Chapter 8, Problem 1P

(a)

To determine

Find the initial current in each branch of the circuit using PSPICE.

Answer to Problem 1P

The initial current through resistor, inductor, and capacitor are $\underset{\_}{200\text{mA}}$, $\underset{\_}{-300\text{mA}}$, and $\underset{\_}{100\text{mA}}$ respectively.

Explanation of Solution

Given data:

Refer to Figure given in the textbook.

The circuit parameters are given as follows:

$\begin{array}{l}R=125\Omega \\ L=200\text{mH}\\ C=5\mu \text{F}\end{array}$

The initial current through the inductor $\left(I_0\right)$ is $-0.3\text{A}$.

The initial voltage across the capacitor $\left(V_0\right)$ is 25 V.

Calculation:

As the inductor, capacitor, and resistor are connected in parallel, the initial voltage across each parallel elements are the same. Therefore,

$v\left(0\right)=25\text{V}$

The initial current flowing through the resistor is,

$i_R\left(0\right)=\frac{v\left(0\right)}{R}$

Substitute 25 for $v\left(0\right)$ and 125 for R in the above equation as follows,

$\begin{array}{c}{i}_R\left(0\right)=\frac{25}{125}\\ =200\text{mA}\end{array}$

The initial current through the inductor is,

$i_L\left(0\right)=-300\text{mA}$

The initial current across the capacitor is,

$i_C\left(0\right)=-i_L\left(0\right)-i_R\left(0\right)$

Substitute $-300$ m for $i_L\left(0\right)$ and 200 m for $i_R\left(0\right)$ in above equation as follows,

$\begin{array}{c}{i}_C\left(0\right)=-\left(-300\times {10}^{-3}\right)-\left(200\times {10}^{-3}\right)\\ =100\text{mA}\end{array}$

Conclusion:

Thus, the initial current through resistor, inductor, and capacitor are $\underset{\_}{200\text{mA}}$, $\underset{\_}{-300\text{mA}}$, and $\underset{\_}{100\text{mA}}$ respectively.

(b)

To determine

Find the value of $v\left(t\right)$ for $t\ge 0$.

Answer to Problem 1P

The value of $v\left(t\right)$ for $t\ge 0$ is $25e^{-800t}\cos 600t+66.67e^{-800t}\sin 600t\text{V}$.

Explanation of Solution

Formula used:

Write the condition for over-damped response for a parallel RLC circuit as follows:

${\omega }_0^2<{\alpha }^2$        (1)

Here,

$\alpha$ is the neper frequency and

${\omega }_0$ is the resonant radian frequency.

Write the condition for under-damped response for a parallel RLC circuit as follows:

${\omega }_0^2>{\alpha }^2$        (2)

Write the condition for critically damped response for a parallel RLC circuit as follows:

${\alpha }^2={\omega }_0^2$        (3)

Write the expression for resonant radian frequency for the given circuit as follows:

${\omega }_0=\frac{1}{\sqrt{LC}}$        (4)

Here,

$L$ is the inductance of the inductor, and

$C$ is the capacitance of the capacitor.

Write the expression for neper frequency for the given circuit as follows:

$\alpha =\frac{1}{2RC}$        (5)

Here,

$R$ is the resistance of the resistor.

Write the expression of required voltage response $v\left(t\right)$ for under-damped response of a parallel RLC circuit as follows,

$v\left(t\right)=B_1{e}^{-\alpha t}\cos {\omega }_{d}t+B_2{e}^{-\alpha t}\sin {\omega }_{d}t,t\ge 0$        (6)

Write the general expression for $\frac{dv}{dt}\left(0\right)$ for under-damped response as follows,

$\frac{dv}{dt}\left(0\right)=-\alpha B_1+{\omega }_d{B}_2=\frac{1}{C}{i}_C\left(0\right)$        (7)

Write the general expression for damping constant ${\omega }_d$ as follows,

${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$        (8)

Write the general expression to find the value of $B_1$ as follows,

$v\left(0\right)=B_1=V_0$        (9)

Calculation:

Substitute 125 for $R$ and $5\mu$ for $C$ in Equation (5) to obtain the value of $\alpha$.

$\begin{array}{c}\alpha =\frac{1}{\left(2\right)\left(125\right)\left(5\times {10}^{-6}\right)}\\ =800\text{rad}/\text{s}\end{array}$

Substitute 200 m for $L$ and $5\mu$ for $C$ in Equation (4) to obtain the value of ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(200\times {10}^{-3}\right)\left(5\times {10}^{-6}\right)}}\\ =1000\text{rad}/\text{s}\end{array}$

Substitute 800 for $\alpha$ and 1000 for ${\omega }_0$ in Equation (1) as follows:

${\left(1000\text{rad}/\text{s}\right)}^2>{\left(800\text{rad}/\text{s}\right)}^2$

The expression (2) is satisfied. Therefore, the response is the under-damped response.

Substitute 1000 for ${\omega }_0$ and 800 for $\alpha$ in equation (8) as follows,

$\begin{array}{c}{\omega }_d=\sqrt{{1000}^2-{800}^2}\\ =600\text{rad}/\text{s}\end{array}$

Substitute 25 V for $V_0$ in equation (9) as follows:

$v\left(0\right)=B_1=25$        (11)

Substitute 800 for $\alpha$, 600 for ${\omega }_d$, 25 for $B_1$, $5\mu$ for $C$, and 100 m for $i_C\left(0\right)$ in equation (7) as follows:

$\begin{array}{l}-\left(800\right)\left(25\right)+600B_2=\frac{1}{5\times {10}^{-6}}\left(100\times {10}^{-3}\right)\\ -20000+600B_2=20000\\ B_2=66.67\end{array}$

Substitute 25 for $B_1$, 66.67 for $B_2$, 800 for $\alpha$, and 600 for ${\omega }_d$ in equation (6) as follows:

$v\left(t\right)=25e^{-800t}\cos 600t+66.67e^{-800t}\sin 600t\text{V},t\ge 0$

For various values of t in the above equation the values are calculated and tabulated in Table 1 as follows.

Table 1

Time t in seconds	Voltage $v_o\left(t\right)$ in volts
0.001	26.186
0.002	14.374
0.003	5.374
0.004	1.084
0.005	$-0.2809$
0.006	$-0.4273$
0.007	$-0.2602$
0.008	$-0.1067$

PSPICE Circuit:

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

Provide the simulation settings as shown in Figure 2.

Now, run the simulation and the output will be as shown below.

Output:

TIME        V(N00117)  

   0.000E+00   2.500E+01

   1.000E-03   2.621E+01

   2.000E-03   1.441E+01

   3.000E-03   5.382E+00

   4.000E-03   1.065E+00

   5.000E-03  -2.932E-01

   6.000E-03  -4.354E-01

   7.000E-03  -2.627E-01

   8.000E-03  -1.067E-01

   9.000E-03  -2.581E-02

The simulated output and the calculated values in Table 1 are approximately equal and verified.

Conclusion:

Thus, the value of $v\left(t\right)$ for $t\ge 0$ is $25e^{-800t}\cos 600t+66.67e^{-800t}\sin 600t\text{V}$.

(c)

To determine

Find the value of $i_L\left(t\right)$ for $t\ge 0$.

Answer to Problem 1P

The value of $i_L\left(t\right)$ for $t\ge 0$ is $\left(-300e^{-800t}\cos 600t-191.7e^{-800t}\sin 600t\right)\text{mA}$.

Explanation of Solution

Formula used:

Write the expression for $i_R\left(t\right)$ for $t\ge 0$ as follows:

$i_R\left(t\right)=\frac{v\left(t\right)}{R}$        (13)

Write the expression for $i_C\left(t\right)$ for $t\ge 0$ as follows:

$i_C\left(t\right)=C\frac{d}{dt}\left[v\left(t\right)\right]$        (14)

Write the expression for $i_L\left(t\right)$ for $t\ge 0$ as follows:

$i_L\left(t\right)=-\left[i_R\left(t\right)+i_C\left(t\right)\right]$        (15)

Calculation:

Substitute $25e^{-800t}\cos 600t+66.67e^{-800t}\sin 600t$ for $v\left(t\right)$ and $125$ for $R$ in Equation (13) to obtain the value of $i_R\left(t\right)$ for $t\ge 0$.

$\begin{array}{c}{i}_R\left(t\right)=\frac{\left[25e^{-800t}\cos 600t+66.67e^{-800t}\sin 600t\right]}{125}\\ =\left(0.2e^{-800t}\cos 600t+0.53336e^{-800t}\sin 600t\right)\times {10}^3\times {10}^{-3}\text{A}\\ =\left(200e^{-800t}\cos 600t+533.36e^{-800t}\sin 600t\right)\text{mA}\end{array}$

Substitute $25e^{-800t}\cos 600t+66.67e^{-800t}\sin 600t$ for $v\left(t\right)$ and $5\mu \text{F}$ for $C$ in Equation (14) to obtain the value of $i_C\left(t\right)$ for $t\ge 0$.

$\begin{array}{c}{i}_C\left(t\right)=\left(5\times {10}^{-6}\right)\left[20000e^{-800t}\cos 600t-68333.33e^{-800t}\sin 600t\right]\\ =\left(0.1e^{-800t}\cos 600t-0.34167e^{-800t}\sin 600t\right)\times {10}^3\times {10}^{-3}\text{A}\\ =\left(\text{100}{e}^{-800t}\cos 600t-341.67e^{-800t}\sin 600t\right)\text{mA}\end{array}$

Substitute $\left(200e^{-800t}\cos 600t+533.36e^{-800t}\sin 600t\right)\text{m}$ for $i_R\left(t\right)$ and $\left(\text{100}{e}^{-800t}\cos 600t-341.67e^{-800t}\sin 600t\right)\text{m}$ for $i_C\left(t\right)$ in Equation (15) to obtain the value of $i_L\left(t\right)$

$\begin{array}{c}{i}_L\left(t\right)=-\left[\left(200e^{-800t}\cos 600t+533.36e^{-800t}\sin 600t\right)\text{m}+\left(\text{100}{e}^{-800t}\cos 600t-341.67e^{-800t}\sin 600t\right)\text{m}\right]\\ =\left(-300e^{-800t}\cos 600t-191.7e^{-800t}\sin 600t\right)\text{mA,}t\ge 0\end{array}$

Conclusion:

Thus, the value of $i_L\left(t\right)$ for $t\ge 0$ is $\left(-300e^{-800t}\cos 600t-191.7e^{-800t}\sin 600t\right)\text{mA}$.