To show that i = i f ( 1 − e − t / T L ) is a correct solution for − i R − L d i d t + ε = 0 .

Question

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Answer 1

Question

Chapter 20, Problem 57E

To determine

To show that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Expert Solution & Answer

Answer to Problem 57E

It is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Explanation of Solution

Write the equation of the circuit.

−iR−Ldidt+ε=0 (I)

Here, i is the current, R is the resistance, L is the inductance, di1dt is the rate of change of current and ε is the EMF.

Write the given solution for the current.

i=if(1−e−t/TL) (II)

Here, if is the final current and TL is the time constant.

Take the time derivative of equation (II).

didt=difdt+ddt(−ife−t/TL)=0+(−if)e−t/TL(−1TL)=ifTLe−t/TL (III)

Write the equation for the final current.

if=εR (IV)

Put equation (IV) in equation (II).

i=εR(1−e−t/TL) (V)

Put equation (IV) in equation (III).

didt=(εR)TLe−t/TL=εRTLe−t/TL (VI)

Put equations (V) and (VI) in equation (I).

−εR(1−e−t/TL)R−LεRTLe−t/TL+ε=0−ε+εe−t/TL−LεRTLe−t/TL+ε=0εe−t/TL−LεRTLe−t/TL=0 (VII)

Write the equation for the time constant.

TL=LR

Put the above equation in the second term on the left hand side of equation (VII).

εe−t/TL−LεR(LR)e−t/TL=0εe−t/TL−LεRe−t/TLRL=0εe−t/TL−εe−t/TL=0

The left hand side of the equation is equal to the right hand side so that the solution is correct.

Conclusion:

Therefore, it is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

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A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).

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simple diagram to illustrate the setup for each law- coulombs law and biot savart law

Answer 2

Question

Chapter 20, Problem 57E

To determine

To show that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Expert Solution & Answer

Answer to Problem 57E

It is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Explanation of Solution

Write the equation of the circuit.

−iR−Ldidt+ε=0 (I)

Here, i is the current, R is the resistance, L is the inductance, di1dt is the rate of change of current and ε is the EMF.

Write the given solution for the current.

i=if(1−e−t/TL) (II)

Here, if is the final current and TL is the time constant.

Take the time derivative of equation (II).

didt=difdt+ddt(−ife−t/TL)=0+(−if)e−t/TL(−1TL)=ifTLe−t/TL (III)

Write the equation for the final current.

if=εR (IV)

Put equation (IV) in equation (II).

i=εR(1−e−t/TL) (V)

Put equation (IV) in equation (III).

didt=(εR)TLe−t/TL=εRTLe−t/TL (VI)

Put equations (V) and (VI) in equation (I).

−εR(1−e−t/TL)R−LεRTLe−t/TL+ε=0−ε+εe−t/TL−LεRTLe−t/TL+ε=0εe−t/TL−LεRTLe−t/TL=0 (VII)

Write the equation for the time constant.

TL=LR

Put the above equation in the second term on the left hand side of equation (VII).

εe−t/TL−LεR(LR)e−t/TL=0εe−t/TL−LεRe−t/TLRL=0εe−t/TL−εe−t/TL=0

The left hand side of the equation is equal to the right hand side so that the solution is correct.

Conclusion:

Therefore, it is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

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Answer 3

Question

Chapter 20, Problem 57E

To determine

To show that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Expert Solution & Answer

Answer to Problem 57E

It is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Explanation of Solution

Write the equation of the circuit.

−iR−Ldidt+ε=0 (I)

Here, i is the current, R is the resistance, L is the inductance, di1dt is the rate of change of current and ε is the EMF.

Write the given solution for the current.

i=if(1−e−t/TL) (II)

Here, if is the final current and TL is the time constant.

Take the time derivative of equation (II).

didt=difdt+ddt(−ife−t/TL)=0+(−if)e−t/TL(−1TL)=ifTLe−t/TL (III)

Write the equation for the final current.

if=εR (IV)

Put equation (IV) in equation (II).

i=εR(1−e−t/TL) (V)

Put equation (IV) in equation (III).

didt=(εR)TLe−t/TL=εRTLe−t/TL (VI)

Put equations (V) and (VI) in equation (I).

−εR(1−e−t/TL)R−LεRTLe−t/TL+ε=0−ε+εe−t/TL−LεRTLe−t/TL+ε=0εe−t/TL−LεRTLe−t/TL=0 (VII)

Write the equation for the time constant.

TL=LR

Put the above equation in the second term on the left hand side of equation (VII).

εe−t/TL−LεR(LR)e−t/TL=0εe−t/TL−LεRe−t/TLRL=0εe−t/TL−εe−t/TL=0

The left hand side of the equation is equal to the right hand side so that the solution is correct.

Conclusion:

Therefore, it is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

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Answer 4

Question

Chapter 20, Problem 57E

To determine

To show that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Expert Solution & Answer

Answer to Problem 57E

It is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Explanation of Solution

Write the equation of the circuit.

−iR−Ldidt+ε=0 (I)

Here, i is the current, R is the resistance, L is the inductance, di1dt is the rate of change of current and ε is the EMF.

Write the given solution for the current.

i=if(1−e−t/TL) (II)

Here, if is the final current and TL is the time constant.

Take the time derivative of equation (II).

didt=difdt+ddt(−ife−t/TL)=0+(−if)e−t/TL(−1TL)=ifTLe−t/TL (III)

Write the equation for the final current.

if=εR (IV)

Put equation (IV) in equation (II).

i=εR(1−e−t/TL) (V)

Put equation (IV) in equation (III).

didt=(εR)TLe−t/TL=εRTLe−t/TL (VI)

Put equations (V) and (VI) in equation (I).

−εR(1−e−t/TL)R−LεRTLe−t/TL+ε=0−ε+εe−t/TL−LεRTLe−t/TL+ε=0εe−t/TL−LεRTLe−t/TL=0 (VII)

Write the equation for the time constant.

TL=LR

Put the above equation in the second term on the left hand side of equation (VII).

εe−t/TL−LεR(LR)e−t/TL=0εe−t/TL−LεRe−t/TLRL=0εe−t/TL−εe−t/TL=0

The left hand side of the equation is equal to the right hand side so that the solution is correct.

Conclusion:

Therefore, it is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

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Answer 5

Chapter 20, Problem 57E

To determine

To show that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Expert Solution & Answer

Answer to Problem 57E

It is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Explanation of Solution

Write the equation of the circuit.

−iR−Ldidt+ε=0 (I)

Here, i is the current, R is the resistance, L is the inductance, di1dt is the rate of change of current and ε is the EMF.

Write the given solution for the current.

i=if(1−e−t/TL) (II)

Here, if is the final current and TL is the time constant.

Take the time derivative of equation (II).

didt=difdt+ddt(−ife−t/TL)=0+(−if)e−t/TL(−1TL)=ifTLe−t/TL (III)

Write the equation for the final current.

if=εR (IV)

Put equation (IV) in equation (II).

i=εR(1−e−t/TL) (V)

Put equation (IV) in equation (III).

didt=(εR)TLe−t/TL=εRTLe−t/TL (VI)

Put equations (V) and (VI) in equation (I).

−εR(1−e−t/TL)R−LεRTLe−t/TL+ε=0−ε+εe−t/TL−LεRTLe−t/TL+ε=0εe−t/TL−LεRTLe−t/TL=0 (VII)

Write the equation for the time constant.

TL=LR

Put the above equation in the second term on the left hand side of equation (VII).

εe−t/TL−LεR(LR)e−t/TL=0εe−t/TL−LεRe−t/TLRL=0εe−t/TL−εe−t/TL=0

The left hand side of the equation is equal to the right hand side so that the solution is correct.

Conclusion:

Therefore, it is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Answer 6

Chapter 20, Problem 57E

To determine

To show that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Expert Solution & Answer

Answer to Problem 57E

It is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Explanation of Solution

Write the equation of the circuit.

−iR−Ldidt+ε=0 (I)

Here, i is the current, R is the resistance, L is the inductance, di1dt is the rate of change of current and ε is the EMF.

Write the given solution for the current.

i=if(1−e−t/TL) (II)

Here, if is the final current and TL is the time constant.

Take the time derivative of equation (II).

didt=difdt+ddt(−ife−t/TL)=0+(−if)e−t/TL(−1TL)=ifTLe−t/TL (III)

Write the equation for the final current.

if=εR (IV)

Put equation (IV) in equation (II).

i=εR(1−e−t/TL) (V)

Put equation (IV) in equation (III).

didt=(εR)TLe−t/TL=εRTLe−t/TL (VI)

Put equations (V) and (VI) in equation (I).

−εR(1−e−t/TL)R−LεRTLe−t/TL+ε=0−ε+εe−t/TL−LεRTLe−t/TL+ε=0εe−t/TL−LεRTLe−t/TL=0 (VII)

Write the equation for the time constant.

TL=LR

Put the above equation in the second term on the left hand side of equation (VII).

εe−t/TL−LεR(LR)e−t/TL=0εe−t/TL−LεRe−t/TLRL=0εe−t/TL−εe−t/TL=0

The left hand side of the equation is equal to the right hand side so that the solution is correct.

Conclusion:

Therefore, it is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Answer 7

Chapter 20, Problem 57E

To determine

To show that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Answer 8

Expert Solution & Answer

Answer to Problem 57E

It is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write the equation of the circuit.

−iR−Ldidt+ε=0 (I)

Here, i is the current, R is the resistance, L is the inductance, di1dt is the rate of change of current and ε is the EMF.

Write the given solution for the current.

i=if(1−e−t/TL) (II)

Here, if is the final current and TL is the time constant.

Take the time derivative of equation (II).

didt=difdt+ddt(−ife−t/TL)=0+(−if)e−t/TL(−1TL)=ifTLe−t/TL (III)

Write the equation for the final current.

if=εR (IV)

Put equation (IV) in equation (II).

i=εR(1−e−t/TL) (V)

Put equation (IV) in equation (III).

didt=(εR)TLe−t/TL=εRTLe−t/TL (VI)

Put equations (V) and (VI) in equation (I).

−εR(1−e−t/TL)R−LεRTLe−t/TL+ε=0−ε+εe−t/TL−LεRTLe−t/TL+ε=0εe−t/TL−LεRTLe−t/TL=0 (VII)

Write the equation for the time constant.

TL=LR

Put the above equation in the second term on the left hand side of equation (VII).

εe−t/TL−LεR(LR)e−t/TL=0εe−t/TL−LεRe−t/TLRL=0εe−t/TL−εe−t/TL=0

The left hand side of the equation is equal to the right hand side so that the solution is correct.

Conclusion:

Therefore, it is showed that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Answer 12

Ch. 20 - Prob. 1RQ Ch. 20 - Prob. 2RQ Ch. 20 - Prob. 3RQ Ch. 20 - Prob. 4RQ Ch. 20 - Prob. 5RQ Ch. 20 - Prob. 6RQ Ch. 20 - Prob. 7RQ Ch. 20 - Prob. 8RQ Ch. 20 - Prob. 9RQ Ch. 20 - Prob. 10RQ

To show that i = i f ( 1 − e − t / T L ) is a correct solution for − i R − L d i d t + ε = 0 .

Videos

To show that i=if(1−e−t/TL) is a correct solution for −iR−Ldidt+ε=0 .

Answer to Problem 57E

Explanation of Solution

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Chapter 20 Solutions