Chapter 32, Problem 32.64AP

(a)

To determine

The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

Answer to Problem 32.64AP

The inductor behaves as the short circuit because of no resistance.

Explanation of Solution

Given info: The induced voltage is $10.0\text{V}$, the resistance of the resistor is $4.00\Omega$ and the inductance of the inductor is $1.00\text{H}$.

The inductor has no resistance. If the switch is closed for a long time, then inductor will reach saturation and voltage passes through the inductor. Hence, it behaves as a short circuit.

Conclusion:

Therefore, the inductor behaves as the short circuit because of no resistance.

(b)

To determine

The current carried by the inductor.

Answer to Problem 32.64AP

The current carried by the inductor is $0.500\text{A}$.

Explanation of Solution

Given info: The induced voltage is $10.0\text{V}$, the resistance of the resistor is $4.00\Omega$ and the inductance of the inductor is $1.00\text{H}$.

The figure of the circuit diagrammed referred from P31.15 is shown below.

Figure (1)

The net resistance for parallel combination is,

$\begin{array}{c}\frac{1}{R\text{'}}=\frac{1}{R}+\frac{1}{2R}\\ R\text{'}=\frac{2R}{3}\end{array}$

Here,

$R\text{'}$ is the net resistance for parallel combination.

$R$ is the resistance of the circuit.

The net resistance is connected in series with $R$ for parallel combination is,

$R_{\text{net}}=R+R\text{'}$ (1)

Here,

$R_{\text{net}}$ is the net resistance is connected in series.

Substitute $\frac{2R}{3}$ for $R\text{'}$ in equation (1).

$R_{\text{net}}=R+\frac{2R}{3}$ (2)

Substitute $4.00\Omega$ for $R$ in equation (2) to find the $R_{\text{net}}$.

$\begin{array}{c}{R}_{\text{net}}=4.00\Omega +\frac{2\left(4.00\Omega \right)}{3}\\ =6.67\Omega \end{array}$

Formula to calculate the current of battery is,

$I_{\text{b}}=\frac{\epsilon }{R_{\text{net}}}$

Here,

$I_{\text{b}}$ is the battery current.

$\epsilon$ is the induced voltage.

Substitute $10.0\text{V}$ for $\epsilon$ and $6.67\Omega$ for $R_{\text{net}}$ in equation (3) to find the $I_{\text{b}}$.

$\begin{array}{c}{I}_{\text{b}}=\frac{10.0\text{V}}{6.67\Omega }\\ =1.499\text{A}\\ \approx \text{1}\text{.50}\text{A}\end{array}$

The voltage across the parallel combination of resistors is,

$V=\epsilon -I_{\text{b}}R$ (6)

Substitute $\text{1}\text{.50}\text{A}$ for $I_{\text{b}}$, $10.0\text{V}$ for $\epsilon$ and $4.00\Omega$ for $R$ in equation (6) to find the $V$.

$\begin{array}{c}V=10.0\text{V}-\left(\text{1}\text{.50}\text{A}\right)\left(4.00\Omega \right)\\ =4.00\text{V}\end{array}$

Formula to calculate the current though the inductor is,

$I_{\text{i}}=\frac{V}{2R}$ (7)

Substitute $4.00\text{V}$ for $V$ and $4.00\Omega$ for $R$ in equation (7) to find the $I_{\text{i}}$.

$\begin{array}{c}{I}_{\text{i}}=\frac{4.00\text{V}}{2\left(4.00\Omega \right)}\\ =0.500\text{A}\end{array}$

Conclusion:

Therefore, the current carried by the inductor for $t<0$ is $0.500\text{A}$.

(c)

To determine

The energy stored in the inductor.

Answer to Problem 32.64AP

The energy stored in the inductor for $t<0$  is $0.125\text{J}$.

Explanation of Solution

Given info: The induced voltage is $10.0\text{V}$, the resistance of the resistor is $4.00\Omega$ and the inductance of the inductor is $1.00\text{H}$.

Formula to calculate the energy stored in the inductor is,

$U_{\text{i}}=\frac{1}{2}L{I}_{\text{i}}{}^2$

Substitute $1.00\text{H}$ for $L$ and $0.500\text{A}$ for $I_{\text{i}}$ to find the $U_{\text{i}}$.

$\begin{array}{c}{U}_{\text{i}}=\frac{1}{2}\left(1.00\text{H}\right){\left(0.500\text{A}\right)}^2\\ =0.125\text{J}\end{array}$

Thus, the energy stored in the inductor for $t<0$ is $0.125\text{J}$.

Conclusion:

Therefore, the energy stored in the inductor for $t<0$ is $0.125\text{J}$.

(d)

To determine

The energy previously stored in the inductor after the switch is opened.

Answer to Problem 32.64AP

The energy becomes $0.125\text{J}$ of the additional internal energy in the $R$ and $2R$. in the middle branch after switch is opened.

Explanation of Solution

Given info: The induced voltage is $10.0\text{V}$, the resistance of the resistor is $4.00\Omega$ and the inductance of the inductor is $1.00\text{H}$.

When switch is opened, the energy stored in the inductor will dissipate through resistor $R$ and $2R$. Thus, the energy becomes $0.125\text{J}$ of the additional internal energy in the $R$ and $2R$. in the middle branch.

Conclusion:

Therefore, the energy becomes $0.125\text{J}$ of the additional internal energy in the $R$ and $2R$. in the middle branch after switch is opened.

(e)

To determine

To draw: The graph of the current in the inductor for $t\ge 0$ with the initial and final values and the time constant.

Answer to Problem 32.64AP

Answer The graph of the current in the inductor for $t\ge 0$ and label the initial and final values and the time constant.

Explanation of Solution

Introduction:

The graph of the current verses time shows the variation of the current in the circuit with time and tells the nature of the current.

Explanation:

Given info: The induced voltage is $10.0\text{V}$, the resistance of the resistor is $4.00\Omega$ and the inductance of the inductor is $1.00\text{H}$.

After time $t=0$, the resistance of the circuit will be,

$\begin{array}{c}{R}_1=R+2R\\ =3R\end{array}$

Formula to calculate the time constant is,

$t=\frac{L}{R_1}$ (8)

Substitute $3R$ for $R_1$ and $1.00\text{H}$ for $L$ to find the $t$.

$t=\frac{1.00\text{H}}{3R}$ (9)

Substitute $4.0\Omega$ for $R$ in equation (9) to find the $t$.

$\begin{array}{c}t=\frac{1.00\text{H}}{3\times 4.0\Omega }\\ =0.083\text{s}\end{array}$

The current flowing through the inductor at time $t$ is,

$I=I_0{e}^{-\frac{R}{L}t}$

Substitute $0.500\text{A}$ for $I_0$ and $12.0\Omega$ for $R_1$ and $0.083\text{s}$ for $t$ in equation (10).

$I=\left(0.500\text{A}\right)e^{-\frac{\left(12.0\Omega \right)}{1.0\text{H}}\left(0.083\text{s}\right)}$

Thus, the graph of the current (the initial and final values) in the inductor for $t\ge 0$ and the time constant from equation (11) is shown below.

Figure (2)

The graph shows that current decays with exponentially with time constant. The current decreases from $0.500\text{A}$ towards zero because time constant goes to zero so, the final value of the current gets zero.