The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

Question

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Answer 1

Question

Chapter 32, Problem 32.64AP

(a)

To determine

The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

(a)

Expert Solution

Answer to Problem 32.64AP

The inductor behaves as the short circuit because of no resistance.

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The inductor has no resistance. If the switch is closed for a long time, then inductor will reach saturation and voltage passes through the inductor. Hence, it behaves as a short circuit.

Conclusion:

Therefore, the inductor behaves as the short circuit because of no resistance.

(b)

To determine

The current carried by the inductor.

(b)

Expert Solution

Answer to Problem 32.64AP

The current carried by the inductor is

$0.500 A$ .

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The figure of the circuit diagrammed referred from P31.15 is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 1

Figure (1)

The net resistance for parallel combination is,

$1R'=1R+12RR'=2R3$

Here,

$R'$ is the net resistance for parallel combination.

$R$ is the resistance of the circuit.

The net resistance is connected in series with $R$ for parallel combination is,

$Rnet=R+R'$ (1)

Here,

$Rnet$ is the net resistance is connected in series.

Substitute $2R3$ for $R'$ in equation (1).

$Rnet=R+2R3$ (2)

Substitute $4.00 Ω$ for $R$ in equation (2) to find the $Rnet$ .

$Rnet=4.00 Ω+2(4.00 Ω)3=6.67 Ω$

Formula to calculate the current of battery is,

$Ib=εRnet$

Here,

$Ib$ is the battery current.

$ε$ is the induced voltage.

Substitute $10.0 V$ for $ε$ and $6.67 Ω$ for $Rnet$ in equation (3) to find the $Ib$ .

$Ib=10.0 V6.67 Ω=1.499 A≈1.50 A$

The voltage across the parallel combination of resistors is,

$V=ε−IbR$ (6)

Substitute $1.50 A$ for $Ib$ , $10.0 V$ for $ε$ and $4.00 Ω$ for $R$ in equation (6) to find the $V$ .

$V=10.0 V−(1.50 A)(4.00 Ω)=4.00 V$

Formula to calculate the current though the inductor is,

$Ii=V2R$ (7)

Substitute $4.00 V$ for $V$ and $4.00 Ω$ for $R$ in equation (7) to find the $Ii$ .

$Ii=4.00 V2(4.00 Ω)=0.500 A$

Conclusion:

Therefore, the current carried by the inductor for $t<0$ is $0.500 A$ .

(c)

To determine

The energy stored in the inductor.

(c)

Expert Solution

Answer to Problem 32.64AP

The energy stored in the inductor for

$t<0$ is

$0.125 J$ .

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

Formula to calculate the energy stored in the inductor is,

$Ui=12LIi2$

Substitute $1.00 H$ for $L$ and $0.500 A$ for $Ii$ to find the $Ui$ .

$Ui=12(1.00 H)(0.500 A)2=0.125 J$

Thus, the energy stored in the inductor for $t<0$ is $0.125 J$ .

Conclusion:

Therefore, the energy stored in the inductor for $t<0$ is $0.125 J$ .

(d)

To determine

The energy previously stored in the inductor after the switch is opened.

(d)

Expert Solution

Answer to Problem 32.64AP

The energy becomes

$0.125 J$ of the additional internal energy in the

$R$ and

$2R$ . in the middle branch after switch is opened.

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

When switch is opened, the energy stored in the inductor will dissipate through resistor $R$ and $2R$ . Thus, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch.

Conclusion:

Therefore, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch after switch is opened.

(e)

To determine

To draw: The graph of the current in the inductor for $t≥0$ with the initial and final values and the time constant.

(e)

Expert Solution

Answer to Problem 32.64AP

Answer The graph of the current in the inductor for $t≥0$ and label the initial and final values and the time constant.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 2

Explanation of Solution

Introduction:

The graph of the current verses time shows the variation of the current in the circuit with time and tells the nature of the current.

Explanation:

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

After time $t=0$ , the resistance of the circuit will be,

$R1=R+2R=3R$

Formula to calculate the time constant is,

$t=LR1$ (8)

Substitute $3R$ for $R1$ and $1.00 H$ for $L$ to find the $t$ .

$t=1.00 H3R$ (9)

Substitute $4.0 Ω$ for $R$ in equation (9) to find the $t$ .

$t=1.00 H3×4.0 Ω=0.083 s$

The current flowing through the inductor at time $t$ is,

$I=I0e−RLt$

Substitute $0.500 A$ for $I0$ and $12.0 Ω$ for $R1$ and $0.083 s$ for $t$ in equation (10).

$I=(0.500 A)e−(12.0 Ω)1.0 H(0.083 s)$

Thus, the graph of the current (the initial and final values) in the inductor for $t≥0$ and the time constant from equation (11) is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 3

Figure (2)

The graph shows that current decays with exponentially with time constant. The current decreases from $0.500 A$ towards zero because time constant goes to zero so, the final value of the current gets zero.

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Answer 2

Question

Chapter 32, Problem 32.64AP

(a)

To determine

The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

(a)

Expert Solution

Answer to Problem 32.64AP

The inductor behaves as the short circuit because of no resistance.

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The inductor has no resistance. If the switch is closed for a long time, then inductor will reach saturation and voltage passes through the inductor. Hence, it behaves as a short circuit.

Conclusion:

Therefore, the inductor behaves as the short circuit because of no resistance.

(b)

To determine

The current carried by the inductor.

(b)

Expert Solution

Answer to Problem 32.64AP

The current carried by the inductor is

$0.500 A$ .

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The figure of the circuit diagrammed referred from P31.15 is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 1

Figure (1)

The net resistance for parallel combination is,

$1R'=1R+12RR'=2R3$

Here,

$R'$ is the net resistance for parallel combination.

$R$ is the resistance of the circuit.

The net resistance is connected in series with $R$ for parallel combination is,

$Rnet=R+R'$ (1)

Here,

$Rnet$ is the net resistance is connected in series.

Substitute $2R3$ for $R'$ in equation (1).

$Rnet=R+2R3$ (2)

Substitute $4.00 Ω$ for $R$ in equation (2) to find the $Rnet$ .

$Rnet=4.00 Ω+2(4.00 Ω)3=6.67 Ω$

Formula to calculate the current of battery is,

$Ib=εRnet$

Here,

$Ib$ is the battery current.

$ε$ is the induced voltage.

Substitute $10.0 V$ for $ε$ and $6.67 Ω$ for $Rnet$ in equation (3) to find the $Ib$ .

$Ib=10.0 V6.67 Ω=1.499 A≈1.50 A$

The voltage across the parallel combination of resistors is,

$V=ε−IbR$ (6)

Substitute $1.50 A$ for $Ib$ , $10.0 V$ for $ε$ and $4.00 Ω$ for $R$ in equation (6) to find the $V$ .

$V=10.0 V−(1.50 A)(4.00 Ω)=4.00 V$

Formula to calculate the current though the inductor is,

$Ii=V2R$ (7)

Substitute $4.00 V$ for $V$ and $4.00 Ω$ for $R$ in equation (7) to find the $Ii$ .

$Ii=4.00 V2(4.00 Ω)=0.500 A$

Conclusion:

Therefore, the current carried by the inductor for $t<0$ is $0.500 A$ .

(c)

To determine

The energy stored in the inductor.

(c)

Expert Solution

Answer to Problem 32.64AP

The energy stored in the inductor for

$t<0$ is

$0.125 J$ .

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

Formula to calculate the energy stored in the inductor is,

$Ui=12LIi2$

Substitute $1.00 H$ for $L$ and $0.500 A$ for $Ii$ to find the $Ui$ .

$Ui=12(1.00 H)(0.500 A)2=0.125 J$

Thus, the energy stored in the inductor for $t<0$ is $0.125 J$ .

Conclusion:

Therefore, the energy stored in the inductor for $t<0$ is $0.125 J$ .

(d)

To determine

The energy previously stored in the inductor after the switch is opened.

(d)

Expert Solution

Answer to Problem 32.64AP

The energy becomes

$0.125 J$ of the additional internal energy in the

$R$ and

$2R$ . in the middle branch after switch is opened.

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

When switch is opened, the energy stored in the inductor will dissipate through resistor $R$ and $2R$ . Thus, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch.

Conclusion:

Therefore, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch after switch is opened.

(e)

To determine

To draw: The graph of the current in the inductor for $t≥0$ with the initial and final values and the time constant.

(e)

Expert Solution

Answer to Problem 32.64AP

Answer The graph of the current in the inductor for $t≥0$ and label the initial and final values and the time constant.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 2

Explanation of Solution

Introduction:

The graph of the current verses time shows the variation of the current in the circuit with time and tells the nature of the current.

Explanation:

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

After time $t=0$ , the resistance of the circuit will be,

$R1=R+2R=3R$

Formula to calculate the time constant is,

$t=LR1$ (8)

Substitute $3R$ for $R1$ and $1.00 H$ for $L$ to find the $t$ .

$t=1.00 H3R$ (9)

Substitute $4.0 Ω$ for $R$ in equation (9) to find the $t$ .

$t=1.00 H3×4.0 Ω=0.083 s$

The current flowing through the inductor at time $t$ is,

$I=I0e−RLt$

Substitute $0.500 A$ for $I0$ and $12.0 Ω$ for $R1$ and $0.083 s$ for $t$ in equation (10).

$I=(0.500 A)e−(12.0 Ω)1.0 H(0.083 s)$

Thus, the graph of the current (the initial and final values) in the inductor for $t≥0$ and the time constant from equation (11) is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 3

Figure (2)

The graph shows that current decays with exponentially with time constant. The current decreases from $0.500 A$ towards zero because time constant goes to zero so, the final value of the current gets zero.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 32, Problem 32.64AP

(a)

To determine

The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

(a)

Expert Solution

Answer to Problem 32.64AP

The inductor behaves as the short circuit because of no resistance.

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The inductor has no resistance. If the switch is closed for a long time, then inductor will reach saturation and voltage passes through the inductor. Hence, it behaves as a short circuit.

Conclusion:

Therefore, the inductor behaves as the short circuit because of no resistance.

(b)

To determine

The current carried by the inductor.

(b)

Expert Solution

Answer to Problem 32.64AP

The current carried by the inductor is

$0.500 A$ .

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The figure of the circuit diagrammed referred from P31.15 is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 1

Figure (1)

The net resistance for parallel combination is,

$1R'=1R+12RR'=2R3$

Here,

$R'$ is the net resistance for parallel combination.

$R$ is the resistance of the circuit.

The net resistance is connected in series with $R$ for parallel combination is,

$Rnet=R+R'$ (1)

Here,

$Rnet$ is the net resistance is connected in series.

Substitute $2R3$ for $R'$ in equation (1).

$Rnet=R+2R3$ (2)

Substitute $4.00 Ω$ for $R$ in equation (2) to find the $Rnet$ .

$Rnet=4.00 Ω+2(4.00 Ω)3=6.67 Ω$

Formula to calculate the current of battery is,

$Ib=εRnet$

Here,

$Ib$ is the battery current.

$ε$ is the induced voltage.

Substitute $10.0 V$ for $ε$ and $6.67 Ω$ for $Rnet$ in equation (3) to find the $Ib$ .

$Ib=10.0 V6.67 Ω=1.499 A≈1.50 A$

The voltage across the parallel combination of resistors is,

$V=ε−IbR$ (6)

Substitute $1.50 A$ for $Ib$ , $10.0 V$ for $ε$ and $4.00 Ω$ for $R$ in equation (6) to find the $V$ .

$V=10.0 V−(1.50 A)(4.00 Ω)=4.00 V$

Formula to calculate the current though the inductor is,

$Ii=V2R$ (7)

Substitute $4.00 V$ for $V$ and $4.00 Ω$ for $R$ in equation (7) to find the $Ii$ .

$Ii=4.00 V2(4.00 Ω)=0.500 A$

Conclusion:

Therefore, the current carried by the inductor for $t<0$ is $0.500 A$ .

(c)

To determine

The energy stored in the inductor.

(c)

Expert Solution

Answer to Problem 32.64AP

The energy stored in the inductor for

$t<0$ is

$0.125 J$ .

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

Formula to calculate the energy stored in the inductor is,

$Ui=12LIi2$

Substitute $1.00 H$ for $L$ and $0.500 A$ for $Ii$ to find the $Ui$ .

$Ui=12(1.00 H)(0.500 A)2=0.125 J$

Thus, the energy stored in the inductor for $t<0$ is $0.125 J$ .

Conclusion:

Therefore, the energy stored in the inductor for $t<0$ is $0.125 J$ .

(d)

To determine

The energy previously stored in the inductor after the switch is opened.

(d)

Expert Solution

Answer to Problem 32.64AP

The energy becomes

$0.125 J$ of the additional internal energy in the

$R$ and

$2R$ . in the middle branch after switch is opened.

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

When switch is opened, the energy stored in the inductor will dissipate through resistor $R$ and $2R$ . Thus, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch.

Conclusion:

Therefore, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch after switch is opened.

(e)

To determine

To draw: The graph of the current in the inductor for $t≥0$ with the initial and final values and the time constant.

(e)

Expert Solution

Answer to Problem 32.64AP

Answer The graph of the current in the inductor for $t≥0$ and label the initial and final values and the time constant.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 2

Explanation of Solution

Introduction:

The graph of the current verses time shows the variation of the current in the circuit with time and tells the nature of the current.

Explanation:

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

After time $t=0$ , the resistance of the circuit will be,

$R1=R+2R=3R$

Formula to calculate the time constant is,

$t=LR1$ (8)

Substitute $3R$ for $R1$ and $1.00 H$ for $L$ to find the $t$ .

$t=1.00 H3R$ (9)

Substitute $4.0 Ω$ for $R$ in equation (9) to find the $t$ .

$t=1.00 H3×4.0 Ω=0.083 s$

The current flowing through the inductor at time $t$ is,

$I=I0e−RLt$

Substitute $0.500 A$ for $I0$ and $12.0 Ω$ for $R1$ and $0.083 s$ for $t$ in equation (10).

$I=(0.500 A)e−(12.0 Ω)1.0 H(0.083 s)$

Thus, the graph of the current (the initial and final values) in the inductor for $t≥0$ and the time constant from equation (11) is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 3

Figure (2)

The graph shows that current decays with exponentially with time constant. The current decreases from $0.500 A$ towards zero because time constant goes to zero so, the final value of the current gets zero.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 32, Problem 32.64AP

(a)

To determine

The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

(a)

Expert Solution

Answer to Problem 32.64AP

The inductor behaves as the short circuit because of no resistance.

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The inductor has no resistance. If the switch is closed for a long time, then inductor will reach saturation and voltage passes through the inductor. Hence, it behaves as a short circuit.

Conclusion:

Therefore, the inductor behaves as the short circuit because of no resistance.

(b)

To determine

The current carried by the inductor.

(b)

Expert Solution

Answer to Problem 32.64AP

The current carried by the inductor is

$0.500 A$ .

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The figure of the circuit diagrammed referred from P31.15 is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 1

Figure (1)

The net resistance for parallel combination is,

$1R'=1R+12RR'=2R3$

Here,

$R'$ is the net resistance for parallel combination.

$R$ is the resistance of the circuit.

The net resistance is connected in series with $R$ for parallel combination is,

$Rnet=R+R'$ (1)

Here,

$Rnet$ is the net resistance is connected in series.

Substitute $2R3$ for $R'$ in equation (1).

$Rnet=R+2R3$ (2)

Substitute $4.00 Ω$ for $R$ in equation (2) to find the $Rnet$ .

$Rnet=4.00 Ω+2(4.00 Ω)3=6.67 Ω$

Formula to calculate the current of battery is,

$Ib=εRnet$

Here,

$Ib$ is the battery current.

$ε$ is the induced voltage.

Substitute $10.0 V$ for $ε$ and $6.67 Ω$ for $Rnet$ in equation (3) to find the $Ib$ .

$Ib=10.0 V6.67 Ω=1.499 A≈1.50 A$

The voltage across the parallel combination of resistors is,

$V=ε−IbR$ (6)

Substitute $1.50 A$ for $Ib$ , $10.0 V$ for $ε$ and $4.00 Ω$ for $R$ in equation (6) to find the $V$ .

$V=10.0 V−(1.50 A)(4.00 Ω)=4.00 V$

Formula to calculate the current though the inductor is,

$Ii=V2R$ (7)

Substitute $4.00 V$ for $V$ and $4.00 Ω$ for $R$ in equation (7) to find the $Ii$ .

$Ii=4.00 V2(4.00 Ω)=0.500 A$

Conclusion:

Therefore, the current carried by the inductor for $t<0$ is $0.500 A$ .

(c)

To determine

The energy stored in the inductor.

(c)

Expert Solution

Answer to Problem 32.64AP

The energy stored in the inductor for

$t<0$ is

$0.125 J$ .

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

Formula to calculate the energy stored in the inductor is,

$Ui=12LIi2$

Substitute $1.00 H$ for $L$ and $0.500 A$ for $Ii$ to find the $Ui$ .

$Ui=12(1.00 H)(0.500 A)2=0.125 J$

Thus, the energy stored in the inductor for $t<0$ is $0.125 J$ .

Conclusion:

Therefore, the energy stored in the inductor for $t<0$ is $0.125 J$ .

(d)

To determine

The energy previously stored in the inductor after the switch is opened.

(d)

Expert Solution

Answer to Problem 32.64AP

The energy becomes

$0.125 J$ of the additional internal energy in the

$R$ and

$2R$ . in the middle branch after switch is opened.

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

When switch is opened, the energy stored in the inductor will dissipate through resistor $R$ and $2R$ . Thus, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch.

Conclusion:

Therefore, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch after switch is opened.

(e)

To determine

To draw: The graph of the current in the inductor for $t≥0$ with the initial and final values and the time constant.

(e)

Expert Solution

Answer to Problem 32.64AP

Answer The graph of the current in the inductor for $t≥0$ and label the initial and final values and the time constant.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 2

Explanation of Solution

Introduction:

The graph of the current verses time shows the variation of the current in the circuit with time and tells the nature of the current.

Explanation:

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

After time $t=0$ , the resistance of the circuit will be,

$R1=R+2R=3R$

Formula to calculate the time constant is,

$t=LR1$ (8)

Substitute $3R$ for $R1$ and $1.00 H$ for $L$ to find the $t$ .

$t=1.00 H3R$ (9)

Substitute $4.0 Ω$ for $R$ in equation (9) to find the $t$ .

$t=1.00 H3×4.0 Ω=0.083 s$

The current flowing through the inductor at time $t$ is,

$I=I0e−RLt$

Substitute $0.500 A$ for $I0$ and $12.0 Ω$ for $R1$ and $0.083 s$ for $t$ in equation (10).

$I=(0.500 A)e−(12.0 Ω)1.0 H(0.083 s)$

Thus, the graph of the current (the initial and final values) in the inductor for $t≥0$ and the time constant from equation (11) is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 3

Figure (2)

The graph shows that current decays with exponentially with time constant. The current decreases from $0.500 A$ towards zero because time constant goes to zero so, the final value of the current gets zero.

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Answer 5

Chapter 32, Problem 32.64AP

(a)

To determine

The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

(a)

Expert Solution

Answer to Problem 32.64AP

The inductor behaves as the short circuit because of no resistance.

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The inductor has no resistance. If the switch is closed for a long time, then inductor will reach saturation and voltage passes through the inductor. Hence, it behaves as a short circuit.

Conclusion:

Therefore, the inductor behaves as the short circuit because of no resistance.

(b)

To determine

The current carried by the inductor.

(b)

Expert Solution

Answer to Problem 32.64AP

The current carried by the inductor is

$0.500 A$ .

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The figure of the circuit diagrammed referred from P31.15 is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 1

Figure (1)

The net resistance for parallel combination is,

$1R'=1R+12RR'=2R3$

Here,

$R'$ is the net resistance for parallel combination.

$R$ is the resistance of the circuit.

The net resistance is connected in series with $R$ for parallel combination is,

$Rnet=R+R'$ (1)

Here,

$Rnet$ is the net resistance is connected in series.

Substitute $2R3$ for $R'$ in equation (1).

$Rnet=R+2R3$ (2)

Substitute $4.00 Ω$ for $R$ in equation (2) to find the $Rnet$ .

$Rnet=4.00 Ω+2(4.00 Ω)3=6.67 Ω$

Formula to calculate the current of battery is,

$Ib=εRnet$

Here,

$Ib$ is the battery current.

$ε$ is the induced voltage.

Substitute $10.0 V$ for $ε$ and $6.67 Ω$ for $Rnet$ in equation (3) to find the $Ib$ .

$Ib=10.0 V6.67 Ω=1.499 A≈1.50 A$

The voltage across the parallel combination of resistors is,

$V=ε−IbR$ (6)

Substitute $1.50 A$ for $Ib$ , $10.0 V$ for $ε$ and $4.00 Ω$ for $R$ in equation (6) to find the $V$ .

$V=10.0 V−(1.50 A)(4.00 Ω)=4.00 V$

Formula to calculate the current though the inductor is,

$Ii=V2R$ (7)

Substitute $4.00 V$ for $V$ and $4.00 Ω$ for $R$ in equation (7) to find the $Ii$ .

$Ii=4.00 V2(4.00 Ω)=0.500 A$

Conclusion:

Therefore, the current carried by the inductor for $t<0$ is $0.500 A$ .

(c)

To determine

The energy stored in the inductor.

(c)

Expert Solution

Answer to Problem 32.64AP

The energy stored in the inductor for

$t<0$ is

$0.125 J$ .

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

Formula to calculate the energy stored in the inductor is,

$Ui=12LIi2$

Substitute $1.00 H$ for $L$ and $0.500 A$ for $Ii$ to find the $Ui$ .

$Ui=12(1.00 H)(0.500 A)2=0.125 J$

Thus, the energy stored in the inductor for $t<0$ is $0.125 J$ .

Conclusion:

Therefore, the energy stored in the inductor for $t<0$ is $0.125 J$ .

(d)

To determine

The energy previously stored in the inductor after the switch is opened.

(d)

Expert Solution

Answer to Problem 32.64AP

The energy becomes

$0.125 J$ of the additional internal energy in the

$R$ and

$2R$ . in the middle branch after switch is opened.

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

When switch is opened, the energy stored in the inductor will dissipate through resistor $R$ and $2R$ . Thus, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch.

Conclusion:

Therefore, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch after switch is opened.

(e)

To determine

To draw: The graph of the current in the inductor for $t≥0$ with the initial and final values and the time constant.

(e)

Expert Solution

Answer to Problem 32.64AP

Answer The graph of the current in the inductor for $t≥0$ and label the initial and final values and the time constant.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 2

Explanation of Solution

Introduction:

The graph of the current verses time shows the variation of the current in the circuit with time and tells the nature of the current.

Explanation:

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

After time $t=0$ , the resistance of the circuit will be,

$R1=R+2R=3R$

Formula to calculate the time constant is,

$t=LR1$ (8)

Substitute $3R$ for $R1$ and $1.00 H$ for $L$ to find the $t$ .

$t=1.00 H3R$ (9)

Substitute $4.0 Ω$ for $R$ in equation (9) to find the $t$ .

$t=1.00 H3×4.0 Ω=0.083 s$

The current flowing through the inductor at time $t$ is,

$I=I0e−RLt$

Substitute $0.500 A$ for $I0$ and $12.0 Ω$ for $R1$ and $0.083 s$ for $t$ in equation (10).

$I=(0.500 A)e−(12.0 Ω)1.0 H(0.083 s)$

Thus, the graph of the current (the initial and final values) in the inductor for $t≥0$ and the time constant from equation (11) is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 3

Figure (2)

The graph shows that current decays with exponentially with time constant. The current decreases from $0.500 A$ towards zero because time constant goes to zero so, the final value of the current gets zero.

Answer 6

Chapter 32, Problem 32.64AP

(a)

To determine

The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

(a)

Expert Solution

Answer to Problem 32.64AP

The inductor behaves as the short circuit because of no resistance.

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The inductor has no resistance. If the switch is closed for a long time, then inductor will reach saturation and voltage passes through the inductor. Hence, it behaves as a short circuit.

Conclusion:

Therefore, the inductor behaves as the short circuit because of no resistance.

Answer 7

Chapter 32, Problem 32.64AP

(a)

To determine

The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

Answer 8

(a)

Expert Solution

Answer to Problem 32.64AP

The inductor behaves as the short circuit because of no resistance.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The inductor has no resistance. If the switch is closed for a long time, then inductor will reach saturation and voltage passes through the inductor. Hence, it behaves as a short circuit.

Conclusion:

Therefore, the inductor behaves as the short circuit because of no resistance.

Answer 13

(b)

To determine

The current carried by the inductor.

(b)

Expert Solution

Answer to Problem 32.64AP

The current carried by the inductor is

$0.500 A$ .

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The figure of the circuit diagrammed referred from P31.15 is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 1

Figure (1)

The net resistance for parallel combination is,

$1R'=1R+12RR'=2R3$

Here,

$R'$ is the net resistance for parallel combination.

$R$ is the resistance of the circuit.

The net resistance is connected in series with $R$ for parallel combination is,

$Rnet=R+R'$ (1)

Here,

$Rnet$ is the net resistance is connected in series.

Substitute $2R3$ for $R'$ in equation (1).

$Rnet=R+2R3$ (2)

Substitute $4.00 Ω$ for $R$ in equation (2) to find the $Rnet$ .

$Rnet=4.00 Ω+2(4.00 Ω)3=6.67 Ω$

Formula to calculate the current of battery is,

$Ib=εRnet$

Here,

$Ib$ is the battery current.

$ε$ is the induced voltage.

Substitute $10.0 V$ for $ε$ and $6.67 Ω$ for $Rnet$ in equation (3) to find the $Ib$ .

$Ib=10.0 V6.67 Ω=1.499 A≈1.50 A$

The voltage across the parallel combination of resistors is,

$V=ε−IbR$ (6)

Substitute $1.50 A$ for $Ib$ , $10.0 V$ for $ε$ and $4.00 Ω$ for $R$ in equation (6) to find the $V$ .

$V=10.0 V−(1.50 A)(4.00 Ω)=4.00 V$

Formula to calculate the current though the inductor is,

$Ii=V2R$ (7)

Substitute $4.00 V$ for $V$ and $4.00 Ω$ for $R$ in equation (7) to find the $Ii$ .

$Ii=4.00 V2(4.00 Ω)=0.500 A$

Conclusion:

Therefore, the current carried by the inductor for $t<0$ is $0.500 A$ .

Answer 14

(b)

To determine

The current carried by the inductor.

Answer 15

(b)

Expert Solution

Answer to Problem 32.64AP

The current carried by the inductor is

$0.500 A$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

The figure of the circuit diagrammed referred from P31.15 is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 1

Figure (1)

The net resistance for parallel combination is,

$1R'=1R+12RR'=2R3$

Here,

$R'$ is the net resistance for parallel combination.

$R$ is the resistance of the circuit.

The net resistance is connected in series with $R$ for parallel combination is,

$Rnet=R+R'$ (1)

Here,

$Rnet$ is the net resistance is connected in series.

Substitute $2R3$ for $R'$ in equation (1).

$Rnet=R+2R3$ (2)

Substitute $4.00 Ω$ for $R$ in equation (2) to find the $Rnet$ .

$Rnet=4.00 Ω+2(4.00 Ω)3=6.67 Ω$

Formula to calculate the current of battery is,

$Ib=εRnet$

Here,

$Ib$ is the battery current.

$ε$ is the induced voltage.

Substitute $10.0 V$ for $ε$ and $6.67 Ω$ for $Rnet$ in equation (3) to find the $Ib$ .

$Ib=10.0 V6.67 Ω=1.499 A≈1.50 A$

The voltage across the parallel combination of resistors is,

$V=ε−IbR$ (6)

Substitute $1.50 A$ for $Ib$ , $10.0 V$ for $ε$ and $4.00 Ω$ for $R$ in equation (6) to find the $V$ .

$V=10.0 V−(1.50 A)(4.00 Ω)=4.00 V$

Formula to calculate the current though the inductor is,

$Ii=V2R$ (7)

Substitute $4.00 V$ for $V$ and $4.00 Ω$ for $R$ in equation (7) to find the $Ii$ .

$Ii=4.00 V2(4.00 Ω)=0.500 A$

Conclusion:

Therefore, the current carried by the inductor for $t<0$ is $0.500 A$ .

Answer 20

(c)

To determine

The energy stored in the inductor.

(c)

Expert Solution

Answer to Problem 32.64AP

The energy stored in the inductor for

$t<0$ is

$0.125 J$ .

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

Formula to calculate the energy stored in the inductor is,

$Ui=12LIi2$

Substitute $1.00 H$ for $L$ and $0.500 A$ for $Ii$ to find the $Ui$ .

$Ui=12(1.00 H)(0.500 A)2=0.125 J$

Thus, the energy stored in the inductor for $t<0$ is $0.125 J$ .

Conclusion:

Therefore, the energy stored in the inductor for $t<0$ is $0.125 J$ .

Answer 21

(c)

To determine

The energy stored in the inductor.

Answer 22

(c)

Expert Solution

Answer to Problem 32.64AP

The energy stored in the inductor for

$t<0$ is

$0.125 J$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

Formula to calculate the energy stored in the inductor is,

$Ui=12LIi2$

Substitute $1.00 H$ for $L$ and $0.500 A$ for $Ii$ to find the $Ui$ .

$Ui=12(1.00 H)(0.500 A)2=0.125 J$

Thus, the energy stored in the inductor for $t<0$ is $0.125 J$ .

Conclusion:

Therefore, the energy stored in the inductor for $t<0$ is $0.125 J$ .

Answer 27

(d)

To determine

The energy previously stored in the inductor after the switch is opened.

(d)

Expert Solution

Answer to Problem 32.64AP

The energy becomes

$0.125 J$ of the additional internal energy in the

$R$ and

$2R$ . in the middle branch after switch is opened.

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

When switch is opened, the energy stored in the inductor will dissipate through resistor $R$ and $2R$ . Thus, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch.

Conclusion:

Therefore, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch after switch is opened.

Answer 28

(d)

To determine

The energy previously stored in the inductor after the switch is opened.

Answer 29

(d)

Expert Solution

Answer to Problem 32.64AP

The energy becomes

$0.125 J$ of the additional internal energy in the

$R$ and

$2R$ . in the middle branch after switch is opened.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

When switch is opened, the energy stored in the inductor will dissipate through resistor $R$ and $2R$ . Thus, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch.

Conclusion:

Therefore, the energy becomes $0.125 J$ of the additional internal energy in the $R$ and $2R$ . in the middle branch after switch is opened.

Answer 34

(e)

To determine

To draw: The graph of the current in the inductor for $t≥0$ with the initial and final values and the time constant.

(e)

Expert Solution

Answer to Problem 32.64AP

Answer The graph of the current in the inductor for $t≥0$ and label the initial and final values and the time constant.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 2

Explanation of Solution

Introduction:

The graph of the current verses time shows the variation of the current in the circuit with time and tells the nature of the current.

Explanation:

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

After time $t=0$ , the resistance of the circuit will be,

$R1=R+2R=3R$

Formula to calculate the time constant is,

$t=LR1$ (8)

Substitute $3R$ for $R1$ and $1.00 H$ for $L$ to find the $t$ .

$t=1.00 H3R$ (9)

Substitute $4.0 Ω$ for $R$ in equation (9) to find the $t$ .

$t=1.00 H3×4.0 Ω=0.083 s$

The current flowing through the inductor at time $t$ is,

$I=I0e−RLt$

Substitute $0.500 A$ for $I0$ and $12.0 Ω$ for $R1$ and $0.083 s$ for $t$ in equation (10).

$I=(0.500 A)e−(12.0 Ω)1.0 H(0.083 s)$

Thus, the graph of the current (the initial and final values) in the inductor for $t≥0$ and the time constant from equation (11) is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 3

Figure (2)

The graph shows that current decays with exponentially with time constant. The current decreases from $0.500 A$ towards zero because time constant goes to zero so, the final value of the current gets zero.

Answer 35

(e)

To determine

To draw: The graph of the current in the inductor for $t≥0$ with the initial and final values and the time constant.

Answer 36

(e)

Expert Solution

Answer to Problem 32.64AP

Answer The graph of the current in the inductor for $t≥0$ and label the initial and final values and the time constant.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 2

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Introduction:

The graph of the current verses time shows the variation of the current in the circuit with time and tells the nature of the current.

Explanation:

Given info: The induced voltage is $10.0 V$ , the resistance of the resistor is $4.00 Ω$ and the inductance of the inductor is $1.00 H$ .

After time $t=0$ , the resistance of the circuit will be,

$R1=R+2R=3R$

Formula to calculate the time constant is,

$t=LR1$ (8)

Substitute $3R$ for $R1$ and $1.00 H$ for $L$ to find the $t$ .

$t=1.00 H3R$ (9)

Substitute $4.0 Ω$ for $R$ in equation (9) to find the $t$ .

$t=1.00 H3×4.0 Ω=0.083 s$

The current flowing through the inductor at time $t$ is,

$I=I0e−RLt$

Substitute $0.500 A$ for $I0$ and $12.0 Ω$ for $R1$ and $0.083 s$ for $t$ in equation (10).

$I=(0.500 A)e−(12.0 Ω)1.0 H(0.083 s)$

Thus, the graph of the current (the initial and final values) in the inductor for $t≥0$ and the time constant from equation (11) is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip 3

Figure (2)

The graph shows that current decays with exponentially with time constant. The current decreases from $0.500 A$ towards zero because time constant goes to zero so, the final value of the current gets zero.

Answer 41

Ch. 32 - A coil with zero resistance has its ends labeled a...Ch. 32 - Prob. 32.2QQ Ch. 32 - Prob. 32.3QQ Ch. 32 - Prob. 32.4QQ Ch. 32 - (i) At an instant of time during the oscillations...Ch. 32 - Prob. 32.1OQ Ch. 32 - Prob. 32.2OQ Ch. 32 - Prob. 32.3OQ Ch. 32 - In Figure OQ32.4, the switch is left in position a...Ch. 32 - Prob. 32.5OQ

The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

Solution Summary: The author explains that the inductor behaves like an open circuit or short circuit, or a resister of some particular resistance before the switch is opened.

Concept explainers

The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

Answer to Problem 32.64AP

Explanation of Solution

The current carried by the inductor.

Answer to Problem 32.64AP

Explanation of Solution

The energy stored in the inductor.

Answer to Problem 32.64AP

Explanation of Solution

The energy previously stored in the inductor after the switch is opened.

Answer to Problem 32.64AP

Explanation of Solution

To draw: The graph of the current in the inductor for $t≥0$ with the initial and final values and the time constant.

Answer to Problem 32.64AP

Explanation of Solution

Want to see more full solutions like this?

Chapter 32 Solutions

The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

Solution Summary: The author explains that the inductor behaves like an open circuit or short circuit, or a resister of some particular resistance before the switch is opened.

Concept explainers

The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

Answer to Problem 32.64AP

Explanation of Solution

The current carried by the inductor.

Answer to Problem 32.64AP

Explanation of Solution

The energy stored in the inductor.

Answer to Problem 32.64AP

Explanation of Solution

The energy previously stored in the inductor after the switch is opened.

Answer to Problem 32.64AP

Explanation of Solution

To draw: The graph of the current in the inductor for t≥0<m:math><m:mrow><m:mi>t</m:mi><m:mo>≥</m:mo><m:mn>0</m:mn></m:mrow> </m:math> with the initial and final values and the time constant.

Answer to Problem 32.64AP

Explanation of Solution

Want to see more full solutions like this?

Chapter 32 Solutions

To draw: The graph of the current in the inductor for $t≥0$ with the initial and final values and the time constant.