Chapter 7, Problem 54P

Obtain the inductor current for both t < 0 and t > 0 in each of the circuits in Fig. 7.120.

To determine

Calculate the value of inductor current $i\left(t\right)$ for both $t<0$ and $t>0$ in the given circuit of Figure 7.120(a).

Answer to Problem 54P

The value of inductor current $i\left(t\right)$ in the circuit of Figure 7.120(a) for $t<0$ is $1\text{A}$ and $t>0$ is $\frac{1}{7}\left(6+e^{-2t}\right)\text{A}$.

Explanation of Solution

Given data:

Refer to Figure 7.120 in the textbook.

The value of inductance L in Figure 7.120(a) is $3.5\text{H}$.

Formula used:

Write the general expression to find the complete response of current for the RL circuit.

$i\left(t\right)=i\left(\infty \right)+\left[i\left(0\right)-i\left(\infty \right)\right]e^{-\frac{t}{\tau }}$ (1)

Here,

$\tau$ is the time constant for the RL circuit,

$i\left(0\right)$ is the initial inductor current, and

$i\left(\infty \right)$ is the final inductor current.

Write the expression to calculate the time constant for the RL circuit.

$\tau =\frac{L}{R_{\text{eq}}}$ (2)

Here,

$R_{\text{eq}}$ is the equivalent resistance of the resistor, and

L is the inductance of the inductor.

Calculation:

Figure 1 shows the modified circuit diagram when $t<0$.

In Figure 1, the switch is kept in open position for all $t<0$. The inductor reaches steady state and acts like a short circuit to dc. The current through the inductor $i\left(0\right)$ is calculated by using current division rule.

$\begin{array}{c}i\left(0\right)=\left(2\text{A}\right)\left(\frac{4\Omega }{4\Omega +4\Omega }\right)\\ =\left(2\text{A}\right)\left(\frac{4}{8}\right)\\ =1\text{A}\end{array}$

Therefore, the inductor current $i\left(t\right)$ for $t<0$ is $1\text{A}$.

Figure 2 shows the modified circuit diagram when the switch is kept in close position for all $t>0$.

Apply Kirchhoff’s current law at node a.

$\begin{array}{l}2=\frac{V_a}{4}+\frac{V_a}{12}+\frac{V_a}{4}\\ 2=V_a\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{4}\right)\\ 2=\left(\frac{7}{12}\right)V_a\end{array}$

Rearrange the equation as follows,

$\begin{array}{c}{V}_a=2\left(\frac{12}{7}\right)\\ =\frac{24}{7}\text{V}\end{array}$

The final inductor current $i\left(\infty \right)$ is calculated by using Ohm’s law.

$i\left(\infty \right)=\frac{V_a}{4\Omega }$

Substitute $\frac{24}{7}\text{V}$ for $V_a$ to find the final inductor current $i\left(\infty \right)$ in amperes.

$\begin{array}{c}i\left(\infty \right)=\frac{\frac{24}{7}\text{V}}{4\Omega }\\ =\frac{24\text{V}}{28\Omega }\\ =\frac{6}{7}\text{A}\left\{\because 1\text{A}=\frac{1\text{V}}{1\Omega }\right\}\end{array}$

Figure 3 shows the equivalent resistance at the inductor terminal.

In Figure 3, the equivalent resistance is calculated as follows.

$\begin{array}{c}{R}_{\text{eq}}=\left(4\Omega \left|\right|12\Omega \right)+4\Omega \\ =\frac{4\Omega \times 12\Omega }{4\Omega +12\Omega }+4\Omega \\ =3\Omega +4\Omega \\ =7\Omega \end{array}$

Substitute $7\Omega$ for $R_{\text{eq}}$ and $3.5\text{H}$ for $L$ in equation (2) to find the time constant $\tau$.

$\tau =\frac{3.5\text{H}}{7\Omega }$ (3)

Substitute the units $\frac{\text{V}}{\text{A}}$ for $\Omega$ and $\frac{\text{V}\cdot \text{s}}{\text{A}}$ for H in equation (3) to find the time constant $\tau$ in seconds.

$\begin{array}{c}\tau =\frac{3.5\frac{\text{V}\cdot \text{s}}{\text{A}}}{7\frac{\text{V}}{\text{A}}}\\ =0.5\text{s}\end{array}$

Substitute $\frac{6}{7}\text{A}$ for $i\left(\infty \right)$, $1\text{A}$ for $i\left(0\right)$, and $0.5\text{s}$ for $\tau$ in equation (1) to find the current response $i\left(t\right)$ in amperes.

$\begin{array}{c}i\left(t\right)=\frac{6}{7}\text{A}+\left[1\text{A}-\frac{6}{7}\text{A}\right]e^{-\frac{t}{0.5\text{s}}}\\ =\frac{6}{7}\text{A}+\left[\frac{1}{7}\text{A}\right]e^{-2t}\\ =\frac{1}{7}\left(6+e^{-2t}\right)\text{A}\end{array}$

Therefore, the inductor current $i\left(t\right)$ for $t>0$ is $\frac{1}{7}\left(6+e^{-2t}\right)\text{A}$.

Conclusion:

Thus, the value of inductor current $i\left(t\right)$ in the circuit of Figure 7.120(a) for $t<0$ is $1\text{A}$ and $t>0$ is $\frac{1}{7}\left(6+e^{-2t}\right)\text{A}$.

To determine

Calculate the value of inductor current $i\left(t\right)$ for both $t<0$ and $t>0$ in given circuit of Figure 7.120(b).

Answer to Problem 54P

The value of inductor current $i\left(t\right)$ in the circuit of Figure 7.120(b) for $t<0$ is $2\text{A}$ and $t>0$ is $3-e^{-\frac{9t}{4}}\text{A}$.

Explanation of Solution

Given data:

Refer to Figure 7.120 in the textbook.

The value of inductance L in Figure 7.120(b) is $2\text{H}$.

Calculation:

Figure 4 shows the modified circuit diagram when switch is kept open for a long time at $t<0$. The inductor reaches steady state and acts like a short circuit to dc.

In Figure 4, $i\left(0\right)$ is calculated by using Kirchhoff’s voltage law.

$\begin{array}{l}10\text{V}=\left(2\Omega \right)i\left(0\right)+\left(3\Omega \right)i\left(0\right)\\ 10\text{V}=\left(2\Omega +3\Omega \right)i\left(0\right)\\ 10\text{V}=\left(5\Omega \right)i\left(0\right)\end{array}$

Rearrange the equation as follows,

$\begin{array}{c}i\left(0\right)=\frac{10\text{V}}{5\Omega }\\ =2\text{A}\left\{\because 1\text{A}=\frac{1\text{V}}{1\Omega }\right\}\end{array}$

Therefore, the inductor current $i\left(t\right)$ for $t<0$ is 2 A.

Figure 5 shows the redrawn circuit when switch is kept in closed position at $t>0$.

In Figure 5, apply Kirchhoff’s current law at node v.

$\begin{array}{l}\frac{10-v}{2}+\frac{24-v}{6}=\frac{v}{3}\\ \frac{10}{2}-\frac{v}{2}+\frac{24}{6}-\frac{v}{6}=\frac{v}{3}\\ \frac{10}{2}+\frac{24}{6}=\frac{v}{3}+\frac{v}{2}+\frac{v}{6}\\ 9=\left(\frac{1}{3}+\frac{1}{2}+\frac{1}{6}\right)v\end{array}$

Rearrange the equation as follows,

$\begin{array}{l}v=\frac{9}{\left(\frac{1}{3}+\frac{1}{2}+\frac{1}{6}\right)}\\ v=9\text{V}\end{array}$

The final inductor current $i\left(\infty \right)$ is calculated by using Ohm’s law.

$i\left(\infty \right)=\frac{v}{3\Omega }$

Substitute $9\text{V}$ for $v$ to find the final inductor current $i\left(\infty \right)$ in amperes.

$\begin{array}{c}i\left(\infty \right)=\frac{9\text{V}}{3\Omega }\\ =3\text{A}\left\{\because 1\text{A}=\frac{1\text{V}}{1\Omega }\right\}\end{array}$

Figure 6 shows the equivalent resistance at the inductor terminal.

In Figure 6, the equivalent resistance is calculated as follows.

$\begin{array}{c}{R}_{\text{eq}}=\left(2\Omega \left|\right|6\Omega \right)+3\Omega \\ =\frac{2\Omega \times 6\Omega }{2\Omega +6\Omega }+3\Omega \\ =1.5\Omega +3\Omega \\ =4.5\Omega \end{array}$

Substitute $4.5\Omega$ for $R_{\text{eq}}$ and $2\text{H}$ for $L$ in equation (2) to find the time constant $\tau$.

$\tau =\frac{2\text{H}}{4.5\Omega }$ (4)

Substitute the units $\frac{\text{V}}{\text{A}}$ for $\Omega$ and $\frac{\text{V}\cdot \text{s}}{\text{A}}$ for H in equation (4) to find the time constant $\tau$ in seconds.

$\begin{array}{c}\tau =\frac{2\frac{\text{V}\cdot \text{s}}{\text{A}}}{4.5\frac{\text{V}}{\text{A}}}\\ =\frac{4}{9}\text{s}\end{array}$

Substitute $3\text{A}$ for $i\left(\infty \right)$, $2\text{A}$ for $i\left(0\right)$, and $\frac{4}{9}\text{s}$ for $\tau$ in equation (1) to find the current response $i\left(t\right)$ in amperes.

$\begin{array}{c}i\left(t\right)=3\text{A}+\left[2\text{A}-3\text{A}\right]e^{-\frac{t}{\frac{4}{9}\text{s}}}\\ =3\text{A}+\left[-1\text{A}\right]e^{-\frac{9t}{4}}\\ =3-e^{-\frac{9t}{4}}\text{A}\end{array}$

Therefore, the inductor current $i\left(t\right)$ for $t>0$ is $3-e^{-\frac{9t}{4}}\text{A}$.

Conclusion:

Thus, the value of inductor current $i\left(t\right)$ in the circuit of Figure 7.120(b) for $t<0$ is $2\text{A}$ and $t>0$ is $3-e^{-\frac{9t}{4}}\text{A}$.