Chapter 7, Problem 1P

a)

To determine

Find the current values $i_o\left(0\right)$ and $i_o\left(\infty \right)$.

Answer to Problem 1P

The current values $i_o\left(0\right)$ and $i_o\left(\infty \right)$ are 0.5 A and 0 A respectively.

Explanation of Solution

PSPICE Circuit:

Refer to Figure P7.1 in the textbook.

Draw the given circuit diagram in PSPICE as shown in Figure 1.

Simulation settings:

Provide the simulation settings as shown in Figure 2.

PSPICE output:

After run the PSPICE circuit a black output screen will be displayed. Right click on the mouse by keeping cursor on the output screen, click the option “Add Trace” and type the expression “I(L1)” in trace expression box.

The current plot $i_o\left(t\right)$ is shown in Figure 3.

From PSPICE output, the initial and final value of output current is,

$\begin{array}{l}{i}_o\left(0\right)=500\text{mA}\left(\text{or}\right)0.5\text{A}\\ i_o\left(\infty \right)=0\text{A}\end{array}$

Conclusion:

Therefore, the current values $i_o\left(0\right)$ and $i_o\left(\infty \right)$ are 0.5 A and 0 A respectively.

b)

To determine

Find the expression $i_o\left(t\right)$ for $t\ge 0$.

Answer to Problem 1P

The expression $i_o\left(t\right)$ for $t\ge 0$ is $-0.5e^{-250t}\text{mA}$.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is closed at $t=0$.

$\begin{array}{c}{R}_{eq}=12\text{Ω}+\text{8}\text{Ω}\\ =20\text{Ω}\end{array}$

Find time constant from the circuit diagram.

$\tau =\frac{L}{R_{eq}}$

Here,

L is the inductance.

$R_{eq}$ is the equivalent resistance across the load.

Substitute $20\text{Ω}$ for $R_{eq}$ and 80 mH for L.

$\begin{array}{c}\tau =\frac{80\text{mH}}{20\text{Ω}}\\ =\frac{80\times {10}^{-3}}{20}\text{ms}\\ =4\text{ms}\end{array}$

The expression $i_o\left(t\right)$ is,

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

Substitute $0\text{A}$ for $i_o\left(\infty \right)$, $0.5\text{A}$ for $i_o\left(0\right)$, and $4\text{ms}$ for $\tau$.

$\begin{array}{c}{i}_o\left(t\right)=0.5\text{A}{e}^{-\frac{t}{4\text{ms}}}\\ =0.5e^{-250t}\text{A}\end{array}$

Conclusion:

Therefore, the expression $i_o\left(t\right)$ for $t\ge 0$ is $-0.5e^{-250t}\text{mA}$.

c)

To determine

Find the time taken to reach the output current of 100 mA after closed the switch.

Answer to Problem 1P

The time required to reach 100 mA of output current is 6.4 ms.

Explanation of Solution

Calculation:

The expression of output current $i_o\left(t\right)$ for $t\ge 0$ is,

$i_o\left(t\right)=0.5e^{-250t}\text{A}$

Substitute 100 mA for $i_o\left(t\right)$.

$\begin{array}{l}100\text{mA}=0.5e^{-250t}\text{A}\\ 0.1=0.5e^{-250t}\\ e^{-250t}=0.2\\ t=6.4\text{ms}\end{array}$

Conclusion:

Therefore, the time required to reach 100 mA of output current is 6.4 ms.