Chapter 7, Problem 1P

a.

To determine

Find the current values $i_1\left(0^{-}\right)$ and $i_2\left(0^{-}\right)$.

Answer to Problem 1P

The current values $i_1\left(0^{-}\right)$ and $i_2\left(0^{-}\right)$ are 5 mA and 15 mA respectively.

Explanation of Solution

PSPICE Circuit:

Refer to the 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(R3)” in trace expression box.

The current plot $i_1\left(t\right)$ is shown in Figure 3.

From PSPICE output, the initial values of current are,

$\begin{array}{l}{i}_1\left(0^{-}\right)=5\text{mA}\\ i_1\left(0^{-}\right)=i_1\left(0^{+}\right)\\ i_1\left(0^{+}\right)=5\text{mA}\end{array}$

Similarly, type the expression “-I(R2)” in trace expression box to obtain the current $i_2\left(t\right)$.

From PSPICE output, the initial values of current are,

$\begin{array}{l}{i}_2\left(0^{-}\right)=15\text{mA}\\ i_2\left(0^{+}\right)=-5\text{mA}\end{array}$

Conclusion:

Therefore, the values of $i_1\left(0^{-}\right)$ and $i_2\left(0^{-}\right)$ are 5 mA and 15 mA respectively.

b.

To determine

Find the current values $i_1\left(0^{+}\right)$ and $i_2\left(0^{+}\right)$.

Answer to Problem 1P

The current values $i_1\left(0^{+}\right)$ and $i_2\left(0^{+}\right)$ are 5 mA and –5 mA respectively.

Explanation of Solution

Calculation:

From Figure 2 and Figure 3, the current values are,

$\begin{array}{l}{i}_1\left(0^{+}\right)=5\text{mA}\\ i_2\left(0^{+}\right)=-5\text{mA}\end{array}$

Conclusion:

Therefore, the current values $i_1\left(0^{+}\right)$ and $i_2\left(0^{+}\right)$ are 5 mA and –5 mA respectively.

c.

To determine

Find the expression $i_1\left(t\right)$ for $t\ge 0$.

Answer to Problem 1P

The expression $i_1\left(t\right)$ for $t\ge 0$ is $5e^{-20,000t}\text{mA}$.

Explanation of Solution

Calculation:

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

$\begin{array}{c}{R}_{eq}=2\text{kΩ}+6\text{kΩ}\\ =8\text{kΩ}\end{array}$

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 $8\text{kΩ}$ for $R_{eq}$ and 400 mH for L.

$\begin{array}{c}\tau =\frac{400\text{mH}}{8\text{kΩ}}\\ =\frac{400\times {10}^{-3}}{8\times {10}^3}\\ =50\text{μs}\end{array}$

The expression $i_1\left(t\right)$ is,

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

Substitute 5 mA for $i_1\left(0^{+}\right)$ and $50\text{μs}$ for $\tau$.

$\begin{array}{c}{i}_1\left(t\right)=5e^{-\frac{t}{50\text{μs}}}\text{mA}\\ =5e^{-20,000t}\text{mA}\end{array}$

Conclusion:

Therefore, the expression $i_1\left(t\right)$ for $t\ge 0$ is $5e^{-20,000t}\text{mA}$.

d.

To determine

Find the expression $i_2\left(t\right)$ for $t\ge 0$.

Answer to Problem 1P

The expression $i_2\left(t\right)$ for $t\ge 0$ is $-5e^{-20,000t}\text{mA}$.

Explanation of Solution

Calculation:

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

$\begin{array}{c}{R}_{eq}=2\text{kΩ}+6\text{kΩ}\\ =8\text{kΩ}\end{array}$

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 $8\text{kΩ}$ for $R_{eq}$ and 400 mH for L.

$\begin{array}{c}\tau =\frac{400\text{mH}}{8\text{kΩ}}\\ =\frac{400\times {10}^{-3}}{8\times {10}^3}\\ =50\text{μs}\end{array}$

The expression $i_2\left(t\right)$ is,

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

Substitute –5 mA for $i_1\left(0^{+}\right)$ and $50\text{μs}$ for $\tau$.

$\begin{array}{c}{i}_2\left(t\right)=-5e^{-\frac{t}{50\text{μs}}}\text{mA}\\ =-5e^{-20,000t}\text{mA}\end{array}$

Conclusion:

Therefore, the expression $i_2\left(t\right)$ for $t\ge 0$ is $-5e^{-20,000t}\text{mA}$.

e.

To determine

Explain the reason for why $i_2\left(0^{-}\right)\ne i_2\left(0^{+}\right)$.

Explanation of Solution

Calculation:

The current in the resistor changes instantaneously. The switching operation makes the current $i_2\left(0^{-}\right)$ is equal to 15 mA and $i_2\left(0^{+}\right)$ is equal to –5 mA.

Conclusion:

Therefore, the reason for why $i_2\left(0^{-}\right)\ne i_2\left(0^{+}\right)$ is stated.