To sketch the super-current I c sin ϕ (t) as a function of t assuming that I I c is slightly greater than 1 and I I c >> 1 , for a Josephson junction in the over-damped limit β = 0 . To sketch the instantaneous voltage V(t) for both the cases Concept Introduction: The Josephson relation in the over-damped limit β = 0 is given by ϕ ˙ = I I c - sin ϕ . Josephson voltage-phase relation is given by V = ℏ 2e ϕ ˙

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Answer 1

Question

Chapter 4.6, Problem 1E

Interpretation Introduction

Interpretation:

To sketch the super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ , for a Josephson junction in the over-damped limit $β = 0$ . To sketch the instantaneous voltage $V(t)$ for both the cases

Concept Introduction:

The Josephson relation in the over-damped limit $β = 0$ is given by $ϕ˙ = IIc - sin ϕ$ .

Josephson voltage-phase relation is given by $V = ℏ2eϕ˙$

Expert Solution & Answer

Answer to Problem 1E

Solution:

a) The super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ for a Josephson junction in the over-damped limit $β = 0$ is sketched.

b) The instantaneous voltage $V(t)$ for both the cases is sketched.

Explanation of Solution

Josephson junctions are superconducting devices that are capable of generating voltage oscillations of extraordinarily high frequency. It consists of two closely spaced superconductors separated by a weak connection.

The Josephson relation in the over-damped limit $β = 0$ is given by

$ϕ˙ = IIc - sin ϕ$

Here, $Ic$ is the critical current.

Assuming that $IIc$ is slightly greater than $1$

Consider $IIc = 1.1$

$ϕ˙ = 1.1 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 1

Assuming that $IIc >> 1$ .

Consider $IIc = 10$ ,

$ϕ′ = 10 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 2

b) Josephson voltage-phase relation is given by

$V = ℏ2eϕ˙$

Substituting $ϕ˙ = IIc - sin ϕ$ , we get

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 1.1$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 3

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 4

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 10$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 6

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Answer 2

Question

Chapter 4.6, Problem 1E

Interpretation Introduction

Interpretation:

To sketch the super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ , for a Josephson junction in the over-damped limit $β = 0$ . To sketch the instantaneous voltage $V(t)$ for both the cases

Concept Introduction:

The Josephson relation in the over-damped limit $β = 0$ is given by $ϕ˙ = IIc - sin ϕ$ .

Josephson voltage-phase relation is given by $V = ℏ2eϕ˙$

Expert Solution & Answer

Answer to Problem 1E

Solution:

a) The super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ for a Josephson junction in the over-damped limit $β = 0$ is sketched.

b) The instantaneous voltage $V(t)$ for both the cases is sketched.

Explanation of Solution

Josephson junctions are superconducting devices that are capable of generating voltage oscillations of extraordinarily high frequency. It consists of two closely spaced superconductors separated by a weak connection.

The Josephson relation in the over-damped limit $β = 0$ is given by

$ϕ˙ = IIc - sin ϕ$

Here, $Ic$ is the critical current.

Assuming that $IIc$ is slightly greater than $1$

Consider $IIc = 1.1$

$ϕ˙ = 1.1 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 1

Assuming that $IIc >> 1$ .

Consider $IIc = 10$ ,

$ϕ′ = 10 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 2

b) Josephson voltage-phase relation is given by

$V = ℏ2eϕ˙$

Substituting $ϕ˙ = IIc - sin ϕ$ , we get

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 1.1$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 3

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 4

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 10$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 6

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Answer 3

Question

Chapter 4.6, Problem 1E

Interpretation Introduction

Interpretation:

To sketch the super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ , for a Josephson junction in the over-damped limit $β = 0$ . To sketch the instantaneous voltage $V(t)$ for both the cases

Concept Introduction:

The Josephson relation in the over-damped limit $β = 0$ is given by $ϕ˙ = IIc - sin ϕ$ .

Josephson voltage-phase relation is given by $V = ℏ2eϕ˙$

Expert Solution & Answer

Answer to Problem 1E

Solution:

a) The super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ for a Josephson junction in the over-damped limit $β = 0$ is sketched.

b) The instantaneous voltage $V(t)$ for both the cases is sketched.

Explanation of Solution

Josephson junctions are superconducting devices that are capable of generating voltage oscillations of extraordinarily high frequency. It consists of two closely spaced superconductors separated by a weak connection.

The Josephson relation in the over-damped limit $β = 0$ is given by

$ϕ˙ = IIc - sin ϕ$

Here, $Ic$ is the critical current.

Assuming that $IIc$ is slightly greater than $1$

Consider $IIc = 1.1$

$ϕ˙ = 1.1 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 1

Assuming that $IIc >> 1$ .

Consider $IIc = 10$ ,

$ϕ′ = 10 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 2

b) Josephson voltage-phase relation is given by

$V = ℏ2eϕ˙$

Substituting $ϕ˙ = IIc - sin ϕ$ , we get

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 1.1$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 3

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 4

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 10$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 6

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Answer 4

Question

Chapter 4.6, Problem 1E

Interpretation Introduction

Interpretation:

To sketch the super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ , for a Josephson junction in the over-damped limit $β = 0$ . To sketch the instantaneous voltage $V(t)$ for both the cases

Concept Introduction:

The Josephson relation in the over-damped limit $β = 0$ is given by $ϕ˙ = IIc - sin ϕ$ .

Josephson voltage-phase relation is given by $V = ℏ2eϕ˙$

Expert Solution & Answer

Answer to Problem 1E

Solution:

a) The super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ for a Josephson junction in the over-damped limit $β = 0$ is sketched.

b) The instantaneous voltage $V(t)$ for both the cases is sketched.

Explanation of Solution

Josephson junctions are superconducting devices that are capable of generating voltage oscillations of extraordinarily high frequency. It consists of two closely spaced superconductors separated by a weak connection.

The Josephson relation in the over-damped limit $β = 0$ is given by

$ϕ˙ = IIc - sin ϕ$

Here, $Ic$ is the critical current.

Assuming that $IIc$ is slightly greater than $1$

Consider $IIc = 1.1$

$ϕ˙ = 1.1 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 1

Assuming that $IIc >> 1$ .

Consider $IIc = 10$ ,

$ϕ′ = 10 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 2

b) Josephson voltage-phase relation is given by

$V = ℏ2eϕ˙$

Substituting $ϕ˙ = IIc - sin ϕ$ , we get

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 1.1$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 3

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 4

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 10$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 6

Want to see more full solutions like this?

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Answer 5

Chapter 4.6, Problem 1E

Interpretation Introduction

Interpretation:

To sketch the super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ , for a Josephson junction in the over-damped limit $β = 0$ . To sketch the instantaneous voltage $V(t)$ for both the cases

Concept Introduction:

The Josephson relation in the over-damped limit $β = 0$ is given by $ϕ˙ = IIc - sin ϕ$ .

Josephson voltage-phase relation is given by $V = ℏ2eϕ˙$

Expert Solution & Answer

Answer to Problem 1E

Solution:

a) The super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ for a Josephson junction in the over-damped limit $β = 0$ is sketched.

b) The instantaneous voltage $V(t)$ for both the cases is sketched.

Explanation of Solution

Josephson junctions are superconducting devices that are capable of generating voltage oscillations of extraordinarily high frequency. It consists of two closely spaced superconductors separated by a weak connection.

The Josephson relation in the over-damped limit $β = 0$ is given by

$ϕ˙ = IIc - sin ϕ$

Here, $Ic$ is the critical current.

Assuming that $IIc$ is slightly greater than $1$

Consider $IIc = 1.1$

$ϕ˙ = 1.1 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 1

Assuming that $IIc >> 1$ .

Consider $IIc = 10$ ,

$ϕ′ = 10 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 2

b) Josephson voltage-phase relation is given by

$V = ℏ2eϕ˙$

Substituting $ϕ˙ = IIc - sin ϕ$ , we get

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 1.1$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 3

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 4

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 10$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 6

Answer 6

Chapter 4.6, Problem 1E

Interpretation Introduction

Interpretation:

To sketch the super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ , for a Josephson junction in the over-damped limit $β = 0$ . To sketch the instantaneous voltage $V(t)$ for both the cases

Concept Introduction:

The Josephson relation in the over-damped limit $β = 0$ is given by $ϕ˙ = IIc - sin ϕ$ .

Josephson voltage-phase relation is given by $V = ℏ2eϕ˙$

Expert Solution & Answer

Answer to Problem 1E

Solution:

a) The super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ for a Josephson junction in the over-damped limit $β = 0$ is sketched.

b) The instantaneous voltage $V(t)$ for both the cases is sketched.

Explanation of Solution

Josephson junctions are superconducting devices that are capable of generating voltage oscillations of extraordinarily high frequency. It consists of two closely spaced superconductors separated by a weak connection.

The Josephson relation in the over-damped limit $β = 0$ is given by

$ϕ˙ = IIc - sin ϕ$

Here, $Ic$ is the critical current.

Assuming that $IIc$ is slightly greater than $1$

Consider $IIc = 1.1$

$ϕ˙ = 1.1 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 1

Assuming that $IIc >> 1$ .

Consider $IIc = 10$ ,

$ϕ′ = 10 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 2

b) Josephson voltage-phase relation is given by

$V = ℏ2eϕ˙$

Substituting $ϕ˙ = IIc - sin ϕ$ , we get

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 1.1$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 3

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 4

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 10$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 6

Answer 7

Chapter 4.6, Problem 1E

Interpretation Introduction

Interpretation:

To sketch the super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ , for a Josephson junction in the over-damped limit $β = 0$ . To sketch the instantaneous voltage $V(t)$ for both the cases

Concept Introduction:

The Josephson relation in the over-damped limit $β = 0$ is given by $ϕ˙ = IIc - sin ϕ$ .

Josephson voltage-phase relation is given by $V = ℏ2eϕ˙$

Answer 8

Expert Solution & Answer

Answer to Problem 1E

Solution:

a) The super-current $Ic sin ϕ (t)$ as a function of $t$ assuming that $IIc$ is slightly greater than $1$ and $IIc >> 1$ for a Josephson junction in the over-damped limit $β = 0$ is sketched.

b) The instantaneous voltage $V(t)$ for both the cases is sketched.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Josephson junctions are superconducting devices that are capable of generating voltage oscillations of extraordinarily high frequency. It consists of two closely spaced superconductors separated by a weak connection.

The Josephson relation in the over-damped limit $β = 0$ is given by

$ϕ˙ = IIc - sin ϕ$

Here, $Ic$ is the critical current.

Assuming that $IIc$ is slightly greater than $1$

Consider $IIc = 1.1$

$ϕ˙ = 1.1 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 1

Assuming that $IIc >> 1$ .

Consider $IIc = 10$ ,

$ϕ′ = 10 - sin ϕ$

The graph of $ϕ$ vs. $t$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 2

b) Josephson voltage-phase relation is given by

$V = ℏ2eϕ˙$

Substituting $ϕ˙ = IIc - sin ϕ$ , we get

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 1.1$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 3

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 4

$V = ℏ2e(IIc - sin ϕ)$

Consider $IIc = 10$

The graph of voltage vs. $ϕ$ is shown below.

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 4.6, Problem 1E , additional homework tip 6

Answer 12

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.2 - Prob. 1E

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Answer to Problem 1E

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