Chapter 23, Problem 47P

(a)

To determine

The elapse in time interval

Answer to Problem 47P

The time interval elapses by $5.66\text{ms}$.

Explanation of Solution

Write the equation for the change in current in the circuit.

    $I=\frac{\epsilon }{R}(1-e^{-t/\tau })$        (I)

Here, $I$ is the current in the circuit, $\epsilon$ is the emf, $R$ is the resistance, time of flow of current and $\tau$ is the time constant of the circuit.

Write the equation for the time constant for the circuit.

    $\tau =\frac{L}{R}$        (II)

Here, $L$ is the inductance and $R$ is the resistance in the circuit. Substitute $0.140\text{H}$ for $L$ and $4.90\Omega$ for $R$.

    $\begin{array}{c}\tau =\frac{0.140\text{H}}{4.90\Omega }\\ =28.6\text{ms}\end{array}$

Conclusion:

Substitute $6.00\text{V}$ for $\epsilon$, $4.90\Omega$ for $R$ and $220\text{mA}$ for $I$ in equation (I).

    $\begin{array}{c}0.220\text{A}=\frac{6.00\text{V}}{4.90\Omega }(1-e^{-t/\tau })\\ 0.220=1.22(1-e^{-t/\tau })\\ e^{-t/\tau }=0.820\\ t=-\tau \ln \left(0.820\right)\end{array}$

Substitute $28.6\text{ms}$ for $\tau$.

    $\begin{array}{c}t=-\left(28.6\text{ms}\right)\ln \left(0.820\right)\\ =\left(0.028\text{s}\right)\ln \left(0.820\right)\\ =5.66\text{ms}\end{array}$

Therefore, the time interval elapses by $5.66\text{ms}$.

(b)

To determine

The current in the inductor

Answer to Problem 47P

The current in the inductor $10.0\text{s}$ after the switch is closed is $1.22\text{A}$.

Explanation of Solution

The change in current in the circuit is given in equation (I).

    $I=\frac{\epsilon }{R}(1-e^{-t/\tau })$

Here, $I$ is the current in the circuit, $\epsilon$ is the emf, $R$ is the resistance, time of flow of current and $\tau$ is the time constant of the circuit.

Conclusion:

Substitute $6.00\text{V}$ for $\epsilon$, $4.90\Omega$ for $R$, $10.0\text{s}$ for $t$ and $28.6\text{ms}$ for $\tau$ to find the current through the inductor.

    $\begin{array}{c}I=\frac{6.00\text{V}}{4.90\Omega }(1-e^{-10.0\text{s}/0.0286\text{s}})\\ =1.22A\left(1-e^{-350}\right)\\ =1.22\text{A}\end{array}$

Therefore, the current in the inductor $10.0\text{s}$ after the switch is closed is $1.22\text{A}$.

(c)

To determine

The elapse in time before the current fall

Answer to Problem 47P

The time elapse before the current fall is $58.1\text{ms}$.

Explanation of Solution

Write the equation for the decay in the current.

    $I=I_{\max }{e}^{-t/\tau }$        (III)

Here, $I$ is the current after the decay, $I_{\max }$ is the current before decay, $t$ is time taken for the decay and $\tau$ is the time constant. Write the equation for the current before decay.

    $I_{\max }=\frac{\epsilon }{R}$

Here, $\epsilon$ is the emf, $R$ is the resistance in the circuit. Substitute $6.00\text{V}$ for $\epsilon$ and $4.90\Omega$ for $R$.

    $\begin{array}{c}{I}_{\max }=\frac{6.00\text{V}}{4.90\Omega }\\ =1.22\text{A}\end{array}$

Conclusion:

Substitute $160\text{mA}$ for $I$ and $1.22\text{A}$ for $I_{\max }$ in equation (III).

    $\begin{array}{c}0.160\text{A}=1.22\text{A}{e}^{-t/\tau }\\ t=-\tau \ln \left(0.131\right)\end{array}$

Substitute $28.6\text{ms}$ for $\tau$.

    $\begin{array}{c}t=-\left(0.286\text{s}\right)\ln \left(0.131\right)\\ =58.1\text{ms}\end{array}$

Therefore, the time elapse before the current in the inductor falls to $160\text{mA}$ is $58.1\text{ms}$.