Chapter 9, Problem 1E

For a certain source-free parallel RLC circuit, R = 1 kΩ, C = 3 μF, and L is such that the circuit response is overdamped. (a) Determine a suitable value of L. (b) Write the equation for the voltage v across the resistor if it is known that v(0⁻) = 9 V and dv/dt|_t=0⁺ = 2 V/s.

To determine

Find inductor $L$.

Answer to Problem 1E

The value of inductor $L$ is $>12\text{H}$.

Explanation of Solution

Given data:

Value of resistance $R$ is $1\text{k}\Omega$ and

Value of capacitor $C$ is $3\mu \text{F}$.

Formula used:

The expression for the two solutions of the characteristic equation of a parallel $RLC$ circuit is as follows:

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

$s_2=-\alpha -\sqrt{{\alpha }^2-{\omega }_0{}^2}$ (2)

Here,

$s_1$ and $s_2$ is the solutions of the characteristic equation of a parallel $RLC$ circuit.

The value of real part of root of natural response of the source free parallel $RLC$ circuit is:

$\alpha =\frac{1}{2RC}$ (3)

The value of imaginary part of root of natural response of the source free parallel $RLC$ circuit is:

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

The expression for over damped source free response of parallel $RLC$ circuit is:

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

Calculation:

Substitute $1\text{k}\Omega$ for $R$ and $3\mu \text{F}$ for $C$ in equation (3):

$\begin{array}{c}\alpha =\frac{1}{2\times 1\text{k}\Omega \times 3\mu \text{F}}\\ =\frac{1}{2\times 1\times {10}^3\Omega \times 3\times {10}^{-6}\text{F}}\left\{\begin{array}{l}\because 1\mu \text{F}={10}^{-6}\text{F}\\ \because \text{ 1 k}\Omega ={10}^3\Omega \end{array}\right\}\end{array}$

$\alpha =\frac{1}{6\times {10}^{-3}}$

Substitute $3\mu \text{F}$ for $C$ in equation (4):

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{L\times 3\mu \text{F}}}\\ =\frac{1}{\sqrt{L\times 3\times {10}^{-6}\text{F}}}\end{array}$

Substitute $\frac{1}{6\times {10}^{-3}}$ for $\alpha$ and $\frac{1}{\sqrt{L\times 3\mu \text{F}}}$ for ${\omega }_0$ in equation (5):

$\begin{array}{l}\frac{1}{6\times {10}^{-3}}>\frac{1}{\sqrt{L\times 3\mu \text{F}}}\\ 6\times {10}^{-3}>\sqrt{L\times 3\mu \text{F}}\end{array}$

Squaring both the sides,

$\begin{array}{l}36\times {10}^{-6}>L\times 3\times {10}^{-6}\text{F}\\ L>\frac{36\times {10}^{-6}}{3\times {10}^{-6}\text{F}}\\ L>12\text{H}\end{array}$

Conclusion:

Thus, the value of inductor $L$ is $>12\text{H}$.

To determine

Find equation for voltage across resistor.

Answer to Problem 1E

The voltage across resistor is $-11.74e^{-120.43t}+20.74e^{-212.91t}\text{V}$.

Explanation of Solution

Given data:

Value of initial voltage $v\left(0^{-}\right)$ is $9\text{V}$.

Value of rate of change of voltage for $t=0^{+}$ is $2\text{V}/\text{s}$.

Calculation:

Choose the value of inductor $L$ as $13\text{H}$.

The general form of natural response for parallel $RLC$ circuit is,

$v\left(t\right)=A_1{e}^{s_{1}t}+A_2{e}^{s_{2}t}$ (6)

Here,

$v\left(t\right)$ is the voltage of parallel $RLC$ circuit,

$s_1$ and $s_2$ are the roots of voltage equation and

$A_1$ and $A_2$ are the arbitrary constants.

Differentiate both sides of equation (6) with respect to time:

$\frac{dv\left(t\right)}{dt}=A_1{s}_1{e}^{s_{1}t}+A_2{s}_2{e}^{s_{2}t}$ (7)

The expression for the voltage across resistor $R$ in parallel $RLC$ circuit is:

$v_R\left(t\right)=A_1{e}^{s_{1}t}+A_2{e}^{s_{2}t}$ (8)

Substitute $0\text{s}$ for $t$ and $9\text{V}$ for $v\left(0^{-}\right)$ in equation (6):

$\begin{array}{l}9\text{V}=A_1{e}^{s_1\times 0}+A_2{e}^{s_2\times 0}\\ 9\text{V}=A_1+A_2\end{array}$

Rearrange for $A_1$,

$A_1=9\text{V}-A_2$ (9)

Substitute $3\mu \text{F}$ for $C$ and $13\text{H}$ for $L$ in equation (4):

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{13\text{H}\times 3\mu \text{F}}}\\ =\frac{1}{\sqrt{13\text{H}\times 3\times {10}^{-6}\text{F}}}\\ =\frac{1}{6.245\times {10}^{-3}}\end{array}$

${\omega }_0=160.128\text{rad}/\text{s}$

Substitute $166.67{\text{s}}^{-1}$ for $\alpha$ and $160.128\text{rad}/\text{s}$ for ${\omega }_0$ in equation (1):

$\begin{array}{c}{s}_1=\left(-166.67+\sqrt{{\left(166.67{\text{s}}^{-1}\right)}^2-{\left(160.128\text{rad/s}\right)}^2}\right){\text{ s}}^{-1}\\ =\left(-166.67+\sqrt{27778.89-25640.97}\right){\text{ s}}^{-1}\\ =\left(-166.67+\sqrt{2137.919}\right){\text{ s}}^{-1}\\ =\left(-166.67+46.238\right){\text{ s}}^{-1}\end{array}$

Solve for $s_1$

$s_1=-120.43{\text{ s}}^{-1}$

Substitute $166.67{\text{s}}^{-1}$ for $\alpha$ and $160.128\text{rad/s}$ for ${\omega }_0$ in equation (2):

$\begin{array}{c}{s}_2=\left(-166.67-\sqrt{{\left(166.67{\text{s}}^{-1}\right)}^2-{\left(160.128\text{rad/s}\right)}^2}\right){\text{ s}}^{-1}\\ =\left(-166.67-\sqrt{27778.89-25640.97}\right){\text{ s}}^{-1}\\ =\left(-166.67-\sqrt{2137.919}\right){\text{ s}}^{-1}\\ =\left(-166.67-46.238\right){\text{ s}}^{-1}\end{array}$

Solve for $s_2$,

$s_2=-212.91{\text{ s}}^{-1}$ (10)

Substitute $2\text{V}/\text{s}$ for ${\frac{dv\left(t\right)}{dt}|}_{t=0}$, $-120.43$ for $s_1$ and $-212.91$ for $s_2$ in equation (7).

$\begin{array}{l}2\text{V}/\text{s}=-120.43A_1{e}^{s_1\times 0}-212.91A_2{e}^{s_2\times 0}\\ 2\text{V}/\text{s}=-120.43A_1-212.91A_2\end{array}$

$-2\text{V/s}=120.43A_1+212.91A_2$ (11)

Substitute $9\text{V}-A_2$ for $A_1$ in equation (11):

$\begin{array}{l}-2\text{V/s}=120.43\left(9\text{V}-A_2\right)+212.91A_2\\ -2\text{V/s}=1083.87\text{V}-120.43A_2+212.91A_2\\ -2\text{V/s}=1083.87\text{V+92}\text{.48}{A}_1\end{array}$

Solve for $A_1$:

$-2\text{V/s}=1083.87\text{V+92}\text{.48}{A}_1$

Rearrange for $A_1$:

$\begin{array}{l}92.48A_1=-1085.87\\ A_1=\frac{-1085.87}{92.48}\end{array}$

$A_1=-11.74$

Substitute $-11.74$ for $A_1$ in equation (9):

$\begin{array}{l}11.74=9\text{V}-A_2\\ A_2=9\text{V+}11.74\end{array}$

$A_2=20.74$

Substitute $-120.43$ for $s_1$ and $-212.91$ for $s_2$, $-11.74$ for $A_1$ and $20.74$ for $A_2$ in equation (6).

$v\left(t\right)=-11.74e^{-120.43t}+20.74e^{-212.91t}\text{V}$

The voltage across resistor is same as the voltage $v\left(t\right)$ due to parallel $RLC$ circuit,

$v_R\left(t\right)=-11.74e^{-120.43t}+20.74e^{-212.91t}\text{V}$

Conclusion:

Thus, the voltage across resistor is $-11.74e^{-120.43t}+20.74e^{-212.91t}\text{V}$.