EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
1st Edition
ISBN: 9780100546714
Author: Katz
Publisher: YUZU
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Chapter 33, Problem 17PQ

(a)

To determine

The expression for the total energy that dissipated by the resistor in one time constant.

(a)

Expert Solution
Check Mark

Answer to Problem 17PQ

The expression for the total energy that dissipated by the resistor in one time constant is, ε2τR(2e12e212).

Explanation of Solution

The following figure shows the given diagram-33.9A.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 33, Problem 17PQ

Figure-(1)

Here, L is inductance, R is resistance, ε is emf, and I(t) is current as function of time.

Write the expression for current pass through the resistor.

    I(t)=εR(1etτ)

Here, t is time during which the current passes through the resistor, and τ is the time constant of the circuit.

Write the expression for power dissipated in the resistor as function of time.

    P(t)=I2(t)R

Here, P is power.

Substitute εR(1etτ) for I(t) in the above equation to calculate P(t).

    P(t)=[εR(1etτ)]2R=ε2R(1etτ)2

Write the expression for energy dissipated (dU(t)) in the resistor in a small time dt.

    dU(t)=P(t)dt

Substitute ε2R(1etτ)2 for P(t) in the above equation to calculate (dU(t)).

    dU(t)=ε2R(1etτ)2dt=ε2R(12etτ+e2tτ)dt

Integrate the above expression between the limits 0 to τ.

    dU(t)=0τε2R(12etτ+e2tτ)dt=ε2R[t+2τetττ2e2tτ]0τ=ε2R[(τ0)+2τ(e11)τ2(e21)]=ε2τR(2e12e212)

Conclusion:

Therefore, the expression for the total energy that dissipated by the resistor in one time constant is ε2τR(2e12e212).

(b)

To determine

The expression for the total charge that passes through the resistor in one time constant.

(b)

Expert Solution
Check Mark

Answer to Problem 17PQ

The expression for the total charge that passes through the resistor in one time constant is, ε2τ2R(11e2).

Explanation of Solution

Write the expression for decaying current.

    I(t)=εRetτ

Write the expression for power dissipated in the resistor as function of time.

    P(t)=I2(t)R

Substitute εRetτ for I(t) in the above equation.

    P(t)=[εRetτ]2R=ε2R(etτ)2=ε2R(e2tτ)

Integrate the above expression between the limits 0 to τ.

    dU(t)=0τε2R(e2tτ)dt=ε2R[τ2e2tτ]0τ=ε2R[τ2(1e2)]=ε2τ2R(11e2)

Conclusion:

Therefore, the expression for the total energy that dissipated by the resistor in one time constant is, ε2τ2R(11e2).

(c)

To determine

The comparison between the above two results and comment.

(c)

Expert Solution
Check Mark

Answer to Problem 17PQ

The energy dissipated in the resistor when the current decays is more than the energy dissipated when the current grows in the circuit.

Explanation of Solution

Write the expression for energy growth that dissipated through the resistor.

    (U)growth=ε2τR(2e12e212)

Here, (U)growth is energy growth.

Write the expression for energy decay that dissipated through the resistor.

    (U)decay=ε2τ2R(11e2)

Here, (U)decay is energy decay.

Take the ratio of both equations.

    (U)growth(U)decay=ε2τR(2e12e212)ε2τ2R(11e2)=4e1e2e21

Conclusion:

Substitute 2.718 for e in the above equation.

    (U)growth(U)decay=(4)(2.718)1(2.718)2(2.718)21=0.39

Therefore, the energy dissipated in the resistor when the current decays is more than the energy dissipated when the current grows in the circuit.

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Chapter 33 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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