Chapter 20, Problem 58P

(a)

To determine

The time taken by the current to reach $67\%$ of its maximum value.

Answer to Problem 58P

The time taken by the current to reach $67\%$ of its maximum value is $5.7\times {10}^{-6}\text{ s}$ .

Explanation of Solution

Write the equation for the current in a $LR$ circuit.

$I=I_f\left(1-e^{-t/\tau }\right)$ (I)

Here, $I$ is the current in the circuit, $I_f$ is the maximum current, $t$ is the time and $\tau$ is the time constant.

Write the equation for the time constant of a $LR$ circuit.

$\tau =\frac{L}{R}$ (II)

Here, $L$ is the inductance and $R$ is the resistance.

Given that the value of the current is $0.67$ times the maximum value of the current.

$I=0.67I_f$ (III)

Put equations (II) and (III) in equation (I) and rewrite the equation for $t$ .

$\begin{array}{c}0.67I_f=I_f\left(1-e^{-tR/L}\right)\\ e^{-tR/L}=1-0.67\\ \frac{-tR}{L}=\ln 0.33\\ t=-\frac{L}{R}\ln 0.33\end{array}$ (IV)

Conclusion:

Given that the value of the inductance is $0.67\text{ mH}$ and the value of the resistance is $130\Omega$ .

Substitute $0.67\text{ mH}$ for $L$ and $130\Omega$ for $R$ in equation (IV) to find $t$ .

$\begin{array}{c}t=-\frac{\left(0.67\text{ mH}\frac{1\text{ H}}{1000\text{ mH}}\right)}{130\Omega }\ln 0.33\\ =5.7\times {10}^{-6}\text{ s}\end{array}$

Therefore, the time taken by the current to reach $67\%$ of its maximum value is $5.7\times {10}^{-6}\text{ s}$ .

(b)

To determine

The maximum energy stored in the conductor.

Answer to Problem 58P

The maximum energy stored in the conductor is $1.1\times {10}^{-5}\text{ J}$ .

Explanation of Solution

Write the equation for the magnetic energy stored in an inductor.

$U=\frac{1}{2}L{I}^2$ (V)

Here, $U$ is the magnetic energy stored in the inductor.

The maximum energy will be stored in the inductor when the current flowing through it is maximum.

Write the equation for the maximum current.

$I=\frac{\epsilon }{R}$

Here, $\epsilon$ is the emf of the battery.

Put the above equation in equation (V).

$\begin{array}{c}U=\frac{1}{2}L{\left(\frac{\epsilon }{R}\right)}^2\\ =\frac{1}{2}L\frac{{\epsilon }^2}{R^2}\end{array}$ (VI)

Conclusion:

Given that the potential difference of the battery is $24\text{ V}$.

Substitute $0.67\text{ mH}$ for $L$ , $24\text{ V}$ for $\epsilon$ and $130\Omega$ for $R$ in equation (VI) to find $U$ .

$\begin{array}{c}U=\frac{1}{2}\left(0.67\text{ mH}\frac{1\text{ H}}{1000\text{ mH}}\right)\frac{{\left(24\text{ V}\right)}^2}{{\left(130\Omega \right)}^2}\\ =\frac{1}{2}\left(0.67\times {10}^{-3}\text{ H}\right)\frac{{\left(24\text{ V}\right)}^2}{{\left(130\Omega \right)}^2}\\ =1.1\times {10}^{-5}\text{ J}\end{array}$

Therefore, the maximum energy stored in the conductor is $1.1\times {10}^{-5}\text{ J}$ .

(c)

To determine

The time taken for the energy stored in the inductor to reach $67\%$ of its maximum value and to compare the answer with result of part (a).

Answer to Problem 58P

The time taken for the energy stored in the inductor to reach $67\%$ of its maximum value is $8.8\times {10}^{-6}\text{ s}$ and the answer is greater than the result of part (a).

Explanation of Solution

According to equation (V), the energy stored is proportional to the square of the current. Apply this on equation (I) to determine the expression for the instantaneous energy stored in the inductor.

$U=U_f{\left(1-e^{-t/\tau }\right)}^2$ (VII)

Here, $U_f$ is the maximum energy stored in the inductor.

Given that the value of the energy stored is $0.67$ times the maximum value of the energy.

$U=0.67U_f$

Put the above equation in equation (VII) and rewrite it for $t$ .

$\begin{array}{c}0.67U_f=U_f{\left(1-e^{-t/\tau }\right)}^2\\ \pm \sqrt{0.67}=1-e^{-t/\tau }\\ \frac{-t}{\tau }=\ln \left(1\pm \sqrt{0.67}\right)\\ t=-\tau \ln \left(1\pm \sqrt{0.67}\right)\end{array}$

Put equation (II) in the above equation.

$t=-\frac{L}{R}\ln \left(1\pm \sqrt{0.67}\right)$ (VIII)

Conclusion:

Substitute $0.67\text{ mH}$ for $L$ and $130\Omega$ for $R$ in equation (VIII) to find $t$ .

$\begin{array}{c}t=-\frac{\left(0.67\text{ mH}\frac{1\text{ H}}{1000\text{ mH}}\right)}{130\Omega }\ln \left(1\pm \sqrt{0.67}\right)\\ =8.8\times {10}^{-6}\text{ s or }-3.1\times {10}^{-6}\text{s}\end{array}$

The negative root has no meaning since time is greater than or equal to zero.

The time calculated is more than the result of part (a). This is because energy stored in the inductor is proportional to the square of the current and it takes longer for the square of the current to be $67\%$ of the maximum square of the current than the current itself to be $67\%$ of the maximum current.

Therefore, the time taken for the energy stored in the inductor to reach $67\%$ of its maximum value is $8.8\times {10}^{-6}\text{ s}$ and the answer is greater than the result of part (a).