The current in the battery, in 100 Ω resistor and inductor immediately after the switch is closed.

Question

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Answer 1

Question

Chapter 28, Problem 66P

(a)

To determine

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed.

(a)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

Explanation of Solution

Given:

The inductance of the inductor is $L=2 H$ .

The resistance of the resistor is $R=100 Ω$ .

The voltage of battery is $εo=10 V$ .

Formula used:

The expression for the Kirchhoff’s of current at the junction given by,

$∑Iin=∑Iout$

Calculation:

The current through the battery and inductor is calculated as,

$∑I in=∑I outIB=IR→100+IL$

As the switch is closed the value of current through the inductor,

$IL=0$

Then,

$IB=IR→100+0IB=IR→100IB=εoRT=10 V( 10 Ω+ 100 Ω)$

Solve further as,

$IB=0.0909 A( 1 mA 10 −3 A)=90.9 mA$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

(b)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed.

(b)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

Explanation of Solution

Calculation:

The current through the resistor is calculated by using the loop rule as:

$(IR→100)×(100 Ω)=2HdILdt$

When the switch remains closed for long time, the current will be steady and $dILdt=0$ .

$(I R→100)×(100 Ω)=2 H×(0)IR→100=0$

The current through the battery is calculated by using the loop rule to right side:

$10 V−IB×10 V−IR→100×(100 Ω)=010 V−IB×10 V−0×(100 Ω)=0IB=1 A$

The current through the inductor is calculated by using the junction rule as:

$IB=IR→100+IL1 A=0+ILIL=1 A$

Conclusion:

Therefore the current in the battery, in $100−Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

(c)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductorimmediately after switch is opened.

(c)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0, −1 A$ and $1 A$ respectively.

Explanation of Solution

Calculation:

When the switch is reopened the current in the battery instantly becomes zero and the current in the inductor changes continuously until it takes the value of $IL=1 A$ . Now using the junction rule,

$IB=IR→100+IL0=IR→100+1 AIR→100=−1 A$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0,−1 A$ and $1 A$ respectively.

(d)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor after a switch is opened for a long time.

(d)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

Explanation of Solution

Calculation:

When the switch is reopened and after some time all the currents will become zero.

$IB=IR→100IB=IL=0$

Conclusion:

Therefore, the current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

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Answer 2

Question

Chapter 28, Problem 66P

(a)

To determine

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed.

(a)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

Explanation of Solution

Given:

The inductance of the inductor is $L=2 H$ .

The resistance of the resistor is $R=100 Ω$ .

The voltage of battery is $εo=10 V$ .

Formula used:

The expression for the Kirchhoff’s of current at the junction given by,

$∑Iin=∑Iout$

Calculation:

The current through the battery and inductor is calculated as,

$∑I in=∑I outIB=IR→100+IL$

As the switch is closed the value of current through the inductor,

$IL=0$

Then,

$IB=IR→100+0IB=IR→100IB=εoRT=10 V( 10 Ω+ 100 Ω)$

Solve further as,

$IB=0.0909 A( 1 mA 10 −3 A)=90.9 mA$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

(b)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed.

(b)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

Explanation of Solution

Calculation:

The current through the resistor is calculated by using the loop rule as:

$(IR→100)×(100 Ω)=2HdILdt$

When the switch remains closed for long time, the current will be steady and $dILdt=0$ .

$(I R→100)×(100 Ω)=2 H×(0)IR→100=0$

The current through the battery is calculated by using the loop rule to right side:

$10 V−IB×10 V−IR→100×(100 Ω)=010 V−IB×10 V−0×(100 Ω)=0IB=1 A$

The current through the inductor is calculated by using the junction rule as:

$IB=IR→100+IL1 A=0+ILIL=1 A$

Conclusion:

Therefore the current in the battery, in $100−Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

(c)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductorimmediately after switch is opened.

(c)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0, −1 A$ and $1 A$ respectively.

Explanation of Solution

Calculation:

When the switch is reopened the current in the battery instantly becomes zero and the current in the inductor changes continuously until it takes the value of $IL=1 A$ . Now using the junction rule,

$IB=IR→100+IL0=IR→100+1 AIR→100=−1 A$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0,−1 A$ and $1 A$ respectively.

(d)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor after a switch is opened for a long time.

(d)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

Explanation of Solution

Calculation:

When the switch is reopened and after some time all the currents will become zero.

$IB=IR→100IB=IL=0$

Conclusion:

Therefore, the current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

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Answer 3

Question

Chapter 28, Problem 66P

(a)

To determine

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed.

(a)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

Explanation of Solution

Given:

The inductance of the inductor is $L=2 H$ .

The resistance of the resistor is $R=100 Ω$ .

The voltage of battery is $εo=10 V$ .

Formula used:

The expression for the Kirchhoff’s of current at the junction given by,

$∑Iin=∑Iout$

Calculation:

The current through the battery and inductor is calculated as,

$∑I in=∑I outIB=IR→100+IL$

As the switch is closed the value of current through the inductor,

$IL=0$

Then,

$IB=IR→100+0IB=IR→100IB=εoRT=10 V( 10 Ω+ 100 Ω)$

Solve further as,

$IB=0.0909 A( 1 mA 10 −3 A)=90.9 mA$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

(b)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed.

(b)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

Explanation of Solution

Calculation:

The current through the resistor is calculated by using the loop rule as:

$(IR→100)×(100 Ω)=2HdILdt$

When the switch remains closed for long time, the current will be steady and $dILdt=0$ .

$(I R→100)×(100 Ω)=2 H×(0)IR→100=0$

The current through the battery is calculated by using the loop rule to right side:

$10 V−IB×10 V−IR→100×(100 Ω)=010 V−IB×10 V−0×(100 Ω)=0IB=1 A$

The current through the inductor is calculated by using the junction rule as:

$IB=IR→100+IL1 A=0+ILIL=1 A$

Conclusion:

Therefore the current in the battery, in $100−Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

(c)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductorimmediately after switch is opened.

(c)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0, −1 A$ and $1 A$ respectively.

Explanation of Solution

Calculation:

When the switch is reopened the current in the battery instantly becomes zero and the current in the inductor changes continuously until it takes the value of $IL=1 A$ . Now using the junction rule,

$IB=IR→100+IL0=IR→100+1 AIR→100=−1 A$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0,−1 A$ and $1 A$ respectively.

(d)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor after a switch is opened for a long time.

(d)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

Explanation of Solution

Calculation:

When the switch is reopened and after some time all the currents will become zero.

$IB=IR→100IB=IL=0$

Conclusion:

Therefore, the current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

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Answer 4

Question

Chapter 28, Problem 66P

(a)

To determine

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed.

(a)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

Explanation of Solution

Given:

The inductance of the inductor is $L=2 H$ .

The resistance of the resistor is $R=100 Ω$ .

The voltage of battery is $εo=10 V$ .

Formula used:

The expression for the Kirchhoff’s of current at the junction given by,

$∑Iin=∑Iout$

Calculation:

The current through the battery and inductor is calculated as,

$∑I in=∑I outIB=IR→100+IL$

As the switch is closed the value of current through the inductor,

$IL=0$

Then,

$IB=IR→100+0IB=IR→100IB=εoRT=10 V( 10 Ω+ 100 Ω)$

Solve further as,

$IB=0.0909 A( 1 mA 10 −3 A)=90.9 mA$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

(b)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed.

(b)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

Explanation of Solution

Calculation:

The current through the resistor is calculated by using the loop rule as:

$(IR→100)×(100 Ω)=2HdILdt$

When the switch remains closed for long time, the current will be steady and $dILdt=0$ .

$(I R→100)×(100 Ω)=2 H×(0)IR→100=0$

The current through the battery is calculated by using the loop rule to right side:

$10 V−IB×10 V−IR→100×(100 Ω)=010 V−IB×10 V−0×(100 Ω)=0IB=1 A$

The current through the inductor is calculated by using the junction rule as:

$IB=IR→100+IL1 A=0+ILIL=1 A$

Conclusion:

Therefore the current in the battery, in $100−Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

(c)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductorimmediately after switch is opened.

(c)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0, −1 A$ and $1 A$ respectively.

Explanation of Solution

Calculation:

When the switch is reopened the current in the battery instantly becomes zero and the current in the inductor changes continuously until it takes the value of $IL=1 A$ . Now using the junction rule,

$IB=IR→100+IL0=IR→100+1 AIR→100=−1 A$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0,−1 A$ and $1 A$ respectively.

(d)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor after a switch is opened for a long time.

(d)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

Explanation of Solution

Calculation:

When the switch is reopened and after some time all the currents will become zero.

$IB=IR→100IB=IL=0$

Conclusion:

Therefore, the current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

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Answer 5

Chapter 28, Problem 66P

(a)

To determine

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed.

(a)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

Explanation of Solution

Given:

The inductance of the inductor is $L=2 H$ .

The resistance of the resistor is $R=100 Ω$ .

The voltage of battery is $εo=10 V$ .

Formula used:

The expression for the Kirchhoff’s of current at the junction given by,

$∑Iin=∑Iout$

Calculation:

The current through the battery and inductor is calculated as,

$∑I in=∑I outIB=IR→100+IL$

As the switch is closed the value of current through the inductor,

$IL=0$

Then,

$IB=IR→100+0IB=IR→100IB=εoRT=10 V( 10 Ω+ 100 Ω)$

Solve further as,

$IB=0.0909 A( 1 mA 10 −3 A)=90.9 mA$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

(b)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed.

(b)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

Explanation of Solution

Calculation:

The current through the resistor is calculated by using the loop rule as:

$(IR→100)×(100 Ω)=2HdILdt$

When the switch remains closed for long time, the current will be steady and $dILdt=0$ .

$(I R→100)×(100 Ω)=2 H×(0)IR→100=0$

The current through the battery is calculated by using the loop rule to right side:

$10 V−IB×10 V−IR→100×(100 Ω)=010 V−IB×10 V−0×(100 Ω)=0IB=1 A$

The current through the inductor is calculated by using the junction rule as:

$IB=IR→100+IL1 A=0+ILIL=1 A$

Conclusion:

Therefore the current in the battery, in $100−Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

(c)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductorimmediately after switch is opened.

(c)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0, −1 A$ and $1 A$ respectively.

Explanation of Solution

Calculation:

When the switch is reopened the current in the battery instantly becomes zero and the current in the inductor changes continuously until it takes the value of $IL=1 A$ . Now using the junction rule,

$IB=IR→100+IL0=IR→100+1 AIR→100=−1 A$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0,−1 A$ and $1 A$ respectively.

(d)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor after a switch is opened for a long time.

(d)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

Explanation of Solution

Calculation:

When the switch is reopened and after some time all the currents will become zero.

$IB=IR→100IB=IL=0$

Conclusion:

Therefore, the current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

Answer 6

Chapter 28, Problem 66P

(a)

To determine

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed.

(a)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

Explanation of Solution

Given:

The inductance of the inductor is $L=2 H$ .

The resistance of the resistor is $R=100 Ω$ .

The voltage of battery is $εo=10 V$ .

Formula used:

The expression for the Kirchhoff’s of current at the junction given by,

$∑Iin=∑Iout$

Calculation:

The current through the battery and inductor is calculated as,

$∑I in=∑I outIB=IR→100+IL$

As the switch is closed the value of current through the inductor,

$IL=0$

Then,

$IB=IR→100+0IB=IR→100IB=εoRT=10 V( 10 Ω+ 100 Ω)$

Solve further as,

$IB=0.0909 A( 1 mA 10 −3 A)=90.9 mA$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

Answer 7

Chapter 28, Problem 66P

(a)

To determine

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed.

Answer 8

(a)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The inductance of the inductor is $L=2 H$ .

The resistance of the resistor is $R=100 Ω$ .

The voltage of battery is $εo=10 V$ .

Formula used:

The expression for the Kirchhoff’s of current at the junction given by,

$∑Iin=∑Iout$

Calculation:

The current through the battery and inductor is calculated as,

$∑I in=∑I outIB=IR→100+IL$

As the switch is closed the value of current through the inductor,

$IL=0$

Then,

$IB=IR→100+0IB=IR→100IB=εoRT=10 V( 10 Ω+ 100 Ω)$

Solve further as,

$IB=0.0909 A( 1 mA 10 −3 A)=90.9 mA$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed is $90.9 mA$ , $90.9 mA$ and $0$ respectively.

Answer 13

(b)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed.

(b)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

Explanation of Solution

Calculation:

The current through the resistor is calculated by using the loop rule as:

$(IR→100)×(100 Ω)=2HdILdt$

When the switch remains closed for long time, the current will be steady and $dILdt=0$ .

$(I R→100)×(100 Ω)=2 H×(0)IR→100=0$

The current through the battery is calculated by using the loop rule to right side:

$10 V−IB×10 V−IR→100×(100 Ω)=010 V−IB×10 V−0×(100 Ω)=0IB=1 A$

The current through the inductor is calculated by using the junction rule as:

$IB=IR→100+IL1 A=0+ILIL=1 A$

Conclusion:

Therefore the current in the battery, in $100−Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

Answer 14

(b)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed.

Answer 15

(b)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

The current through the resistor is calculated by using the loop rule as:

$(IR→100)×(100 Ω)=2HdILdt$

When the switch remains closed for long time, the current will be steady and $dILdt=0$ .

$(I R→100)×(100 Ω)=2 H×(0)IR→100=0$

The current through the battery is calculated by using the loop rule to right side:

$10 V−IB×10 V−IR→100×(100 Ω)=010 V−IB×10 V−0×(100 Ω)=0IB=1 A$

The current through the inductor is calculated by using the junction rule as:

$IB=IR→100+IL1 A=0+ILIL=1 A$

Conclusion:

Therefore the current in the battery, in $100−Ω$ resistor and in the inductor a long time after switch is closed is $1 A, 0$ and $1 A$ respectively.

Answer 20

(c)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductorimmediately after switch is opened.

(c)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0, −1 A$ and $1 A$ respectively.

Explanation of Solution

Calculation:

When the switch is reopened the current in the battery instantly becomes zero and the current in the inductor changes continuously until it takes the value of $IL=1 A$ . Now using the junction rule,

$IB=IR→100+IL0=IR→100+1 AIR→100=−1 A$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0,−1 A$ and $1 A$ respectively.

Answer 21

(c)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductorimmediately after switch is opened.

Answer 22

(c)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0, −1 A$ and $1 A$ respectively.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

When the switch is reopened the current in the battery instantly becomes zero and the current in the inductor changes continuously until it takes the value of $IL=1 A$ . Now using the junction rule,

$IB=IR→100+IL0=IR→100+1 AIR→100=−1 A$

Conclusion:

Therefore, the current in the battery, in $100 Ω$ resistor and in the inductor immediately after switch is opened is $0,−1 A$ and $1 A$ respectively.

Answer 27

(d)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor after a switch is opened for a long time.

(d)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

Explanation of Solution

Calculation:

When the switch is reopened and after some time all the currents will become zero.

$IB=IR→100IB=IL=0$

Conclusion:

Therefore, the current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

Answer 28

(d)

To determine

The current in the battery, in $100 Ω$ resistor and in the inductor after a switch is opened for a long time.

Answer 29

(d)

Expert Solution

Answer to Problem 66P

The current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation:

When the switch is reopened and after some time all the currents will become zero.

$IB=IR→100IB=IL=0$

Conclusion:

Therefore, the current in the battery, in $100−Ω$ resistor and in the inductor after a switch is opened for a long time is zero.

Answer 34

Ch. 28 - Prob. 1P Ch. 28 - Prob. 2P Ch. 28 - Prob. 3P Ch. 28 - Prob. 4P Ch. 28 - Prob. 5P Ch. 28 - Prob. 6P Ch. 28 - Prob. 7P Ch. 28 - Prob. 8P Ch. 28 - Prob. 9P Ch. 28 - Prob. 10P

The current in the battery, in 100 Ω resistor and inductor immediately after the switch is closed.

Concept explainers

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed.

Answer to Problem 66P

Explanation of Solution

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed.

Answer to Problem 66P

Explanation of Solution

The current in the battery, in $100 Ω$ resistor and in the inductorimmediately after switch is opened.

Answer to Problem 66P

Explanation of Solution

The current in the battery, in $100 Ω$ resistor and in the inductor after a switch is opened for a long time.

Answer to Problem 66P

Explanation of Solution

Want to see more full solutions like this?

Chapter 28 Solutions

The current in the battery, in 100 Ω resistor and inductor immediately after the switch is closed.

Concept explainers

The current in the battery, in 100 Ω<m:math><m:mrow><m:mn>100</m:mn><m:mtext> </m:mtext><m:mi>Ω</m:mi></m:mrow></m:math> resistor and inductor immediately after the switch is closed.

Answer to Problem 66P

Explanation of Solution

The current in the battery, in 100 Ω<m:math><m:mrow><m:mn>100</m:mn><m:mtext> </m:mtext><m:mi>Ω</m:mi></m:mrow></m:math> resistor and in the inductor a long time after switch is closed.

Answer to Problem 66P

Explanation of Solution

The current in the battery, in 100 Ω<m:math><m:mrow><m:mn>100</m:mn><m:mtext> </m:mtext><m:mi>Ω</m:mi></m:mrow></m:math> resistor and in the inductorimmediately after switch is opened.

Answer to Problem 66P

Explanation of Solution

The current in the battery, in 100 Ω<m:math><m:mrow><m:mn>100</m:mn><m:mtext> </m:mtext><m:mi>Ω</m:mi></m:mrow></m:math> resistor and in the inductor after a switch is opened for a long time.

Answer to Problem 66P

Explanation of Solution

Want to see more full solutions like this?

Chapter 28 Solutions

The current in the battery, in $100 Ω$ resistor and inductor immediately after the switch is closed.

The current in the battery, in $100 Ω$ resistor and in the inductor a long time after switch is closed.

The current in the battery, in $100 Ω$ resistor and in the inductorimmediately after switch is opened.

The current in the battery, in $100 Ω$ resistor and in the inductor after a switch is opened for a long time.