Chapter 32, Problem 31P

(a)

To determine

The time interval elapsed before the current reaches $220\text{mA}$ when switch is closed.

Answer to Problem 31P

The time interval elapsed before the current reaches $220\text{mA}$ when switch is closed is $5.66\text{ms}$.

Explanation of Solution

Write the expression to obtain the time taken when current reaches $220\text{mA}$.

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

Here, current in the circuit is $I$, $\epsilon$ is the voltage in the circuit, $R$ is the resistance of the resistor, $L$ is the inductance of the inductor and $t$ is the duration of time.

Conclusion:

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

    $\begin{array}{c}220\text{mA}=\frac{6.00\text{V}}{4.90\text{Ω}}\left(1-e^{\frac{-\left(4.90\text{Ω}\right)t}{140\text{mH}}}\right)\\ \left(1-e^{\frac{-\left(4.90\text{Ω}\right)t}{140\text{mH}\times \frac{1\text{H}}{{10}^3\text{mH}}}}\right)=\frac{\left(220\text{mA}\times \frac{1\text{A}}{{10}^3\text{mA}}\right)\left(4.90\text{Ω}\right)}{6.00\text{V}}\\ e^{\frac{-\left(4.90\text{Ω}\right)t}{140\text{mH}\times \frac{1\text{H}}{{10}^3\text{mH}}}}=1-\frac{\left(220\text{mA}\times \frac{1\text{A}}{{10}^3\text{mA}}\right)\left(4.90\text{Ω}\right)}{6.00\text{V}}\\ e^{\frac{-\left(4.90\text{Ω}\right)t}{140\text{mH}\times \frac{1\text{H}}{{10}^3\text{mH}}}}=0.8203\end{array}$

Further solve the above equation.

    $\begin{array}{c}{e}^{\frac{-\left(4.90\text{Ω}\right)t}{140\text{mH}\times \frac{1\text{H}}{{10}^3\text{mH}}}}=\ln \left(0.8203\right)\\ -35t=-0.1980\\ t=5.659\times {10}^{-3}\text{s}\times \frac{{10}^3\text{ms}}{1\text{s}}\\ =5.66\text{ms}\end{array}$

Therefore, the time interval elapsed before the current reaches $220\text{mA}$ when switch is closed is

$5.66\text{ms}$.

(b)

To determine

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

Answer to Problem 31P

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

Explanation of Solution

Write the expression to obtain the current in the inductor $10.0\text{s}$ after switch is closed.

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

Here, current in the circuit is $I$, $\epsilon$ is the voltage in the circuit, $R$ is the resistance of the resistor, $L$ is the inductance of the inductor and $t$ is the duration of time.

Conclusion:

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

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

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

(c)

To determine

The time interval elapsed before the current in the inductor reaches $160\text{mA}$ when switch is closed.

Answer to Problem 31P

The time interval elapsed before the current in the inductor reaches $220\text{mA}$ when switch is closed is $58.1\text{ms}$.

Explanation of Solution

Write the expression to obtain the time taken when current reaches $220\text{mA}$.

    $I=I_0\left(1-e^{\frac{-Rt}{L}}\right)$

Here, current in the circuit is $I$, $I_0$ is the current in the inductor, $R$ is the resistance of the resistor, $L$ is the inductance of the inductor and $t$ is the duration of time.

Conclusion:

Substitute $160\text{mA}$ for $I$, $4.90\text{Ω}$ for $R$, $140\text{mH}$ for $L$ and $1.22\text{A}$ for $I_0$ in the above equation to calculate $t$.

    $\begin{array}{c}160\text{mA}=\left(1.22\text{A}\right)\left(1-e^{\frac{-\left(4.90\text{Ω}\right)t}{140\text{mH}}}\right)\\ \left(1-e^{\frac{-\left(4.90\text{Ω}\right)t}{140\text{mH}\times \frac{1\text{H}}{{10}^3\text{mH}}}}\right)=\frac{\left(160\text{mA}\times \frac{1\text{A}}{{10}^3\text{mA}}\right)}{1.22\text{A}}\\ e^{\frac{-\left(4.90\text{Ω}\right)t}{140\text{mH}\times \frac{1\text{H}}{{10}^3\text{mH}}}}=1-\frac{\left(160\text{mA}\times \frac{1\text{A}}{{10}^3\text{mA}}\right)}{1.22\text{A}}\\ e^{\frac{-\left(4.90\text{Ω}\right)t}{140\text{mH}\times \frac{1\text{H}}{{10}^3\text{mH}}}}=0.1311\end{array}$

Further solve the above equation.

    $\begin{array}{c}{e}^{\frac{-\left(4.90\text{Ω}\right)t}{140\text{mH}\times \frac{1\text{H}}{{10}^3\text{mH}}}}=\ln \left(0.1311\right)\\ -35t=-2.0314\\ t=0.05804\text{s}\times \frac{{10}^3\text{ms}}{1\text{s}}\\ =58.1\text{ms}\end{array}$

Therefore, the time interval elapsed before the current reaches $220\text{mA}$ when switch is closed is

$58.1\text{ms}$.