Chapter 7, Problem 68E

(a)

To determine

Find the energy stored in the element at time $t=0$ for the circuit shown in Figure 7.82.

Answer to Problem 68E

The energy stored in the element at time $t=0$ for the circuit shown in Figure 7.82 is $3\text{nJ}$.

Explanation of Solution

Given data:

Write a general expression to calculate the energy stored in an inductor.

$w=\frac{1}{2}L{i}^2\left(t\right)$        (1)

Here,

$L$ is the value of inductance, and

$i\left(t\right)$ is the current flow through the inductor.

Given data:

Refer to Figure 7.82 in the textbook.

The value of initial current through inductor $i_L\left(0\right)=i\left(0\right)$ is $1\text{mA}$.

Calculation:

Substitute $0$ for $t$ in equation (1) to find $w$ at time $t=0$.

$w=\frac{1}{2}L{\left(i\left(0\right)\right)}^2$

Substitute $1\text{mA}$ for $i\left(0\right)$, and $6\text{mH}$ for $L$ in above equation to find $w$.

$\begin{array}{c}w=\frac{1}{2}\left(6\text{mH}\right){\left(1\text{mA}\right)}^2\\ =\frac{1}{2}\left(6\times {10}^{-3}\text{H}\right){\left(1\times {10}^{-3}\text{A}\right)}^2\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{1}{2}\left(6\times {10}^{-3}\text{H}\right)\left(1\times {10}^{-6}{\text{A}}^2\right)\\ =\frac{1}{2}\left(6\times {10}^{-3}\Omega \cdot \text{s}\right)\left(1\times {10}^{-6}\frac{{\text{C}}^2}{{\text{s}}^2}\right)\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s}\\ \text{1}\text{A}=\frac{1\text{C}}{1\text{s}}\end{array}\right\}\end{array}$

Simplify the above equation to find $w$.

$\begin{array}{c}w=3\times {10}^{-9}\frac{{\text{C}}^2\cdot \left(\frac{\text{V}}{\text{A}}\right)}{\text{s}}\left\{\because 1\Omega =\frac{\text{1}\text{V}}{\text{1}\text{A}}\right\}\\ =3\times {10}^{-9}\frac{{\text{C}}^2\cdot \text{V}}{\frac{\text{C}}{\text{s}}\cdot \text{s}}\left\{\because \text{1}\text{A}=\frac{1\text{C}}{1\text{s}}\right\}\\ =3\times {10}^{-9}\text{C}\cdot \text{V}\\ =3\text{nJ}\left\{\begin{array}{l}\because 1\text{J}=1\text{C}\cdot \text{1}\text{V}\\ \text{1}\text{n}={10}^{-9}\end{array}\right\}\end{array}$

Conclusion:

Thus, the energy stored in the element at time $t=0$ for the circuit shown in Figure 7.82 is $3\text{nJ}$.

(b)

To determine

Determine the value of $i_L$ at time $t=0$, $t=130\text{ns}$, $t=260\text{ns}$ and $t=500\text{ns}$ using transient simulation.

Answer to Problem 68E

The value of $i_L$ at time $t=0$ is $999.99\mu \text{A}$, $t=130\text{ns}$ is $372.96\mu \text{A}$, $t=260\text{ns}$ is $136.81\mu \text{A}$ and $t=500\text{ns}$ is $21.40\mu \text{A}$.

Explanation of Solution

Calculation:

Create the new schematic in LTspice with series connected resistor and inductor of given circuit as shown in Figure 1.

Using SPICE Directive mention the command .ic I(L1)=1m as shown in Figure 2.

Enter the stop time as 500ns, time to start saving data as 0, and maximum Timestep as 25ns in Edit simulation Cmd as shown in Figure 3.

After adding the Spice directives the circuit shows as in Figure 4.

Now run the simulation and place the probe at inductor, the plot of the current through inductor with respect to time is shown as shown in Figure 5.

By placing the cursor on the graph, we obtain the current values for different time as shown in below.

For time $t=0,i_L=999.99\mu \text{A}$

For time $t=130\text{ns},i_L=372.96\mu \text{A}$

For time $t=260\text{ns},i_L=136.81\mu \text{A}$

For time $t=500\text{ns},i_L=21.40\mu \text{A}$

Conclusion:

Thus, the value of $i_L$ at time $t=0$ is $999.99\mu \text{A}$, $t=130\text{ns}$ is $372.96\mu \text{A}$, $t=260\text{ns}$ is $136.81\mu \text{A}$ and $t=500\text{ns}$ is $21.40\mu \text{A}$.

(c)

To determine

Find the value of initial energy remains in the inductor at time $t=130\text{ns}$ and $t=500\text{ns}$.

Answer to Problem 68E

The value of initial energy remains in the inductor at time $t=130\text{ns}$ and $t=500\text{ns}$ is

Explanation of Solution

Calculation:

Refer to part (b), the value of current at time $t=130\text{ns}$ is $i_L\left(130\text{ns}\right)=372.96\mu \text{A}$ and the value of current at time $t=500\text{ns}$ is $i_L\left(500\text{ns}\right)=21.40\mu \text{A}$.

Substitute $130\text{ns}$ for $t$ in equation (1) to find $w$ at time $t=130\text{ns}$.

$w\left(130\text{ns}\right)=\frac{1}{2}L{\left(i_L\left(130\text{ns}\right)\right)}^2$

Substitute $372.96\mu \text{A}$ for $i_L\left(130\text{ns}\right)$, and $6\text{mH}$ for $L$ in above equation to find $w\left(130\text{ns}\right)$.

$\begin{array}{c}w\left(130\text{ns}\right)=\frac{1}{2}\left(6\text{mH}\right){\left(372.96\mu \text{A}\right)}^2\\ =\frac{1}{2}\left(6\times {10}^{-3}\text{H}\right){\left(372.96\times {10}^{-6}\text{A}\right)}^2\left\{\begin{array}{l}\because 1\text{m}={10}^{-3}\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\frac{1}{2}\left(6\times {10}^{-3}\text{H}\right)\left(139.10\times {10}^{-9}{\text{A}}^2\right)\\ =\frac{1}{2}\left(6\times {10}^{-3}\Omega \cdot \text{s}\right)\left(139.10\times {10}^{-9}\frac{{\text{C}}^2}{{\text{s}}^2}\right)\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s}\\ \text{1}\text{A}=\frac{1\text{C}}{1\text{s}}\end{array}\right\}\end{array}$

Simplify the above equation to find $w\left(130\text{ns}\right)$.

$\begin{array}{c}w\left(130\text{ns}\right)=417.3\times {10}^{-12}\frac{{\text{C}}^2\cdot \left(\frac{\text{V}}{\text{A}}\right)}{\text{s}}\left\{\because 1\Omega =\frac{\text{1}\text{V}}{\text{1}\text{A}}\right\}\\ =417.3\times {10}^{-12}\frac{{\text{C}}^2\cdot \text{V}}{\frac{\text{C}}{\text{s}}\cdot \text{s}}\left\{\because \text{1}\text{A}=\frac{1\text{C}}{1\text{s}}\right\}\\ =417.3\times {10}^{-12}\text{C}\cdot \text{V}\\ =417.3\text{pJ}\left\{\begin{array}{l}\because 1\text{J}=1\text{C}\cdot \text{1}\text{V}\\ \text{1}\text{p}={10}^{-12}\end{array}\right\}\end{array}$

Substitute $500\text{ns}$ for $t$ in equation (1) to find $w$ at time $t=500\text{ns}$.

$w\left(500\text{ns}\right)=\frac{1}{2}L{\left(i_L\left(500\text{ns}\right)\right)}^2$

Substitute $21.40\mu \text{A}$ for $i_L\left(500\text{ns}\right)$$i\left(0\right)$, and $6\text{mH}$ for $L$ in above equation to find $w\left(500\text{ns}\right)$.

$\begin{array}{c}w\left(500\text{ns}\right)=\frac{1}{2}\left(6\text{mH}\right){\left(21.40\mu \text{A}\right)}^2\\ =\frac{1}{2}\left(6\times {10}^{-3}\text{H}\right){\left(21.40\times {10}^{-6}\text{A}\right)}^2\left\{\begin{array}{l}\because 1\text{m}={10}^{-3}\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\frac{1}{2}\left(6\times {10}^{-3}\text{H}\right)\left(0.45796\times {10}^{-9}{\text{A}}^2\right)\\ =\frac{1}{2}\left(6\times {10}^{-3}\Omega \cdot \text{s}\right)\left(0.45796\times {10}^{-9}\frac{{\text{C}}^2}{{\text{s}}^2}\right)\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s}\\ \text{1}\text{A}=\frac{1\text{C}}{1\text{s}}\end{array}\right\}\end{array}$

Simplify the above equation to find $w\left(500\text{ns}\right)$.

$\begin{array}{c}w\left(500\text{ns}\right)=1.38\times {10}^{-12}\frac{{\text{C}}^2\cdot \left(\frac{\text{V}}{\text{A}}\right)}{\text{s}}\left\{\because 1\Omega =\frac{\text{1}\text{V}}{\text{1}\text{A}}\right\}\\ =1.38\times {10}^{-12}\frac{{\text{C}}^2\cdot \text{V}}{\frac{\text{C}}{\text{s}}\cdot \text{s}}\left\{\because \text{1}\text{A}=\frac{1\text{C}}{1\text{s}}\right\}\\ =1.38\times {10}^{-12}\text{C}\cdot \text{V}\\ =1.38\text{pJ}\left\{\begin{array}{l}\because 1\text{J}=1\text{C}\cdot \text{1}\text{V}\\ \text{1}\text{p}={10}^{-12}\end{array}\right\}\end{array}$

Conclusion:

Thus, the value of initial energy remains in the inductor at time $t=130\text{ns}$ is $417.3\text{pJ}$ and $t=500\text{ns}$ is $1.38\text{pJ}$.