Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 28, Problem 66P

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 66P

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

Explanation of Solution

Given:

The inductance of the inductor is L=2H .

The resistance of the resistor is R=100Ω .

The voltage of battery is εo=10V .

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=IR100+IL

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

  IL=0

Then,

  IB=IR100+0IB=IR100IB=εoRT=10V( 10Ω+100Ω)

Solve further as,

  IB=0.0909A( 1mA 10 3 A)=90.9mA

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution
Check Mark

Answer to Problem 66P

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

Explanation of Solution

Calculation:

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

  (IR100)×(100Ω)=2HdILdt

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

  (I R100)×(100Ω)=2H×(0)IR100=0

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

  10VIB×10VIR100×(100Ω)=010VIB×10V0×(100Ω)=0IB=1A

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

  IB=IR100+IL1A=0+ILIL=1A

Conclusion:

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

(c)

To determine

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

(c)

Expert Solution
Check Mark

Answer to Problem 66P

The current in the battery, in 100Ω resistor and in the inductor immediately after switch is opened is 0,1A and 1A 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=1A . Now using the junction rule,

  IB=IR100+IL0=IR100+1AIR100=1A

Conclusion:

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

(d)

To determine

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

(d)

Expert Solution
Check Mark

Answer to Problem 66P

The current in the battery, in 100Ω 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=IR100IB=IL=0

Conclusion:

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

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Suppose there are two transformers between your house and the high-voltage transmission line that distributes the power. In addition, assume your house is the only one using electric power. At a substation the primary of a step-down transformer (turns ratio = 1:23) receives the voltage from the high-voltage transmission line. Because of your usage, a current of 51.1 mA exists in the primary of the transformer. The secondary is connected to the primary of another step-down transformer (turns ratio = 1:36) somewhere near your house, perhaps up on a telephone pole. The secondary of this transformer delivers a 240-V emf to your house. How much power is your house using? Remember that the current and voltage given in this problem are rms values.

Chapter 28 Solutions

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