=0 V.

(a)

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=1500000(−32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

Explanation of Solution

Result used:

From Kirchhoff’s voltage law, RI+LdIdt+1CQ−E(t)=0.

From the above equation, the system of differential equations {dQdt=IdIdt=−1LCQ−RLI+1LE(t).

The initial condition Q(0)=Q0 and I(0)=I0 on charge and current, respectively.

Here, I(t)=current(dQ(t)dt), Q(t)=charge, R is resistance, C is capacitance, E is voltage and L is inductance.

Calculation:

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=0 V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11(0)

Simplify the system of equations as follows.

{dQdt=IdIdt=−23Q−53I

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(00)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further as,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Note that,

X(0)=(Q(0)I(0))=(10−60)

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(00)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+0}=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)

Simplify further as,

X(t)=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)=((3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)=(1500000(32e−23t−e−t)1500000(−e−23t+e−t))

The solution of one-loop L-R-C circuit is, Q(t)=1500000(32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

b.

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=e−t V.

b.

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

Explanation of Solution

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=e−t V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11e−t

On further simplification.

{dQdt=IdIdt=−23Q−53I+e−t

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(0e−t)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Here, the matrix X(0)=(10−60).

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(0e−u)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−53u3)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)}

Simplify further as,

X(t)=(−3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)=((−3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)+(−95e−23t(e−53t−1)−3te−t−15e−23t(6e−53t−6)−3te−t)=((−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t)

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

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Answer 5

Chapter 7, Problem 1RE

(a)

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=0 V.

(a)

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=1500000(−32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

Explanation of Solution

Result used:

From Kirchhoff’s voltage law, RI+LdIdt+1CQ−E(t)=0.

From the above equation, the system of differential equations {dQdt=IdIdt=−1LCQ−RLI+1LE(t).

The initial condition Q(0)=Q0 and I(0)=I0 on charge and current, respectively.

Here, I(t)=current(dQ(t)dt), Q(t)=charge, R is resistance, C is capacitance, E is voltage and L is inductance.

Calculation:

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=0 V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11(0)

Simplify the system of equations as follows.

{dQdt=IdIdt=−23Q−53I

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(00)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further as,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Note that,

X(0)=(Q(0)I(0))=(10−60)

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(00)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+0}=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)

Simplify further as,

X(t)=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)=((3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)=(1500000(32e−23t−e−t)1500000(−e−23t+e−t))

The solution of one-loop L-R-C circuit is, Q(t)=1500000(32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

b.

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=e−t V.

b.

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

Explanation of Solution

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=e−t V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11e−t

On further simplification.

{dQdt=IdIdt=−23Q−53I+e−t

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(0e−t)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Here, the matrix X(0)=(10−60).

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(0e−u)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−53u3)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)}

Simplify further as,

X(t)=(−3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)=((−3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)+(−95e−23t(e−53t−1)−3te−t−15e−23t(6e−53t−6)−3te−t)=((−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t)

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

Answer 6

Chapter 7, Problem 1RE

(a)

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=0 V.

(a)

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=1500000(−32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

Explanation of Solution

Result used:

From Kirchhoff’s voltage law, RI+LdIdt+1CQ−E(t)=0.

From the above equation, the system of differential equations {dQdt=IdIdt=−1LCQ−RLI+1LE(t).

The initial condition Q(0)=Q0 and I(0)=I0 on charge and current, respectively.

Here, I(t)=current(dQ(t)dt), Q(t)=charge, R is resistance, C is capacitance, E is voltage and L is inductance.

Calculation:

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=0 V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11(0)

Simplify the system of equations as follows.

{dQdt=IdIdt=−23Q−53I

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(00)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further as,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Note that,

X(0)=(Q(0)I(0))=(10−60)

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(00)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+0}=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)

Simplify further as,

X(t)=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)=((3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)=(1500000(32e−23t−e−t)1500000(−e−23t+e−t))

The solution of one-loop L-R-C circuit is, Q(t)=1500000(32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

Answer 7

Chapter 7, Problem 1RE

(a)

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=0 V.

Answer 8

(a)

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=1500000(−32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Result used:

From Kirchhoff’s voltage law, RI+LdIdt+1CQ−E(t)=0.

From the above equation, the system of differential equations {dQdt=IdIdt=−1LCQ−RLI+1LE(t).

The initial condition Q(0)=Q0 and I(0)=I0 on charge and current, respectively.

Here, I(t)=current(dQ(t)dt), Q(t)=charge, R is resistance, C is capacitance, E is voltage and L is inductance.

Calculation:

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=0 V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11(0)

Simplify the system of equations as follows.

{dQdt=IdIdt=−23Q−53I

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(00)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further as,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Note that,

X(0)=(Q(0)I(0))=(10−60)

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(00)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+0}=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)

Simplify further as,

X(t)=(3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)=((3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)=(1500000(32e−23t−e−t)1500000(−e−23t+e−t))

The solution of one-loop L-R-C circuit is, Q(t)=1500000(32e−23t−e−t) and I(t)=1500000(−e−23t+e−t)_.

Answer 13

b.

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=e−t V.

b.

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

Explanation of Solution

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=e−t V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11e−t

On further simplification.

{dQdt=IdIdt=−23Q−53I+e−t

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(0e−t)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Here, the matrix X(0)=(10−60).

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(0e−u)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−53u3)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)}

Simplify further as,

X(t)=(−3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)=((−3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)+(−95e−23t(e−53t−1)−3te−t−15e−23t(6e−53t−6)−3te−t)=((−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t)

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

Answer 14

b.

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=e−t V.

Answer 15

b.

Expert Solution

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=e−t V.

The initial condition is Q(0)=10−6C and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=−1LCQ−RLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=−1(1)(32)Q−(53)1I+11e−t

On further simplification.

{dQdt=IdIdt=−23Q−53I+e−t

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(01−23−53)(QI)+(0e−t)

Consider the vector [01−23−53], compute eigenvalues of the matrix as follows.

|01−23−53|−λ|1001|=|01−23−53|−|λ00λ|=|−λ1−23−53−λ|=−λ(−53−λ)−(−23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further,

(3λ+2)(λ+1)=0λ=−23 and λ=−1

Thus, the corresponding eigenvectors are v1=(−321) and v2=(−11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(−32e−23t−e−te−23te−t).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|−32e−23t−e−te−23te−t|=(−32e−23t)(e−t)−(−e−t)(e−23t)=−32e−53t+e−53t=−12e−53t

Now compute inverse matrix as follows.

Φ−1(t)=1(−12e−53t)adj(−32e−23t−e−te−23te−t)=−2e53t(e−te−t−e−23t−32e−23t)=(−2e23t−2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du.

Φ−1(0)=(−2e23(0)−2e23(0)2e(0)3e(0))=(−2−223)

Here, the matrix X(0)=(10−60).

X(t)=Φ(t)Φ−1(0)X(0)+Φ(t)∫0tΦ−1(u)F(u)du={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−23u−2e−23u2eu3eu)(0e−u)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)∫0t(−2e−53u3)du}={(−32e−23t−e−te−23te−t)(−2−223)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)}

Simplify further as,

X(t)=(−3e−23t−2e−t3e−23t−3e−t−2e−23t+2e−t−2e−23t+3e−t)(10−60)+(−32e−23t−e−te−23te−t)(65e−53t−653t)=((−3e−23t−2e−t)10−6+0(−2e−23t+2e−t)10−6+0)+(−95e−23t(e−53t−1)−3te−t−15e−23t(6e−53t−6)−3te−t)=((−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t)

The solution of one-loop L-R-C circuit is, Q(t)=(−3e−23t−2e−t)10−6−95e−23t(e−53t−1)−3te−t_ and I(t)=(−2e−23t+2e−t)10−6−15e−23t(6e−53t−6)−3te−t_.

Answer 20

Ch. 7.1 - Prob. 1E Ch. 7.1 - Prob. 2E Ch. 7.1 - Prob. 13E Ch. 7.1 - Prob. 14E Ch. 7.1 - Prob. 15E Ch. 7.1 - Prob. 16E Ch. 7.1 - Prob. 17E Ch. 7.1 - Prob. 18E Ch. 7.1 - Prob. 19E Ch. 7.1 - Prob. 20E

The solution of one loop circuit with L = 1 H , R = 5 3 Ω , C = 3 2 F , Q = 10 − 6 C, I ( 0 ) = 0 A and E ( t ) = 0 V .

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The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=0 V.

Answer to Problem 1RE

Explanation of Solution

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=10−6C, I(0)=0 A and E(t)=e−t V.

Answer to Problem 1RE

Explanation of Solution

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