Chapter 20, Problem 57E

To determine

To show that $i=i_f\left(1-e^{-t/T_L}\right)$ is a correct solution for $-iR-L\frac{di}{dt}+\epsilon =0$ .

Answer to Problem 57E

It is showed that $i=i_f\left(1-e^{-t/T_L}\right)$ is a correct solution for $-iR-L\frac{di}{dt}+\epsilon =0$ .

Explanation of Solution

Write the equation of the circuit.

    $-iR-L\frac{di}{dt}+\epsilon =0$        (I)

Here, $i$ is the current, $R$ is the resistance, $L$ is the inductance, $\frac{d{i}_1}{dt}$ is the rate of change of current and $\epsilon$ is the EMF.

Write the given solution for the current.

  $i=i_f\left(1-e^{-t/T_L}\right)$        (II)

Here, $i_f$ is the final current and $T_L$ is the time constant.

Take the time derivative of equation (II).

  $\begin{array}{c}\frac{di}{dt}=\frac{d{i}_f}{dt}+\frac{d}{dt}\left(-i_f{e}^{-t/T_L}\right)\\ =0+\left(-i_f\right)e^{-t/T_L}\left(\frac{-1}{T_L}\right)\\ =\frac{i_f}{T_L}{e}^{-t/T_L}\end{array}$        (III)

Write the equation for the final current.

  $i_f=\frac{\epsilon }{R}$        (IV)

Put equation (IV) in equation (II).

  $i=\frac{\epsilon }{R}\left(1-e^{-t/T_L}\right)$        (V)

Put equation (IV) in equation (III).

  $\begin{array}{c}\frac{di}{dt}=\frac{\left(\frac{\epsilon }{R}\right)}{T_L}{e}^{-t/T_L}\\ =\frac{\epsilon }{R{T}_L}{e}^{-t/T_L}\end{array}$        (VI)

Put equations (V) and (VI) in equation (I).

  $\begin{array}{c}-\frac{\epsilon }{R}\left(1-e^{-t/T_L}\right)R-L\frac{\epsilon }{R{T}_L}{e}^{-t/T_L}+\epsilon =0\\ -\epsilon +\epsilon e^{-t/T_L}-L\frac{\epsilon }{R{T}_L}{e}^{-t/T_L}+\epsilon =0\\ \epsilon e^{-t/T_L}-L\frac{\epsilon }{R{T}_L}{e}^{-t/T_L}=0\end{array}$        (VII)

Write the equation for the time constant.

  $T_L=\frac{L}{R}$

Put the above equation in the second term on the left hand side of equation (VII).

  $\begin{array}{c}\epsilon e^{-t/T_L}-L\frac{\epsilon }{R\left(\frac{L}{R}\right)}{e}^{-t/T_L}=0\\ \epsilon e^{-t/T_L}-\frac{L\epsilon R{e}^{-t/T_L}}{RL}=0\\ \epsilon e^{-t/T_L}-\epsilon e^{-t/T_L}=0\end{array}$

The left hand side of the equation is equal to the right hand side so that the solution is correct.

Conclusion:

Therefore, it is showed that $i=i_f\left(1-e^{-t/T_L}\right)$ is a correct solution for $-iR-L\frac{di}{dt}+\epsilon =0$ .