Chapter 8, Problem 1P

a.

To determine

Find the roots of the characteristic equation that describes the voltage response of the circuit.

Answer to Problem 1P

The roots of the characteristic equation are $\underset{\_}{-100\text{rad}/\text{s}}$ and $\underset{\_}{-400\text{rad}/\text{s}}$.

Explanation of Solution

Given data:

The values of resistance, inductance, and capacitance of a parallel RLC circuit are given as follows:

$\begin{array}{l}R=1\text{k}\Omega \\ L=12.5\text{H}\\ C=2\mu \text{F}\end{array}$

Formula used:

Write the expression for roots of characteristic equation for a parallel RLC circuit as follows:

$s_1,s_2=-\alpha \pm \sqrt{{\alpha }^2-{\omega }_0^2}$        (1)

Here,

$\alpha$ is the neper frequency and

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

Write the expression for Neper frequency as follows:

$\alpha =\frac{1}{2RC}$        (2)

Here,

$R$ is the value of resistance and

$C$ is the value of capacitance.

Write the expression for resonant radian frequency as follows:

${\omega }_0=\frac{1}{\sqrt{LC}}$        (3)

Here,

$L$ is the value of inductance.

Calculation:

Substitute $1\text{k}\Omega$ for $R$ and $2\mu \text{F}$ for $C$ in Equation (2) to obtain the neper frequency.

$\begin{array}{c}\alpha =\frac{1}{\left(2\right)\left(1\text{k}\Omega \right)\left(2\mu \text{F}\right)}\\ =\frac{1}{\left(2\right)\left(1000\Omega \right)\left(2\times {10}^{-6}\text{F}\right)}\\ =250\text{rad}/\text{s}\end{array}$

Substitute 12.5 H for $L$ and $2\mu \text{F}$ for $C$ in Equation (3) to obtain the resonant radian frequency.

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(12.5\text{H}\right)\left(2\mu \text{F}\right)}}\\ =\frac{1}{\sqrt{\left(12.5\text{H}\right)\left(2\times {10}^{-6}\text{F}\right)}}\\ =\frac{1000}{5}\text{rad}/\text{s}\\ =200\text{rad}/\text{s}\end{array}$

Substitute $250\text{rad}/\text{s}$ for $\alpha$ and $200\text{rad}/\text{s}$ for ${\omega }_0$ in Equation (1) to obtain the roots of characteristic equation as follows:

$\begin{array}{c}{s}_1,s_2=-\left(250\text{rad}/\text{s}\right)\pm \sqrt{{\left(250\text{rad}/\text{s}\right)}^2-{\left(200\text{rad}/\text{s}\right)}^2}\\ =-\left(250\text{rad}/\text{s}\right)\pm \sqrt{62500{\left(\text{rad}/\text{s}\right)}^2-40000{\left(\text{rad}/\text{s}\right)}^2}\\ =-\left(250\text{rad}/\text{s}\right)\pm \left(150\text{rad}/\text{s}\right)\end{array}$

The roots are determined as follows:

$\begin{array}{c}{s}_1=-\left(250\text{rad}/\text{s}\right)+\left(150\text{rad}/\text{s}\right)\\ =-100\text{rad}/\text{s}\\ s_2=-\left(250\text{rad}/\text{s}\right)-\left(150\text{rad}/\text{s}\right)\\ =-400\text{rad}/\text{s}\end{array}$

Conclusion:

Thus, the roots of the characteristic equation are $\underset{\_}{-100\text{rad}/\text{s}}$ and $\underset{\_}{-400\text{rad}/\text{s}}$.

b.

To determine

Find whether the response is over-, under-, or critically damped response.

Answer to Problem 1P

The response for the given system is an over-damped response.

Explanation of Solution

Given data:

The value of capacitance is adjusted to give a neper frequency of $5\text{krad}/\text{s}$.

$\alpha =5\text{krad}/\text{s}$

Formula used:

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

${\omega }_0^2<{\alpha }^2$        (4)

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

${\omega }_0^2>{\alpha }^2$        (5)

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

${\alpha }^2={\omega }_0^2$        (6)

Calculation:

From Part (a), the values of $\alpha$ and ${\omega }_0$ are written as follows:

$\begin{array}{l}\alpha =250\text{rad}/\text{s}\\ {\omega }_0=200\text{rad}/\text{s}\end{array}$

Substitute $250\text{rad}/\text{s}$ for $\alpha$ and $200\text{rad}/\text{s}$ for ${\omega }_0$ in Equation (4) as follows:

$\begin{array}{l}{\left(200\text{rad}/\text{s}\right)}^2<{\left(250\text{rad}/\text{s}\right)}^2\\ 40000{\left(\text{rad}/\text{s}\right)}^2<62500{\left(\text{rad}/\text{s}\right)}^2\end{array}$

The expression is satisfied. Therefore, the response is the over-damped response. As the Equation (4) is satisfied, it is not required to check the Equations (5) and (6).

Conclusion:

Thus, the response for the given system is an over-damped response.

c.

To determine

Calculate the value of resistance that yields a damped frequency of $120\text{rad}/\text{s}$.

Answer to Problem 1P

The value of resistance that yields a damped frequency of $120\text{rad}/\text{s}$ is $\underset{\_}{1562.5\Omega }$.

Explanation of Solution

Given data:

${\omega }_d=120\text{rad}/\text{s}$

Formula used:

Write the expression for damped frequency for a parallel RLC circuit as follows:

${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$        (7)

Calculation:

Rearrange the expression in Equation (7) as follows:

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

Substitute $200\text{rad}/\text{s}$ for ${\omega }_0$ and $120\text{rad}/\text{s}$ for ${\omega }_d$ as follows:

$\begin{array}{c}\alpha =\sqrt{{\left(200\text{rad}/\text{s}\right)}^2-{\left(120\text{rad}/\text{s}\right)}^2}\\ =\sqrt{40000{\left(\text{rad}/\text{s}\right)}^2-14400{\left(\text{rad}/\text{s}\right)}^2}\\ =160\text{rad}/\text{s}\end{array}$

Rearrange the expression in Equation (2) as follows:

$R=\frac{1}{2\alpha C}$

Substitute $160\text{rad}/\text{s}$ for $\alpha$ and $2\mu \text{F}$ for $C$ to obtain the required value of resistance.

$\begin{array}{c}R=\frac{1}{\left(2\right)\left(160\text{rad}/\text{s}\right)\left(2\mu \text{F}\right)}\\ =\frac{1}{\left(2\right)\left(160\text{rad}/\text{s}\right)\left(2\times {10}^{-6}\text{F}\right)}\\ =1562.5\Omega \end{array}$

Conclusion:

Thus, the value of resistance that yields a damped frequency of $120\text{rad}/\text{s}$ is $\underset{\_}{1562.5\Omega }$.

d.

To determine

Find the roots of the characteristic equation for the value of resistance obtained in Part (c).

Answer to Problem 1P

The roots of the characteristic equation are $\underset{\_}{\left(-160+j120\right)\text{rad}/\text{s}}$ and $\underset{\_}{\left(-160-j120\right)\text{rad}/\text{s}}$.

Explanation of Solution

Calculation:

From Part (c), the value of resistance is obtained as $1562.5\Omega$.

$R=1562.5\Omega$

Substitute $1562.5\Omega$ for $R$ and $2\mu \text{F}$ for $C$ in Equation (2) to obtain the neper frequency.

$\begin{array}{c}\alpha =\frac{1}{\left(2\right)\left(1562.5\Omega \right)\left(2\mu \text{F}\right)}\\ =\frac{1}{\left(2\right)\left(1562.5\Omega \right)\left(2\times {10}^{-6}\text{F}\right)}\\ =160\text{rad}/\text{s}\end{array}$

Substitute $160\text{rad}/\text{s}$ for $\alpha$ and $200\text{rad}/\text{s}$ for ${\omega }_0$ in Equation (1) to obtain the roots of characteristic equation as follows:

$\begin{array}{c}{s}_1,s_2=-\left(160\text{rad}/\text{s}\right)\pm \sqrt{{\left(160\text{rad}/\text{s}\right)}^2-{\left(200\text{rad}/\text{s}\right)}^2}\\ =-\left(160\text{rad}/\text{s}\right)\pm \sqrt{25600{\left(\text{rad}/\text{s}\right)}^2-40000{\left(\text{rad}/\text{s}\right)}^2}\\ =-\left(160\text{rad}/\text{s}\right)\pm \sqrt{j^{2}14400{\left(\text{rad}/\text{s}\right)}^2}\left\{\because j^2=-1\right\}\\ =\left(-160\pm j120\right)\text{rad}/\text{s}\end{array}$

The roots are determined as follows:

$\begin{array}{c}{s}_1=\left(-160+j120\right)\text{rad}/\text{s}\\ s_2=\left(-160-j120\right)\text{rad}/\text{s}\end{array}$

Conclusion:

Thus, the roots of the characteristic equation are $\underset{\_}{\left(-160+j120\right)\text{rad}/\text{s}}$ and $\underset{\_}{\left(-160-j120\right)\text{rad}/\text{s}}$.

e.

To determine

Calculate the value of resistance $\left(R\right)$ that makes the voltage response critically damped.

Answer to Problem 1P

The value of resistance that makes the voltage response critically damped is $\underset{\_}{1250\Omega }$.

Explanation of Solution

Calculation:

From Equations (2) and (3), substitute $\left(\frac{1}{2RC}\right)$ for $\alpha$ and $\left(\frac{1}{\sqrt{LC}}\right)$ for ${\omega }_0$ in Equation (6) as follows:

$\begin{array}{l}{\left(\frac{1}{2RC}\right)}^2={\left(\frac{1}{\sqrt{LC}}\right)}^2\\ \frac{1}{4R^2{C}^2}=\frac{1}{LC}\end{array}$

Rearrange the expression for resistance of the capacitor.

$\begin{array}{l}{R}^2=\frac{L}{4C}\\ R=\frac{1}{2}\sqrt{\frac{L}{C}}\end{array}$

Substitute 12.5 H for $L$ and $2\mu \text{F}$ for $C$ to obtain the value of resistance.

$\begin{array}{c}R=\frac{1}{2}\sqrt{\frac{12.5\text{H}}{2\mu \text{F}}}\\ =\frac{1}{2}\sqrt{\frac{12.5\text{H}}{2\times {10}^{-6}\text{F}}}\\ =\frac{2500}{2}\Omega \\ =1250\Omega \end{array}$

Conclusion:

Thus, the value of resistance that makes the voltage response critically damped is $\underset{\_}{1250\Omega }$.