EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Question
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Chapter 28, Problem 59P

(a)

To determine

The current and its rate of change at t=0 .

(a)

Expert Solution
Check Mark

Answer to Problem 59P

The current and its rate of change at t=0 is 0and25kA/s respectively.

Explanation of Solution

Given:

The resistance and self-inductance of the coil is 8.00Ωand4.00mH respectively.

The voltage of the battery is 100V .

Formula used:

The expression for the current is given by,

  I=If×(1e t τ )

Here,

  If=εoR

And.

  τ=LR

Here If is the final current and τ is time constant.

Calculation:

The final current in a coil is calculated as,

  If=εoR=100V8.00Ω=12.5A

The time constant is calculated as,

  τ=LR=4.00mH8.00Ω=0.5ms

The current is calculated as,

  I=If×(1e t τ )I0=(12.5A)×(1e 0 0.5 )=0

The rate of current with time is calculated as,

  dIdt=ddt(12.5A)×(1e t 0.5ms )dIdt t=0=(12.5A)×(1e 0 0.5ms )×10.5ms=25×103A/s( 10 3 kA 1A)=25kA/s

Conclusion:

Thereforethe current and its rate of change at t=0 is 0 and 25kA/s respectively.

(b)

To determine

The current and its rate of change at t=0.100ms

(b)

Expert Solution
Check Mark

Answer to Problem 59P

The current and its rate of change at t=0.100ms is 2.27Aand20.5kA/s .

Explanation of Solution

Formula used:

The expression for the current is given by,

  I=If×(1e t τ )

Here,

  If=εoR

And.

  τ=LR

Here If is the final current and τ is time constant.

Calculation:

The current at time t=0.100ms is calculated as,

  I=If×(1e t τ )I0.1=(12.5A)×(1e -0.1ms 0.5ms )=2.27A

The rate of the current with time is calculated as,

  dIdt=ddt(12.5A)×(1e t 0.5ms )dIdt t=0.1=(12.5A)×(e -0.1ms 0.5ms )×10.5ms=20.5kA/s

Conclusion:

Therefore,the current and its rate of change at t=0.1ms is 2.27A and 20.5kA/s respectively.

(c)

To determine

The current and its rate of change at t=0.5ms

(c)

Expert Solution
Check Mark

Answer to Problem 59P

The current and its rate of change at t=0.5ms is 7.9Aand9.2kA/s respectively.

Explanation of Solution

Formula used:

The expression for the current is given by,

  I=If×(1e t τ )

Here,

  If=εoR

And.

  τ=LR

Here If is the final current and τ is time constant.

Calculation:

The current at time t=0.500ms is calculated as

  I=If×(1e t τ )I0.5=(12.5A)×(1e -0.5ms 0.5ms )=7.9A

The rate of the current with time is calculated as:

  dIdt=ddt(12.5A)×(1e t 0.5ms )dIdt t=0.5=(12.5A)×(e -0.5ms 0.5ms )×10.5ms=9.2kA/s

Conclusion:

Therefore, the current and its rate of change at t=0.5ms is 7.9Aand9.2kA/s respectively.

(d)

To determine

The current and its rate of change at t=1.0ms

(d)

Expert Solution
Check Mark

Answer to Problem 59P

The current and its rate of change at t=1.0ms is 10.8Aand3.38kA/s respectively.

Explanation of Solution

Formula used:

The expression for the current is given by,

  I=If×(1e t τ )

Here,

  If=εoR

And.

  τ=LR

Here If is the final current and τ is time constant.

Calculation:

The current at time t=1.0ms is calculated as

  I=If×(1e t τ )I1.0=(12.5A)×(1e -1.0ms 0.5ms )=10.8A

The rate of the current with time is calculated as:

  dIdt=ddt(12.5A)×(1e t 0.5ms )dIdt t=1.0=(12.5A)×(e -1.0ms 0.5ms )×10.5ms=3.38kA/s

Conclusion:

Therefore, the current and its rate of change at t=1.0ms is 10.8Aand3.38kA/s respectively.

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Chapter 28 Solutions

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