Chapter 32, Problem 32.31P

(a)

To determine

The time interval the current reach the value $220\text{mA}$.

Answer to Problem 32.31P

The time interval the current reach the value $220\text{mA}$ is $5.66\text{ms}$.

Explanation of Solution

Given Info: The inductance in the circuit is $140\text{mH}$, the resistance in the circuit is $4.90\Omega$ and the emf of the battery is $6.00\text{V}$.

The equation for the current rise in a $LR$ circuit is,

$I=\frac{\epsilon }{R}\left(1-\exp \left(-\frac{Rt}{L}\right)\right)$

Here,

$\epsilon$ is the emf of the battery.

$R$ is the resistance in the circuit.

$L$ is the inductance.

$t$ is the time interval.

Substitute $140\text{mH}$ for $L$, $4.90\Omega$ for $R$, $6.00\text{V}$ for $\epsilon$ and $220\text{mA}$ for $I$ in the above equation.

$\begin{array}{c}220\text{mA}=\frac{6.00\text{V}}{4.90\Omega }\left(1-\exp \left(-\frac{\left(4.90\Omega \right)t}{140\text{mH}}\right)\right)\\ 220\text{mA}\times \left(\frac{1.00\text{A}}{{10}^3\text{mA}}\right)=1.22\left(1-\exp \left(-\frac{\left(4.90\Omega \right)t}{140\text{mH}\left(\frac{1.00\text{H}}{{10}^3\text{mH}}\right)}\right)\right)\\ 0.220\text{A}=1.22\left(1-\exp \left(-\frac{\left(4.90\Omega \right)t}{0.140\text{H}}\right)\right)\\ \frac{0.220\text{A}}{1.22}=\left(1-\exp \left(-35t\right)\right)\end{array}$

Further solve the above equation.

$\begin{array}{c}\frac{.220\text{A}}{1.22}=\left(1-\exp \left(-35t\right)\right)\\ 1-\frac{.220\text{A}}{1.22}=\exp \left(-35t\right)\\ 0.82=\exp \left(-35t\right)\end{array}$

Take the logarithms of both sides.

$\begin{array}{c}\log \left(0.82\right)=-35t\\ -0.1984=-35t\\ t=\frac{0.1984}{35}\\ =5.67\times {10}^{-3}\text{s}\end{array}$

Further solve the above equation.

$\begin{array}{c}t=5.67\times {10}^{-3}\text{s}\\ \text{=5}\text{.67}\text{ms}\\ \approx \text{5}\text{.66}\text{ms}\end{array}$

Conclusion:

Therefore, the time interval the current reach the value $220\text{mA}$ is $5.66\text{ms}$.

(b)

To determine

The current in the inductor after $10.0\text{s}$.

Answer to Problem 32.31P

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

Explanation of Solution

Given Info: The inductance in the circuit is $140\text{mH}$, the resistance in the circuit is $4.90\Omega$ and the emf of the battery is $6.00\text{V}$.

The equation for the current rise in a $LR$ circuit is,

$I=\frac{\epsilon }{R}\left(1-\exp \left(-\frac{Rt}{L}\right)\right)$

Substitute $140\text{mH}$ for $L$, $4.90\Omega$ for $R$, $6.00\text{V}$ for $\epsilon$ and $10.0\text{s}$ for $t$ in the above equation.

$\begin{array}{c}I=\frac{6.00\text{V}}{4.90\Omega }\left(1-\exp \left(-\frac{4.90\Omega \left(10.0\text{s}\right)}{140\text{mH}}\right)\right)140\text{mH}\left(\frac{1.00\text{H}}{{10}^3\text{mH}}\right)\\ =1.22\left(1-\exp \left(-\frac{4.90\Omega \left(10.0\text{s}\right)}{140\text{mH}\left(\frac{1.00\text{H}}{{10}^3\text{mH}}\right)}\right)\right)\\ =1.22\left(1-\exp \left(-350\right)\right)\\ =1.22\text{A}\end{array}$

Conclusion:

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

(c)

To determine

The time interval the current falls to $160\text{mA}$.

Answer to Problem 32.31P

The time interval the current falls to $160\text{mA}$ is $58.1\text{ms}$.

Explanation of Solution

Given Info: The inductance in the circuit is $140\text{mH}$, the resistance in the circuit is $4.90\Omega$ and the emf of the battery is $6.00\text{V}$.

The equation for the current drop in a $LR$ circuit as the battery is detached is,

$I=\frac{\epsilon }{R}\left(\exp \left(-\frac{Rt}{L}\right)\right)$

Substitute $140\text{mH}$ for $L$, $4.90\Omega$ for $R$, $6.00\text{V}$ for $\epsilon$ and $160\text{mA}$ for $I$ in the above equation.

$\begin{array}{c}160\text{mA}=\frac{6.00\text{V}}{4.90\Omega }\left(\exp \left(-\frac{\left(4.90\Omega \right)t}{140\text{mH}}\right)\right)\\ 160\text{mA}\times \left(\frac{1.00\text{A}}{{10}^3\text{mA}}\right)=1.22\left(\exp \left(-\frac{\left(4.90\Omega \right)t}{140\text{mH}\left(\frac{1.00\text{H}}{{10}^3\text{mH}}\right)}\right)\right)\\ .160\text{A}=1.22\left(\exp \left(-\frac{\left(4.90\Omega \right)t}{0.140\text{H}}\right)\right)\\ \frac{.160\text{A}}{1.22}=\left(\exp \left(-35t\right)\right)\end{array}$

Further solve the above equation.

$\begin{array}{c}\frac{.160\text{A}}{1.22}=\left(\exp \left(-35t\right)\right)\\ \frac{.160\text{A}}{1.22}=\exp \left(-35t\right)\\ 0.131=\exp \left(-35t\right)\end{array}$

Take the logarithms of both sides.

$\begin{array}{c}\log \left(0.131\right)=-35t\\ -2.032=-35t\\ t=\frac{2.032}{35}\\ =58.07\times {10}^{-3}\text{s}\end{array}$

Further solve the above equation.

$\begin{array}{c}t=58.07\times {10}^{-3}\text{s}\\ \text{=58}\text{.07}\text{ms}\\ \approx \text{58}\text{.1}\text{ms}\end{array}$

Conclusion:

Therefore, the time interval the current falls to $160\text{mA}$ is $58.1\text{ms}$.