Chapter 20, Problem 56P

(a)

To determine

How much is the energy stored in the inductor initially?

Answer to Problem 56P

The energy stored in the inductor initially is $38\text{mJ}$.

Explanation of Solution

Write the equation to find the energy stored in an inductor.

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

Here, $U$ is the energy stored in inductor, $L$ is the inductance, $I$ is the current

Write the equation to find the current flowing through the inductor.

$I=\frac{\epsilon }{R}$ (II)

Here, $I$ is the current, $\epsilon$ is the emf, $R$ is the resistance of inductor

Substitute equation (II) in (I) to get $U$.

$U=\frac{1}{2}L{\left(\frac{\epsilon }{R}\right)}^2$ (III)

Conclusion

Substitute $6\text{V}$ for $\epsilon$ , $12\Omega$ for $R$ , $0.30\text{H}$ for $L$ in equation (III) to get $U$.

$\begin{array}{c}U=\frac{1}{2}\left(0.30\text{H}\right){\left(\frac{6.0\text{V}}{12\Omega }\right)}^2\\ =38\text{mJ}\end{array}$

Therefore, The energy stored in the inductor initially is $38\text{mJ}$.

(b)

To determine

What is the instantaneous rate of change of inductor’s energy initially?

Answer to Problem 56P

The instantaneous rate of change of inductor’s energy initially is $-7.5\text{W}$.

Explanation of Solution

The instantaneous rate of change of energy is otherwise called the power of the inductor.

Write the equation to find the power of the inductor.

$P=I\epsilon$ (I)

Here, $P$ is the power, $I$ is the current, $\epsilon$ is the emf of the inductor

Write the equation to find the emf in the inductor.

$\epsilon =IR$ (II)

Here, $I$ is the current, $\epsilon$ is the emf of the inductor, $R$ is the resistance of inductor

Conclusion:

Substitute $-0.50\text{A}$ for $I$ , and $30\Omega$ for $R$ in equation (II) to get $\epsilon$

$\begin{array}{c}\epsilon =\left(0.50\text{A}\right)\left(30\Omega \right)\\ =15\text{V}\end{array}$

Substitute $15\text{V}$ for $\epsilon$ and $-0.50\text{A}$ in equation (I) to get $P$

$\begin{array}{c}P=-\left(0.50\text{A}\right)\left(15\text{V}\right)\\ =-7.5\text{W}\end{array}$

Therefore, The instantaneous rate of change of inductor’s energy initially is $-7.5\text{W}$.

(c)

To determine

What is the average rate of change of the inductor energy between time $0\text{s}\text{to 1}\text{.0}\text{s}$?

Answer to Problem 56P

The average rate of change of the inductor energy between time $0\text{s}\text{to 1}\text{.0}\text{s}$ is $-38\text{mW}$.

Explanation of Solution

Write the equation to find the average power between time $0\text{s}\text{to 1}\text{.0}\text{s}$.

$P_{av}=\frac{U_f-U_i}{t_f-t_i}$ (I)

Here, $P_{av}$ is the average power, $U_f$ is the final energy , $U_i$ is the initial energy, $t_f$ is the final time, $t_i$ is the initial time

Conclusion:

Substitute $0\text{J}$ for $U_f$ , $38\text{mJ}$ for $U_i$ , $1.0\text{s}$ for $t_f$ , and $0\text{s}$ for $t_i$ in equation (I) to get $P_{av}$.

$\begin{array}{c}{P}_{av}=\frac{0-38\times {10}^{-3}\text{J}}{1.0\text{s}}\\ =-38\text{mJ}\end{array}$

Therefore, The average rate of change of the inductor energy between time $0\text{s}\text{to 1}\text{.0}\text{s}$ is $-38\text{mW}$.

(d)

To determine

How long does it take for the current in the inductor to reach $0.0010\text{times}$ its initial value?

Answer to Problem 56P

It lakes $69\text{ms}$ for the current in the inductor to reach $0.0010\text{times}$ its initial value.

Explanation of Solution

Let $t$ be time taken for the current $I$ to become $0.0010\text{times }{I}_0$.

Write the equation to find the current after time $t$.

$I=I_0{e}^{-t/\tau }$ (I)

Here, $I$ is the current after time $t$ , $I_0$ is the initial current, $\tau$ is the decay constant

Take log and simplify equation (I) to get $t$.

$\begin{array}{c}\ln e^{t/\tau }=\ln \frac{I_0}{I}\\ t=\tau \ln \frac{I_0}{I}\\ =\frac{L}{R_1+R_2}\ln \frac{I_0}{I}\end{array}$ (II)

Conclusion:

Substitute $0.30\text{H}$ for $L$ , $18\Omega$ for $R_1$ , $12\Omega$ for $R_2$ , $0.0010{\text{I}}_0$ for $I$ in equation (II) to get $t$.

$\begin{array}{c}t=-\frac{0.30\text{H}}{\left(18\Omega +12\Omega \right)}\ln \frac{I_0}{0.0010I_0}\\ =69\text{ms}\end{array}$

Therefore, it lakes $69\text{s}$ for the current in the inductor to reach $0.0010\text{times}$ its initial value.