Chapter 20, Problem 68P

(a)

To determine

The maximum current flows through the circuit.

Answer to Problem 68P

The maximum current flows through the circuit is $45\text{mA}$.

Explanation of Solution

Write an expression for the maximum current flows through the circuit.

  $I_0=\frac{\epsilon }{R}$                                                                                                                     (I)

Here, $I_0$ is the maximum current flow through the circuit, $\epsilon$ is the emf and $R$ is the resistance.

Conclusion:

Substitute $9.0\text{V}$ for $\epsilon$ and $200.0\Omega$ for $R$ in equation (I) to find $I_0$.

  $\begin{array}{c}{I}_0=\frac{9.0\text{V}}{200.0\Omega }\\ =\left(45\times {10}^{-3}\text{A}\right)\left(\frac{1\text{mA}}{{10}^{-3}\text{A}}\right)\\ =45\text{mA}\end{array}$

Thus, the maximum current flows through the circuit is $45\text{mA}$.

(b)

To determine

The time taken for reaching the maximum current.

Answer to Problem 68P

The time taken for reaching the maximum current is $1.0\text{ms}$.

Explanation of Solution

Write an expression for the instantaneous current.

  $I\text{'}=I_0\left(1-\exp \left(\frac{-t}{\tau }\right)\right)$                                                                                              (II)

Here, $I\text{'}$ is the instantaneous current, $t$ is the time and $\tau$ is the time constant.

Rearrange equation (II) to find $t$.

  $t=-\tau \ln \left(1-\frac{I\text{'}}{I_0}\right)$                                                                                                   (III)

Write an expression for the time constant.

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

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

Rearrange the equation (III) using equation (IV).

  $t=-\frac{L}{R}\ln \left(1-\frac{I\text{'}}{I_0}\right)$                                                                                                  (V)

Conclusion:

Substitute $0.30\text{H}$ for $L$, $200.0\Omega$ for $R$ and $I_0/2$ for $I\text{'}$  in equation (V) to find $t$.

  $\begin{array}{c}t=-\frac{0.30\text{H}}{200.0\Omega }\ln \left(1-\frac{I_0/2}{I_0}\right)\\ =-\frac{0.30\text{H}}{200.0\Omega }\ln \left(1-\frac{1}{2}\right)\\ =\left(1.0\times {10}^{-3}\text{s}\right)\left(\frac{1\text{ms}}{{10}^{-3}\text{s}}\right)\\ =1.0\text{ms}\end{array}$

Thus, the time taken for reaching the maximum current is $1.0\text{ms}$.

(c)

To determine

The energy stored in the inductor and the rate at which energy is dissipated in the resistor.

Answer to Problem 68P

The energy stored in the inductor is $76\text{μJ}$ and the rate at which energy is dissipated in the resistor is $0.10\text{W}$.

Explanation of Solution

Write an expression for the energy stored in the inductor.

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

Here, $U$ is the energy.

The current is half of the maximum current. Thus, rewrite equation (VI).

  $U=\frac{1}{2}L{\left(I_0/2\right)}^2$                                                                                                   (VII)

Write an expression for the rate at which energy is converted in the resistor.

  $P={I^{\prime }}^{2}R$                                                                                                         (VIII)

The current is half of the maximum current. Thus, rewrite equation (VIII).

  $P={\left(I_0/2\right)}^{2}R$                                                                                                    (IX)

Conclusion:

Substitute $0.30\text{H}$ for $L$, $200.0\Omega$ for $R$ and $45\text{mA}$ for $I_0$  in equation (IX) to find $U$.

  $\begin{array}{c}U=\frac{1}{2}\left(0.30\text{H}\right){\left(\frac{\left(45\text{mA}\right)\left(\frac{1\text{A}}{{10}^3\text{mA}}\right)}{2}\right)}^2\\ =\frac{1}{2}\left(0.30\text{H}\right){\left(\frac{0.045\text{m}}{2}\right)}^2\\ =\left(76\times {10}^{-6}\text{J}\right)\left(\frac{1\text{μJ}}{{10}^{-6}\text{J}}\right)\\ =76\text{μJ}\end{array}$

Substitute $0.30\text{H}$ for $L$, $200.0\Omega$ for $R$ and $45\text{mA}$ for $I_0$  in equation (IX) to find $P$.

  $\begin{array}{c}P={\left(\left(45\text{mA}\right)\left(\frac{1\text{A}}{{10}^3\text{mA}}\right)/2\right)}^2\left(200.0\Omega \right)\\ ={\left(45\times {10}^{-3}\text{A}/2\right)}^2\left(200.0\Omega \right)\\ =0.10\text{W}\end{array}$

Thus, the energy stored in the inductor is $76\text{μJ}$ and the rate at which energy is dissipated in the resistor is $0.10\text{W}$.

(d)

To determine

The maximum current flows through the circuit and the time taken for reaching the maximum current.

Answer to Problem 68P

The energy stored is $110\text{kJ}$.

Explanation of Solution

Write an expression for equivalent resistance.

    $R_{eq}=R+r_L+r$                                                                                                       (X)

Here, $R_{eq}$ is the equivalent resistance, $r_L$  is the internal resistance of inductor and $r$ is the internal resistance of battery.

Write an expression for the maximum current flows through the circuit.

    $I_{\max }=\frac{\epsilon }{R_{eq}}$                                                       (XI)

Write an expression for the time taken for reaching the maximum current.

    $t=-\frac{L}{R_{eq}}\ln \left(1-\frac{I\text{'}}{I_0}\right)$       (XII)

Conclusion:

Substitute $75\Omega$ for $r_L$, $20\Omega$ for $r$ and $200.0\Omega$ for $R$ in equation (X) to find $R_{eq}$.

    $\begin{array}{c}{R}_{eq}=75\Omega +20\Omega +200.0\Omega \\ =295\Omega \end{array}$

Substitute $9.0\text{V}$ for $\epsilon$ and $295\Omega$ for $R_{eq}$ in equation (XI) to find $I_{\max }$.

    $\begin{array}{c}{I}_0=\frac{9.0\text{V}}{295\Omega }\\ =\left(31\times {10}^{-3}\text{A}\right)\left(\frac{1\text{mA}}{{10}^{-3}\text{A}}\right)\\ =31\text{mA}\end{array}$

Substitute $0.30\text{H}$ for $L$, $295\Omega$ for $R_{eq}$ and $I_0/2$ for $I\text{'}$  in equation (XII) to find $t$.

    $\begin{array}{c}t=-\frac{0.30\text{H}}{295\Omega }\ln \left(1-\frac{I_0/2}{I_0}\right)\\ =-\frac{0.30\text{H}}{295\Omega }\ln \left(1-\frac{1}{2}\right)\\ =\left(0.70\times {10}^{-3}\text{s}\right)\left(\frac{1\text{ms}}{{10}^{-3}\text{s}}\right)\\ =0.70\text{ms}\end{array}$

Thus, the maximum current flows through the circuit is $31\text{mA}$ and the time taken for reaching the maximum current is $0.70\text{ms}$.