Chapter 6, Problem 65P

The inductors in Fig. 6.87 are initially charged and are connected to the black box at t = 0. If i₁(0) = 4 A, i₂(0) = −2 A, and v(t) = 50e^−200t mV, t ≥ 0, find:

1. (a) the energy initially stored in each inductor,
2. (b) the total energy delivered to the black box from t = 0 to t = ∞,
3. (c) i₁(t) and i₂(t), t ≥ 0,
4. (d) i(t), t ≥ 0.

Figure 6.87

For Prob. 6.65.

To determine

Calculate the initial energy stored in each inductor for the given initial conditions.

Answer to Problem 65P

The energy stored initially in each inductor $5\text{ H}$ and $20\text{ H}$ is $\underset{\_}{40\text{ J}}$ and $\underset{\_}{40\text{ J}}$ respectively

Explanation of Solution

Given data:

The Black box connects across the initially charged inductors at $t=0$.

The initial current of inductor $5\text{ H}$ is $i_1\left(0\right)=4\text{ A}$.

The initial current of inductor $20\text{ H}$ is $i_2\left(0\right)=-2\text{ A}$.

The voltage across the inductors and black box is same. That is,

$v\left(t\right)=50e^{-200t}\text{ mV}$ for $t\ge 0$.

Formula used:

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

$W=\frac{1}{2}L{i}^2$        (1)

Calculation:

Re-draw the given figure as shown in Figure 1.

Using the formula in equation (1), the energy stored initially in inductor $L_1=5\text{ H}$ is,

$W_1=\frac{1}{2}{L}_1{\left(i_1\left(0\right)\right)}^2$

Substitute $5\text{ H}$ for $L_1$ and $4\text{ A}$ for $i_1\left(0\right)$ in above equation.

$\begin{array}{l}{W}_1=\frac{1}{2}\left(5\text{ H}\right){\left(4\text{ A}\right)}^2\\ W_1=\frac{1}{2}\left(5\right)\left(16\right)\text{J}\left\{\because \text{H}\cdot {\text{A}}^2\text{=J}\right\}\\ W_1=40\text{ J}\end{array}$

Using the formula in equation (1), the energy stored initially in inductor $L_2=20\text{ H}$ is,

$W_2=\frac{1}{2}{L}_2{\left(i_2\left(0\right)\right)}^2$

Substitute $20\text{ H}$ for $L_2$ and $-2\text{ A}$ for $i_2\left(0\right)$ in above equation.

$\begin{array}{l}{W}_2=\frac{1}{2}\left(\text{20 H}\right){\left(-\text{2 A}\right)}^2\\ W_2=\frac{1}{2}\left(20\right)\left(4\right)\text{J}\left\{\because \text{H}\cdot {\text{A}}^2\text{=J}\right\}\\ W_2=40\text{ J}\end{array}$

Conclusion:

Thus, the energy stored initially in each inductor $5\text{ H}$ and $20\text{ H}$ is $\underset{\_}{40\text{ J}}$ and $\underset{\_}{40\text{ J}}$ respectively.

To determine

Calculate the total energy delivered to the black box by the inductors for $0<t<\infty$.

Answer to Problem 65P

The total energy delivered to the black box by the inductors is $\underset{\_}{80\text{ J}}$

Explanation of Solution

Given data:

Refer to Part (a).

Formula used:

Write the formula to find the total energy delivered to the black box by the inductors from $0<t<\infty$.

$W=W_1+W_2$        (2)

Here,

$W_1$ is the initial energy in inductor $5\text{ H}$.

$W_2$ is the initial energy in inductor $\text{20 H}$.

Calculation:

The total energy delivered to the black box in period of $0<t<\infty$ is equals to the total initial energy stored in an inductors.

Substitute $40\text{ J}$ for $W_1$ and $40\text{ J}$ for $W_2$ in above equation (2) to find the total energy delivered to the black box.

$\begin{array}{l}W=40\text{ J}+40\text{ J}\\ W=80\text{ J}\end{array}$

Conclusion:

Thus, the total energy delivered to the black box by the inductors is $\underset{\_}{80\text{ J}}$

To determine

Calculate the currents in each inductor for the period of $t\ge 0$.

Answer to Problem 65P

The currents $i_1\left(t\right)$ and $i_2\left(t\right)$ in inductor $5\text{ H}$ and $20\text{ H}$ are $\underset{\_}{\left(5\times {10}^{-5}\left(e^{-200t}-1\right)+4\right)\text{ A }}$ and $\underset{\_}{\left(1.25\times {10}^{-5}\left(e^{-200t}-1\right)-2\right)\text{ A }}$ respectively for the period of $t\ge 0$.

Explanation of Solution

Given data:

Refer to Part (a).

Formula used:

Write the formula to find the current through an inductor.

$i\left(t\right)=\frac{1}{L}{\displaystyle \int v\left(t\right)dt+i\left(0\right)}$ for $t\ge 0$        (3)

Here,

$i\left(0\right)$ is the initial current of an inductor.

$v\left(t\right)$ is the voltage across an inductor.

Calculation:

Using the formula in equation (3), the current through an inductor $L_1=5\text{ H}$ is,

$i_1\left(t\right)=\frac{1}{L_1}{\displaystyle \int v_1\left(t\right)dt+i_1\left(0\right)}$ for $t\ge 0$        (4)

Since the black box and both inductors are in parallel,

$v_1\left(t\right)=v_2\left(t\right)=v\left(t\right)$

Substitute $v\left(t\right)$ for $v_1\left(t\right)$ in equation (4).

$i_1\left(t\right)=\frac{1}{L_1}{\displaystyle \int v\left(t\right)dt+i_1\left(0\right)}$

In the Figure 1, currents $i_1\left(t\right)$ and $i_2\left(t\right)$ are flows in opposite direction to the supposed direction according with black box voltage, to keep the both in same direction change the polarities of voltage as shown in Figure 2.

From Figure 2, the current $i_1\left(t\right)$ for $t\ge 0$ can be written as,

$i_1\left(t\right)=\frac{1}{L_1}{\displaystyle \int -v\left(t\right)dt+i_1\left(0\right)}$

Substitute $5\text{ H}$ for $L_1$, $4\text{ A}$ for $i_1\left(0\right)$ and $50e^{-200t}\text{ mV}$ for $v\left(t\right)$ in above equation.

$\begin{array}{l}{i}_1\left(t\right)=\left(\frac{1}{5}{\displaystyle \int \left(-50e^{-200t}\times {10}^{-3}\right)dt+4}\right)\text{ A}\\ i_1\left(t\right)=\left(\frac{1}{5}{\left[-50\left(-\frac{1}{200}\right)e^{-200t}\times {10}^{-3}\right]}_0^t+4\right)\text{ A }\left\{\because {\displaystyle \int e^{-at}dt=}-\frac{1}{a}{e}^{-at}\right\}\\ i_1\left(t\right)=\left(\frac{1}{5}{\left[\frac{1}{4}{e}^{-200t}\times {10}^{-3}\right]}_0^t+4\right)\text{ A }\\ i_1\left(t\right)=\left(\frac{1}{5}\left(\frac{1}{4}{e}^{-200t}\times {10}^{-3}-\frac{1}{4}{e}^{-200\left(0\right)}\times {10}^{-3}\right)+4\right)\text{ A}\end{array}$

Reduce the equation as follows.

$\begin{array}{l}{i}_1\left(t\right)=\left(\frac{1}{5}\left(\frac{1}{4}{e}^{-200\left(t\right)}\times {10}^{-3}-\frac{1}{4}{e}^{-200\left(0\right)}\times {10}^{-3}\right)+4\right)\text{ A }\\ i_1\left(t\right)=\left(\frac{1}{5}\left(0.25e^{-200t}\times {10}^{-3}-0.25\times {10}^{-3}\right)+4\right)\text{ A}\\ i_1\left(t\right)=\left(0.05e^{-200t}\times {10}^{-3}-0.05\times {10}^{-3}+4\right)\text{ A}\\ i_1\left(t\right)=\left(5\times {10}^{-5}\left(e^{-200t}-1\right)+4\right)\text{ A }\end{array}$

Using the formula in equation (3), the current $i_2\left(t\right)$ for $t\ge 0$ is written as,

$i_2\left(t\right)=\frac{1}{L_2}{\displaystyle \int v_2\left(t\right)dt+i_2\left(0\right)}$ for $t\ge 0$

Substitute $v\left(t\right)$ for $v_2\left(t\right)$ in above equation.

$i_2\left(t\right)=\frac{1}{L_2}{\displaystyle \int v\left(t\right)dt+i_2\left(0\right)}$

Consider reversing polarities for voltage $v\left(t\right)$ as shown in Figure 2.

$i_2\left(t\right)=\frac{1}{L_2}{\displaystyle \int -v\left(t\right)dt+i_2\left(0\right)}$

Substitute $\text{20 H}$ for $L_1$, $-2\text{ A}$ for $i_1\left(0\right)$ and $50e^{-200t}\text{ mV}$ for $v\left(t\right)$ in above equation.

$\begin{array}{l}{i}_2\left(t\right)=\left(\frac{1}{20}{\displaystyle \int \left(-50e^{-200t}\times {10}^{-3}\right)dt-2}\right)\text{ A}\\ i_2\left(t\right)=\left(\frac{1}{20}{\left[-50\left(-\frac{1}{200}\right)e^{-200t}\times {10}^{-3}\right]}_0^t-2\right)\text{ A }\left\{\because {\displaystyle \int e^{-at}dt=}-\frac{1}{a}{e}^{-at}\right\}\\ i_2\left(t\right)=\left(\frac{1}{20}{\left[\frac{1}{4}{e}^{-200t}\times {10}^{-3}\right]}_0^t+4\right)\text{ A }\\ i_2\left(t\right)=\left(\frac{1}{20}\left(\frac{1}{4}{e}^{-200t}\times {10}^{-3}-\frac{1}{4}{e}^{-200\left(0\right)}\times {10}^{-3}\right)-2\right)\text{ A}\end{array}$

Reduce the equation as follows.

$\begin{array}{l}{i}_2\left(t\right)=\left(\frac{1}{20}\left(\frac{1}{4}{e}^{-200\left(t\right)}\times {10}^{-3}-\frac{1}{4}{e}^{-200\left(0\right)}\times {10}^{-3}\right)-2\right)\text{ A }\\ i_2\left(t\right)=\left(\frac{1}{20}\left(0.25e^{-200t}\times {10}^{-3}-0.25\times {10}^{-3}\right)-2\right)\text{ A}\\ i_2\left(t\right)=\left(0.0125e^{-200t}\times {10}^{-3}-0.0125\times {10}^{-3}-2\right)\text{ A}\\ i_2\left(t\right)=\left(1.25\times {10}^{-5}\left(e^{-200t}-1\right)-2\right)\text{ A }\end{array}$

Conclusion:

Thus, the currents $i_1\left(t\right)$ and $i_2\left(t\right)$ in inductor $5\text{ H}$ and $20\text{ H}$ are  $\underset{\_}{\left(5\times {10}^{-5}\left(e^{-200t}-1\right)+4\right)\text{ A }}$ and $\underset{\_}{\left(1.25\times {10}^{-5}\left(e^{-200t}-1\right)-2\right)\text{ A }}$ respectively for the period of $t\ge 0$.

To determine

Find the current $t\left(t\right)$ for $t>0$.

Answer to Problem 65P

The current $t\left(t\right)$ for $t>0$ is $\underset{\_}{6.25\times {10}^{-5}\left(e^{-200t}-1\right)+2\text{ A}}$.

Explanation of Solution

Given data:

Refer to part (a).

Formula used:

Write the formula for the current $i\left(t\right)$ for $t>0$ using Kirchhoff’s current law in Figure 2.

$i\left(t\right)=i_1\left(t\right)+i_2\left(t\right)$        (5)

Here,

$i_1\left(t\right)$ is the in inductor $5\text{ H}$ current.

$i_2\left(t\right)$ is the in inductor $\text{20 H}$ current.

Calculation:

Refer to part (c), the currents $i_1\left(t\right)$ and $i_2\left(t\right)$ are,

$i_1\left(t\right)=\left(5\times {10}^{-5}\left(e^{-200t}-1\right)+4\right)\text{ A }$

$i_2\left(t\right)=\left(1.25\times {10}^{-5}\left(e^{-200t}-1\right)-2\right)\text{ A }$

Substitute $\left(5\times {10}^{-5}\left(e^{-200t}-1\right)+4\right)\text{ A}$ for $i_1\left(t\right)$ and $\left(1.25\times {10}^{-5}\left(e^{-200t}-1\right)-2\right)\text{ A}$ for $i_2\left(t\right)$ in equation (5).

$\begin{array}{l}i\left(t\right)=\left(5\times {10}^{-5}\left(e^{-200t}-1\right)+4\right)+\left(1.25\times {10}^{-5}\left(e^{-200t}-1\right)-2\right)\text{ A}\\ i\left(t\right)=\left(5\times {10}^{-5}+1.25\times {10}^{-5}\right)\left(e^{-200t}-1\right)+4-2\text{ A}\\ i\left(t\right)=6.25\times {10}^{-5}\left(e^{-200t}-1\right)+2\text{ A}\end{array}$

Conclusion:

Thus, the current $t\left(t\right)$ for $t>0$ is $\underset{\_}{6.25\times {10}^{-5}\left(e^{-200t}-1\right)+2\text{ A}}$.