Chapter 7, Problem 57P

To determine

Calculate the current response $i_1\left(t\right)$ and $i_2\left(t\right)$ for $t>0$ in the given circuit of Figure 7.123.

Answer to Problem 57P

The current response $i_1\left(t\right)$ is $2.4e^{-2t}u\left(t\right)\text{A}$ and $i_2\left(t\right)$ is $600e^{-5t}u\left(t\right)\text{mA}$ for all $t>0$.

Explanation of Solution

Given data:

Refer to Figure 7.123 in the textbook.

The value of inductance $L_1$ is $2.5\text{H}$.

The value of inductance $L_2$ is $4\text{H}$.

Formula used:

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

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

Here,

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

$i\left(0\right)$ is the initial 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.

Write the general expression for the unit step function.

$u(t)=\{\begin{array}{l}0,t<0\\ 1,t>0\end{array}$ (3)

Calculation:

For $t<0$:

The given Figure 7.123 is redrawn as shown in Figure 1. At this condition, the inductor reaches steady state and acts like a short circuit to DC.

In Figure 1, apply Kirchhoff’s current law at node $v$.

$\begin{array}{l}5\text{A}=\frac{v}{6\Omega }+\frac{v}{5\Omega }+\frac{v}{20\Omega }\\ 5\text{A}=v\left(\frac{1}{6\Omega }+\frac{1}{5\Omega }+\frac{1}{20\Omega }\right)\\ 5=0.4167v\\ v=\frac{5}{0.4167}\end{array}$

Simplify the equation as follows,

$\begin{array}{c}v=11.999\text{V}\\ \cong 12\text{V}\end{array}$

Now find the initial inductor current $i_1\left(0\right)$ in Figure 1 using Ohm’s law. Therefore,

$i_1\left(0\right)=\frac{v}{5\Omega }$ (4)

Substitute $12\text{V}$ for $v$ in equation (4) to find the current $i_1\left(0\right)$ in amperes.

$\begin{array}{c}{i}_1\left(0\right)=\frac{12\text{V}}{5\text{Ω}}\\ =2.4\frac{\text{V}}{\text{Ω}}\\ =2.4\text{A}\left\{\because 1\text{A}=\frac{\text{1}\text{V}}{1\text{Ω}}\right\}\end{array}$

Now find the initial inductor current $i_2\left(0\right)$ in Figure 1 using Ohm’s law. Therefore,

$i_2\left(0\right)=\frac{v}{20\Omega }$ (5)

Substitute $12\text{V}$ for $v$ in equation (5) to find the current $i_2\left(0\right)$ in amperes.

$\begin{array}{c}{i}_2\left(0\right)=\frac{12\text{V}}{20\text{Ω}}\\ =0.6\frac{\text{V}}{\text{Ω}}\\ =0.6\text{A}\left\{\because 1\text{A}=\frac{\text{1}\text{V}}{1\text{Ω}}\right\}\end{array}$

For $t>0$:

Figure 2 shows the modified circuit diagram when the switch is kept in closed position.

Consider the value of resistances $R_1$ and $R_2$ are $5\text{Ω}$ and $20\text{Ω}$ respectively. Also consider the value of inductances $L_1$ and $L_2$ are $2.5\text{H}$ and $4\text{H}$ respectively.

In Figure 2, the switch is closed so that the energies in the inductance $L_1$ and $L_2$ flows through the closed switch and become dissipated in the resistors $5\text{Ω}$ and $20\text{Ω}$.

The current response $i_1\left(t\right)$ through the RL circuit can be determined by the equation (1) as follows.

$i_1\left(t\right)=i_1\left(0\right)e^{-\frac{t}{{\tau }_1}}$ (6)

The time constant in equation (6) can be rewritten as like in equation (2) as follows.

${\tau }_1=\frac{L_1}{R_1}$ (7)

Substitute $5\text{Ω}$ for $R_1$ and $2.5\text{H}$ for $L_1$ in equation (7) to find the time constant ${\tau }_1$.

${\tau }_1=\frac{2.5\text{H}}{5\text{Ω}}$ (8)

Substitute the units $\frac{\text{V}}{\text{A}}$ for $\Omega$ and $\frac{\text{V}\cdot \text{s}}{\text{A}}$ for H in equation (8) to find the time constant ${\tau }_1$ in seconds.

$\begin{array}{c}{\tau }_1=\frac{2.5\frac{\text{V}\cdot \text{s}}{\text{A}}}{5\frac{\text{V}}{\text{A}}}\\ =0.5\text{s}\end{array}$

Substitute $2.4\text{A}$ for $i_1\left(0\right)$ and $0.5\text{s}$ for ${\tau }_1$ in equation (6) to find the current response $i_1\left(t\right)$ in amperes.

$i_1\left(t\right)=\left(2.4\text{A}\right)e^{-\frac{t}{0.5\text{s}}}$

$i_1\left(t\right)=2.4e^{-2t}\text{A}$ (9)

Apply the unit step function in equation (3) to equation (9).

$\begin{array}{c}{i}_1\left(t\right)=\left(2.4e^{-2t}\text{A}\right)u\left(t\right)\\ =2.4e^{-2t}u\left(t\right)\text{A}\end{array}$

The current response $i_2\left(t\right)$ through the RL circuit can be determined by the equation (1) as follows.

$i_2\left(t\right)=i_2\left(0\right)e^{-\frac{t}{{\tau }_2}}$ (10)

The time constant in equation (10) can be rewritten as like in equation (2) as follows.

${\tau }_2=\frac{L_2}{R_2}$ (11)

Substitute $20\text{Ω}$ for $R_2$ and $4\text{H}$ for $L_2$ in equation (11) to find the time constant ${\tau }_2$.

${\tau }_2=\frac{4\text{H}}{20\text{Ω}}$ (12)

Substitute the units $\frac{\text{V}}{\text{A}}$ for $\Omega$ and $\frac{\text{V}\cdot \text{s}}{\text{A}}$ for H in equation (12) to find the time constant ${\tau }_2$ in seconds.

$\begin{array}{c}{\tau }_2=\frac{4\frac{\text{V}\cdot \text{s}}{\text{A}}}{20\frac{\text{V}}{\text{A}}}\\ =0.2\text{s}\end{array}$

Substitute $0.6\text{A}$ for $i_2\left(0\right)$ and $0.2\text{s}$ for ${\tau }_2$ in equation (10) to find the current response $i_2\left(t\right)$ in amperes.

$i_2\left(t\right)=\left(0.6\text{A}\right)e^{-\frac{t}{0.2\text{s}}}$

$i_2\left(t\right)=0.6e^{-5t}\text{A}$ (13)

Apply the unit step function in equation (3) to equation (13).

$\begin{array}{c}{i}_2\left(t\right)=\left(0.6e^{-5t}\text{A}\right)u\left(t\right)\\ =0.6e^{-5t}u\left(t\right)\text{A}\end{array}$

Convert the unit A to mA.

$\begin{array}{c}{i}_2\left(t\right)=0.6e^{-5t}u\left(t\right)\times {10}^3\times {10}^{-3}\text{A}\\ =600e^{-5t}u\left(t\right)\times {10}^{-3}\text{A}\\ =600e^{-5t}u\left(t\right)\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Conclusion:

Thus, the current response $i_1\left(t\right)$ is $2.4e^{-2t}u\left(t\right)\text{A}$ and $i_2\left(t\right)$ is $600e^{-5t}u\left(t\right)\text{mA}$ for all $t>0$.