Chapter 31, Problem 16P

The switch in Figure P31.15 is open for t < 0 and is then thrown closed at time t = 0. Find (a) the current in the inductor and (b) the current in the switch as functions of time thereafter.

To determine

The current in the inductor.

Answer to Problem 16P

The current in inductor in the terms of time is $\frac{\epsilon }{5R}\left(1-e^{-\frac{2.5R}{L}t}\right)$.

Explanation of Solution

Given info: The value of resistance $R$ is $4.00\Omega$, inductance of the circuit is $1.00\text{H}$ and emf of the battery is $10.0\text{V}$.

Explanation:

Formula to calculate current in a loop as per Kirchhoff law is,

$\begin{array}{c}{\displaystyle \sum i}=0\\ I_s-I-I_L=0\\ I=I_s-I_L\end{array}$ (1)

Here,

$I_s$ is current flowing through switch $\text{s}$.

$I$ is current flowing through resistance $\text{R}$.

$I_l$ is the current flowing through the inductance $L$.

Write the expression for net voltage in loop 1,

$\epsilon -R{I}_s-RI=0$ (2)

Write the expression to calculate net voltage in loop 2,

$\epsilon -R{I}_s-2R{I}_L-L\frac{d{I}_L}{dt}=0$ (3)

Here,

$R$ is resistance of circuit.

$L$ is inductance of circuit.

$\frac{d{I}_L}{dt}$ rate of change of current in inductance.

Substitute $I_s-I_L$ for $I$ in equation (II).

$\begin{array}{c}\epsilon -R{I}_s-R\left(I_s-I_L\right)=0\\ \epsilon -2R{I}_s-R{I}_L=0\\ I_{\text{s}}=\frac{\epsilon }{2R}+\frac{I_L}{2}\end{array}$ (4)

Substitute $\frac{\epsilon }{2R}+\frac{I_L}{2}$ for $I_s$ in equation (3).

$\begin{array}{c}\epsilon -R\left(\frac{\epsilon }{2R}+\frac{I_L}{2}\right)-2R{I}_L-L\frac{d{I}_L}{dt}=0\\ \epsilon -\left(\frac{\epsilon }{2}+\frac{R{I}_L}{2}\right)-2R{I}_L-L\frac{d{I}_L}{dt}=0\\ \frac{\epsilon }{2}-\frac{R{I}_L}{2}-2R{I}_L-L\frac{d{I}_L}{dt}=0\\ \frac{\epsilon }{2}-\frac{5R{I}_L}{2}-L\frac{d{I}_L}{dt}=0\end{array}$

Arrange the terms of above equation to simplify for integration.

$\begin{array}{c}L\frac{d{I}_L}{dt}=\frac{\epsilon }{2}-2.5R{I}_L\\ \frac{L}{2.5R}\frac{d{I}_L}{dt}=\frac{\epsilon }{5R}-I_L\\ -\frac{L}{2.5R}\frac{d{I}_L}{dt}=\left(-\frac{\epsilon }{5R}+I_L\right)\\ \frac{d{I}_L}{\left(-\frac{\epsilon }{5R}+I_L\right)}=-\frac{2.5R}{L}dt\end{array}$

On integrate,

${\displaystyle \int \frac{d{I}_L}{-\frac{\epsilon }{5R}+I_L}}={\displaystyle \int \left(-\frac{2.5R}{L}\right)}dt$ (5)

Assume $-\frac{\epsilon }{5R}+I_L=T$.

Differentiate above equation,

$d{I}_L=dT$

Substitute $dT$ for $d{I}_L$ and $T$ for $-\frac{\epsilon }{5R}+I_L$ in equation (5).

$\begin{array}{l}{\displaystyle \int \frac{d{I}_L}{T}}={\displaystyle \int \left(-\frac{2.5R}{L}\right)}dt\\ {\displaystyle \int \frac{d{I}_L}{T}}=-\frac{2.5R}{L}{\displaystyle \int dt}\\ \ln T=-\frac{2.5R}{L}t+c\end{array}$

Substitute $-\frac{\epsilon }{5R}+I_L$ for $T$ in above equation,

$\ln \left(-\frac{\epsilon }{5R}+I_L\right)=-\frac{2.5R}{L}t+c$ (6)

Apply boundary condition,

Substitute $0$ for $t$ and $0$ for $I_L$ in above equation.

$\begin{array}{c}\ln \left(-\frac{\epsilon }{5R}+0\right)=-\frac{2.5R}{L}\times 0+c\\ c=\ln \left(-\frac{\epsilon }{5R}\right)\end{array}$

Substitute $\ln \left(-\frac{\epsilon }{5R}\right)$ for $c$ in equation (VI)

$\begin{array}{c}\ln \left(-\frac{\epsilon }{5R}+I_L\right)=-\frac{2.5R}{L}t+\ln \left(-\frac{\epsilon }{5R}\right)\\ \ln \left(-\frac{\epsilon }{5R}+I_L\right)-\ln \left(-\frac{\epsilon }{5R}\right)=-\frac{2.5R}{L}t\\ \ln \frac{\left(-\frac{\epsilon }{5R}+I_L\right)}{\left(-\frac{\epsilon }{5R}\right)}=-\frac{2.5R}{L}t\\ \ln \left(1-\frac{I_L}{\frac{\epsilon }{5R}}\right)=-\frac{2.5R}{L}t\end{array}$

Further solve the above expression.

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

Thus, the current in inductor in the terms of time is $I_L=\frac{\epsilon }{5R}\left(1-e^{-\frac{2.5R}{L}t}\right)$.

Conclusion:

Therefore, the current in inductor in the terms of time is $I_L=\frac{\epsilon }{5R}\left(1-e^{-\frac{2.5R}{L}t}\right)$.

To determine

The current in the switch as a function of time.

Answer to Problem 16P

The current in the switch as the function of time is $\frac{3\epsilon }{5R}-\frac{\epsilon }{10R}{e}^{-\frac{2.5R}{L}t}$.

Explanation of Solution

Formula to calculate current in inductor, from equation (VII)

$I_L=\frac{\epsilon }{5R}\left(1-e^{-\frac{2.5R}{L}t}\right)$

Formula to calculate current in switch, from equation, from equation (IV)

$I_{\text{s}}=\frac{\epsilon }{2R}+\frac{I_L}{2}$

Substitute $\frac{\epsilon }{5R}\left(1-e^{-\frac{2.5R}{L}t}\right)$ for $I_L$ in above equation to calculate $I_{\text{s}}$.

$\begin{array}{c}{I}_{\text{s}}=\frac{\epsilon }{2R}+\frac{1}{2}\left(\frac{\epsilon }{5R}\right)\left(1-e^{-\frac{2.5R}{L}t}\right)\\ =\frac{\epsilon }{2R}+\frac{\epsilon }{10R}\left(1-e^{-10.0t}\right)\\ =\frac{\epsilon }{2R}+\frac{\epsilon }{10R}+\frac{\epsilon }{10R}{e}^{-10.0t}\\ =\frac{3\epsilon }{5R}-\frac{\epsilon }{10R}{e}^{-\frac{2.5R}{L}t}\end{array}$

Thus, current in switch is $\frac{3\epsilon }{5R}-\frac{\epsilon }{10R}{e}^{-\frac{2.5R}{L}t}$.

Conclusion:

Therefore, the current in switch is $\frac{3\epsilon }{5R}-\frac{\epsilon }{10R}{e}^{-\frac{2.5R}{L}t}$.