Chapter 20, Problem 58E

(a)

To determine

The equation of the current for $t>0$.

Answer to Problem 58E

The equation of current for $t>0$ is satisfied by $i$.

Explanation of Solution

Write the expression for voltage drop across resistance at position 2.

    $V_R=iR$

Here, $V_R$ is the voltage drop across $R$, $R$ is the resistance and $i$ is the current.

Write the expression for voltage drop across inductor at position 2.

    $V_L=-L\frac{di}{dt}$

Here, $V_L$ is the voltage drop across $L$, $L$ is the inductance and $\frac{di}{dt}$ is the current change.

Apply Kirchoff’s volatge law at point 2.

    $V_L=V_R$        (1)

Conclusion:

Substitute $\left(-L\frac{di}{dt}\right)$ for $V_L$ and $iR$ for $V_R$ in the equation (1).

    $\left(-L\frac{di}{dt}\right)=iR$

Rearrange the above equation.

    $-iR-L\frac{di}{dt}=0$

Thus, the equation for current is given above.

(b)

To determine

The value of current for $t>0$.

Answer to Problem 58E

The value of the current is $\left(\frac{\epsilon }{R}{e}^{-Rt/L}\right)$.

Explanation of Solution

Write the equation of current for position 2.

    $-iR-L\frac{di}{dt}=0$        (2)

Write the expression for the initial current at position 2.

    $i_0=\frac{\epsilon }{R}$

Here, $i_0$ is the initial current and $\epsilon$ is the battery voltage.

Conclusion:

Rearrange equation (2).

    $\frac{di}{i}=-\frac{R}{L}dt$

Integrate both side of the above equation to find the current $i$.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{i}{\int }}\frac{di}{i}}=-\frac{R}{L}{\displaystyle \underset{0}{\overset{t}{\int }}dt}\\ {\left[ln\left(i\right)\right]}_{i_0}^i=-\frac{R}{L}{\left[t\right]}_0^t\\ ln\left(\frac{i}{i_0}\right)=-\frac{Rt}{L}\end{array}$

Solve the above equation.

    $i=i_0{e}^{-Rt/L}$

Substitute $\left(\frac{\epsilon }{R}\right)$ for $i_0$ in the above equation.

    $i=\frac{\epsilon }{R}{e}^{-Rt/L}$

Thus, the value of the current is $\left(\frac{\epsilon }{R}{e}^{-Rt/L}\right)$.

(c)

To determine

The value of current for $0\text{s}$.

Answer to Problem 58E

The current for $0\text{s}$ is $\left(\frac{\epsilon }{R}\right)$.

Explanation of Solution

Write the expression for the current in $RL$ circuit for the position 2.

    $i=\frac{\epsilon }{R}{e}^{-Rt/L}$        (3)

Conclusion:

Substitute $0$ for $t$ in equation (3).

    $i=\frac{\epsilon }{R}{e}^{-R\left(0\right)/L}$

Solve the above equation.

    $i=\frac{\epsilon }{R}$

Thus, the current for $0\text{s}$ is $\left(\frac{\epsilon }{R}\right)$.