Chapter 32, Problem 26P

(a)

To determine

The current in the inductor as a function of time.

Answer to Problem 26P

The current in the inductor as a function of time is $500\text{mA}\left(1-e^{-10t}\right)$.

Explanation of Solution

Write the expression based on junction rule.

    ${\displaystyle \sum _{\text{junction}}I}=0$

Here, ${\displaystyle \sum _{\text{junction}}I}$ is the sum current at the junction.

Write the expression to obtain the loop rule.

    ${\displaystyle \sum _{\text{closed}\text{loop}}\Delta V}=0$

Here, ${\displaystyle \sum _{\text{closed}\text{loop}}\Delta V}$ is the potential drop across each element in a closed circuit.

Write the expression to obtain the current in the circuit.

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

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.

The current flow in the circuit is as shown in the figure below.

Figure-(1)

Write the expression to obtain the voltage the loop $ACDFA$.

    $\epsilon =R_1{I}_1+R_2{I}_2+L\frac{d{I}_2}{dt}$                                                                                 (I)

Here, $\epsilon$ is the voltage in the battery, $R_1$ is the resistor across $AB$ $I_1$ is the current across resistor $R_1$, $R_2$ is the resistor across $BC$, $I_2$ is the current across the resistor $R_2$, $L$ is the inductor across $CD$.

Write the expression to obtain the loop $ABEFA$.

    $\epsilon =I_1{R}_1+\left(I_1-I_2\right)R_3$

Here, $R_3$ is the resistor across $BC$.

Re-write the above equation.

    $\begin{array}{c}\epsilon =I_1\left(R_1+R_3\right)-I_2{R}_3\\ \epsilon +I_2{R}_3=I_1\left(R_1+R_3\right)\\ I_1=\frac{\epsilon +I_2{R}_3}{R_1+R_3}\end{array}$                                                                                (II)

Substitute $\frac{\epsilon +I_2{R}_3}{R_1+R_3}$ for $I_1$ in equation (I).

    $\begin{array}{l}\epsilon =R_1\left(\frac{\epsilon +I_2{R}_3}{R_1+R_3}\right)+R_2{I}_2+L\frac{d{I}_2}{dt}\\ L\frac{d{I}_2}{dt}+I_2{R}_2=\epsilon -R_1\left(\frac{\epsilon +I_2{R}_3}{R_1+R_3}\right)\\ L\frac{d{I}_2}{dt}+I_2{R}_2=\frac{\epsilon R_1+\epsilon R_3-\epsilon R_1-I_2{R}_1{R}_3}{R_1+R_3}\\ L\frac{d{I}_2}{dt}=\frac{\epsilon R_3-I_2{R}_1{R}_3}{R_1+R_3}-I_2{R}_2\end{array}$

Further solve the above equation.

    $\begin{array}{l}L\frac{d{I}_2}{dt}=\frac{\epsilon R_3}{R_1+R_3}-\frac{I_2{R}_1{R}_3}{R_1+R_3}-I_2{R}_2\\ L\frac{d{I}_2}{dt}=\frac{\epsilon R_3}{R_1+R_3}-I_2\left(\frac{R_1{R}_3}{R_1+R_3}+R_2\right)\\ \frac{d{I}_2}{dt}=\frac{\epsilon R_3}{L\left(R_1+R_3\right)}-\frac{I_2}{L}\left(R_2+\frac{R_1{R}_3}{R_1+R_3}\right)\\ \frac{d{I}_2}{dt}+\frac{I_2}{L}\left(R_2+\frac{R_1{R}_3}{R_1+R_3}\right)=\frac{\epsilon R_3}{L\left(R_1+R_3\right)}\end{array}$

The general solution of the linear differential equation.

    $I_2{e}^{\frac{1}{L}\left(R_2+\frac{R_1{R}_3}{R_1+R_3}\right)t}=\frac{\epsilon R_3}{\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)}{e}^{\frac{1}{L}\left(R_2+\frac{R_1{R}_3}{R_1+R_3}\right)t}+c$                                         (III)

Here, $c$ is the constant term.

Substitute $0$ for $I_2$ and $t$ in the above equation to find $c$.

    $\begin{array}{c}\left(0\right)e^{\frac{1}{L}\left(R_2+\frac{R_1{R}_3}{R_1+R_3}\right)\left(0\right)}=\frac{\epsilon R_3}{\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)}{e}^{\frac{1}{L}\left(R_2+\frac{R_1{R}_3}{R_1+R_3}\right)\left(0\right)}+c\\ 0=\frac{\epsilon R_3}{\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)}+c\\ c=-\frac{\epsilon R_3}{\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)}\end{array}$

Substitute $-\frac{\epsilon R_3}{\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)}$ for $c$ in equation (III) to find $I_2$.

    $\begin{array}{c}{I}_2{e}^{\frac{1}{L}\left(R_2+\frac{R_1{R}_3}{R_1+R_3}\right)t}=\frac{\epsilon R_3}{\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)}{e}^{\frac{1}{L}\left(R_2+\frac{R_1{R}_3}{R_1+R_3}\right)t}-\frac{\epsilon R_3}{\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)}\\ I_2=\frac{\frac{\epsilon R_3}{\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)}{e}^{\frac{1}{L}\left(R_2+\frac{R_1{R}_3}{R_1+R_3}\right)t}}{e^{\frac{1}{L}\left(R_2+\frac{R_1{R}_3}{R_1+R_3}\right)t}}-\frac{\frac{\epsilon R_3}{\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)}}{e^{\frac{1}{L}\left(R_2+\frac{R_1{R}_3}{R_1+R_3}\right)t}}\\ =\frac{\epsilon R_3}{\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)}-\frac{\epsilon R_3}{\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)}{e}^{-\frac{1}{L}\left(R_2+\frac{R_1{R}_3}{R_1+R_3}\right)t}\\ =\frac{\epsilon R_3}{\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)}\left(1-e^{-\frac{1}{L}\left(R_2+\frac{R_1{R}_3}{R_1+R_3}\right)t}\right)\end{array}$

Substitute $R$ for $R_1$ and $R_3$, $2R$ for $R_2$ in the above equation.

    $\begin{array}{c}{I}_2=\frac{\epsilon R}{\left(2R^2+2R^2+R^2\right)}\left(1-e^{-\frac{1}{L}\left(2R+\frac{R^2}{R+R}\right)t}\right)\\ =\frac{\epsilon R}{\left(5R^2\right)}\left(1-e^{-\frac{1}{L}\left(2R+\frac{R^2}{2R}\right)t}\right)\\ =\frac{\epsilon }{5R}\left(1-e^{-\frac{R}{L}\left(2.5\right)t}\right)\end{array}$

Conclusion:

Substitute $10.0\text{V}$ for $\epsilon$, $1.00\text{H}$ for $L$ and $4.00\text{Ω}$ for $R$ in the above equation to calculate $I_2$.

    $\begin{array}{c}{I}_2=\frac{10.0\text{V}}{5\left(4.00\text{Ω}\right)}\left(1-e^{-\frac{4.00\text{Ω}}{1.00\text{H}}\left(2.5\right)t}\right)\\ =\left(500\times {10}^3\text{A}\times \frac{1\text{mA}}{{10}^{-3}\text{A}}\right)\left(1-e^{-10t}\right)\\ =500\text{mA}\left(1-e^{-10t}\right)\end{array}$

Therefore, the current in the inductor as a function of time is $500\text{mA}\left(1-e^{-10t}\right)$.

(b)

To determine

The current in the switch as a function of time.

Answer to Problem 26P

The current in the switch as a function of time is $1.5\text{A}-\left(0.25\text{A}\right)e^{-10t}$.

Explanation of Solution

Consider equation (II) to obtain the current in the switch as a function of time.

    $I_1=\frac{\epsilon +I_2{R}_3}{R_1+R_3}$

Substitute $R$ for $R_1$ and $R_2$ in the above equation.

    $\begin{array}{c}{I}_1=\frac{\epsilon +I_{2}R}{R+R}\\ =\frac{\epsilon +I_{2}R}{2R}\\ =\frac{\epsilon }{2R}+\frac{I_2}{2}\end{array}$

Further substitute $\frac{\epsilon }{5R}\left(1-e^{-\frac{R}{L}\left(2.5\right)t}\right)$ for $I_2$ in the above equation.

    $\begin{array}{c}{I}_1=\frac{\epsilon }{2R}+\frac{1}{2}\left(\frac{\epsilon }{5R}\left(1-e^{-\frac{R}{L}\left(2.5\right)t}\right)\right)\\ =\frac{\epsilon }{2R}+\frac{\epsilon }{10R}\left(1-e^{-\frac{R}{L}\left(2.5\right)t}\right)\end{array}$

Conclusion:

Substitute $10.0\text{V}$ for $\epsilon$, $1.00\text{H}$ for $L$ and $4.00\text{Ω}$ for $R$ in the above equation to calculate $I_2$.

    $\begin{array}{c}{I}_1=\frac{10.0\text{V}}{2\left(4.00\text{Ω}\right)}+\frac{10.0\text{V}}{10\left(4.00\text{Ω}\right)}\left(1-e^{-\frac{4.00\text{Ω}}{1.00\text{H}}\left(2.5\right)t}\right)\\ =1.5\text{A}-\left(0.25\text{A}\right)e^{-10t}\end{array}$

Therefore, the current in the switch as a function of time is $1.5\text{A}-\left(0.25\text{A}\right)e^{-10t}$.