Chapter 16, Problem 7E

(a)

To determine

The determinant and the input impedance of the network when $\omega =100\pi \text{rad}/\text{s}$.

Answer to Problem 7E

The determinant of the network is ${\Delta }_Y=524.12\angle {89.95}^{\text{ο}}$ and the input admittance of the network is $Y_{\text{in}}=0.031\angle -{79.29}^{\text{ο}}\text{S}$.

Explanation of Solution

Given data:

The given diagram is shown in Figure 1.

Calculation:

The conversion of $\text{mH}$ into $\text{H}$ is given by,

$1\text{mH}=1\times {10}^{-3}\text{H}$

The conversion of $\text{100}\text{mH}$ into $\text{H}$ is given by,

$100\text{mH}=0.1\text{H}$

The conversion of $\text{mH}$ into $\text{H}$ is given by,

$1\text{mH}=1\times {10}^{-3}\text{H}$

The conversion of $50\text{mH}$ into $\text{H}$ is given by,

$50\text{mH}=0.050\text{H}$

The conversion of $\text{nF}$ into $\text{F}$ is given by,

$1\text{nF}=1\times {10}^{-9}\text{F}$

The conversion of $20\text{nF}$ into $\text{F}$ is given by,

$20\text{nF}=20\times {10}^{-9}\text{F}$

For $0.1\text{H}$ inductor,

The admittance of inductor is given by,

$Y_{\text{L}}=\frac{1}{j\omega L}$

Substitute $0.1\text{H}$ for $L$ in the above expression.

$\begin{array}{c}{Y}_{\text{L}}=\frac{1}{j\omega \left(0.1\text{H}\right)}\\ =\frac{10}{j\omega }\text{S}\\ =\frac{10}{j100\pi }\text{S}\\ =-\text{0}\text{.032}j\text{S}\end{array}$

For $0.05\text{H}$ inductor,

The admittance of inductor is given by,

$Y_{\text{L}}=\frac{1}{j\omega L}$

Substitute $0.05\text{H}$ for $L$ in the above expression.

$\begin{array}{c}{Y}_{\text{L}}=\frac{1}{j\omega \left(0.05\text{H}\right)}\\ =\frac{20}{j\omega }\text{S}\\ =\frac{20}{j100\pi }\text{S}\\ =-j0.064\text{S}\end{array}$

For $20\times {10}^{-9}\text{F}$ capacitor,

The admittance of capacitor is given by,

$Y_{\text{C}}=j\omega C$

Substitute $20\times {10}^{-9}\text{F}$ for $C$ in the above expression.

$\begin{array}{c}{Y}_{\text{C}}=j\omega \left(20\times {10}^{-9}\text{F}\right)\\ =j\omega \left(2\times {10}^{-8}\right)\text{S}\\ =j100\pi \left(2\times {10}^{-8}\right)\text{S}\\ =j6.28\times {10}^{-6}\text{S}\end{array}$

For $6\Omega$ resistor,

The admittance of inductor is given by,

$Y_{\text{R}}=\frac{1}{R}$

Substitute $6\Omega$ for $R$ in the above expression.

$\begin{array}{c}{Y}_{\text{R}}=\frac{1}{6\Omega }\\ =0.166\text{S}\end{array}$

The required diagram is shown in Figure 2.

The inductor and the capacitor are in series.

The modified diagram is shown in Figure 3.

The expression for $Y_1$ is given by,

$\begin{array}{c}{Y}_1=\frac{10}{j\omega }\parallel \left\{j\omega \left(2\times {10}^{-8}\right)\right\}\\ =\frac{\frac{10}{j\omega }\left\{j\omega \left(2\times {10}^{-8}\right)\right\}}{\frac{10}{j\omega }+\left\{j\omega \left(2\times {10}^{-8}\right)\right\}}\\ =\frac{j\omega \left(20\times {10}^{-8}\right)}{10-{\omega }^2\left(2\times {10}^{-8}\right)}\text{S}\end{array}$

The admittance $Y_1$ and resistance are connected in parallel.

The modified diagram is shown in Figure 4.

The expression of $Y_2$ is given by,

$\begin{array}{c}{Y}_2=0.1666+\frac{j\omega \left(20\times {10}^{-8}\right)}{10-{\omega }^2\left(2\times {10}^{-8}\right)}\\ =\frac{0.1666\left(10-{\omega }^2\left(2\times {10}^{-8}\right)\right)+j\omega \left(20\times {10}^{-8}\right)}{10-{\omega }^2\left(2\times {10}^{-8}\right)}\\ =\frac{1.666-{\omega }^2\left(3.332\times {10}^{-9}\right)+j\omega \left(20\times {10}^{-8}\right)}{10-{\omega }^2\left(2\times {10}^{-8}\right)}\end{array}$

The admittance $Y_2$ and inductor are connected in series.

The modified diagram is shown in Figure 5.

The expression of $Y_3$ is given by,

$\frac{1}{Y_3}=\frac{j\omega }{10}+\frac{1}{Y_2}$

Substitute $\frac{1.666-{\omega }^2\left(3.332\times {10}^{-9}\right)+j\omega \left(20\times {10}^{-8}\right)}{10-{\omega }^2\left(2\times {10}^{-8}\right)}$ for $Y_2$ in the above expression.

$\begin{array}{c}\frac{1}{Y_3}=0.1j\omega +\frac{10-{\omega }^2\left(2\times {10}^{-8}\right)}{1.666-{\omega }^2\left(3.332\times {10}^{-9}\right)+j\omega \left(20\times {10}^{-8}\right)}\\ =\frac{0.1j\omega \left(1.666-{\omega }^2\left(3.332\times {10}^{-9}\right)+j\omega \left(20\times {10}^{-8}\right)\right)+10-{\omega }^2\left(2\times {10}^{-8}\right)}{1.666-{\omega }^2\left(3.332\times {10}^{-9}\right)+j\omega \left(20\times {10}^{-8}\right)}\\ =\frac{\left(0.1666j\omega -j{\omega }^3\left(3.332\times {10}^{-10}\right)+j\omega \left(20\times {10}^{-8}\right)\right)+10-{\omega }^2\left(2\times {10}^{-8}\right)}{1.666-{\omega }^2\left(3.332\times {10}^{-9}\right)+j\omega \left(20\times {10}^{-8}\right)}\\ =\frac{\left(10+0.1666j\omega -j{\omega }^3\left(3.332\times {10}^{-10}\right)+{\omega }^2\left(4\times {10}^{-8}\right)\right)}{1.666-{\omega }^2\left(3.332\times {10}^{-9}\right)+j\omega \left(20\times {10}^{-8}\right)}\end{array}$

Further solve as,

$Y_3=\frac{1.666-{\omega }^2\left(3.332\times {10}^{-9}\right)+j\omega \left(20\times {10}^{-8}\right)}{10+0.1666j\omega -j{\omega }^3\left(3.332\times {10}^{-10}\right)+{\omega }^2\left(4\times {10}^{-8}\right)}$

The input admittance is the parallel combination of the inductor and $Y_3$.

The expression of the input admittance $Y_{\text{in}}$ is given by,

$Y_{\text{in}}=\frac{1}{20j\omega }+Y_3$

Substitute $\frac{1.666-{\omega }^2\left(3.332\times {10}^{-9}\right)+j\omega \left(20\times {10}^{-8}\right)}{10+0.1666j\omega -j{\omega }^3\left(3.332\times {10}^{-10}\right)+{\omega }^2\left(4\times {10}^{-8}\right)}$ for $Y_3$ in the above expression.

$\begin{array}{c}{Y}_{\text{in}}=\frac{1}{20j\omega }+\frac{1.666-{\omega }^2\left(3.332\times {10}^{-9}\right)+j\omega \left(20\times {10}^{-8}\right)}{10+0.1666j\omega -j{\omega }^3\left(3.332\times {10}^{-10}\right)+{\omega }^2\left(4\times {10}^{-8}\right)}\\ =\frac{\left\{\begin{array}{c}10+0.1666j\omega -j{\omega }^3\left(3.332\times {10}^{-10}\right)+{\omega }^2\left(4\times {10}^{-8}\right)\\ +20j\omega \left(1.666-{\omega }^2\left(3.332\times {10}^{-9}\right)+j\omega \left(20\times {10}^{-8}\right)\right)\end{array}\right\}}{20j\omega \left(10+0.1666j\omega -j{\omega }^3\left(3.332\times {10}^{-10}\right)+{\omega }^2\left(4\times {10}^{-8}\right)\right)}\\ =\frac{\begin{array}{l}10+0.1666j\omega -j{\omega }^3\left(3.332\times {10}^{-10}\right)+{\omega }^2\left(4\times {10}^{-8}\right)\\ +33.2j\omega -j{\omega }^3\left(66.64\times {10}^{-9}\right)-{\omega }^2\left(4\times {10}^{-6}\right)\end{array}}{200j\omega -3.332{\omega }^2+{\omega }^4\left(66.64\times {10}^{-10}\right)-j{\omega }^3\left(80\times {10}^{-8}\right)}\\ =\frac{10+33.366j\omega -{\omega }^2\left(4.04\times {10}^{-6}\right)-j{\omega }^3\left(6.697\times {10}^{-8}\right)}{200j\omega -3.332{\omega }^2+{\omega }^4\left(66.64\times {10}^{-10}\right)-j{\omega }^3\left(80\times {10}^{-8}\right)}\end{array}$

Substitute $100\pi$ for $\omega$ in the above expression.

$\begin{array}{c}{Y}_{\text{in}}=\frac{\left\{\begin{array}{l}10+33.366j\left(100\pi \right)-{\left(100\pi \right)}^2\left(4.04\times {10}^{-6}\right)\\ -j{\left(100\pi \right)}^3\left(6.697\times {10}^{-8}\right)\end{array}\right\}}{\left\{\begin{array}{l}200j\left(100\pi \right)-3.332{\left(100\pi \right)}^2\\ +{\left(100\pi \right)}^4\left(66.64\times {10}^{-10}\right)-j{\left(100\pi \right)}^3\left(80\times {10}^{-8}\right)\end{array}\right\}}\\ =\frac{9.602+j10480.124}{-328851.97+j62807.09}\\ =\frac{10480.12\angle {89.9}^{\text{ο}}}{334795.98\angle {169.87}^{\text{ο}}}\\ =0.031\angle -{79.29}^{\text{ο}}\end{array}$

Mark the node voltages as $V_1$ and $V_2$ in Figure 1.

The required diagram is shown in Figure 6.

The equation at node voltage $V_1$ using Kirchhoff’s current law is given by,

$\begin{array}{l}{V}_1\left(\frac{1}{20j\omega }\right)+\left(V_1-V_2\right)\left(\frac{10}{j\omega }\right)=0\\ 0.05V_1+\left(V_1-V_2\right)10=0\\ 10.05V_1-10V_2=0\end{array}$        (1)

The equation at node voltage $V_2$ using Kirchhoff’s current law is given by,

$\begin{array}{l}\left(V_2-V_1\right)\left(\frac{10}{j\omega }\right)+0.166V_2+V_2\left(\frac{10}{j\omega }\parallel \left\{j\omega \left(2\times {10}^{-8}\right)\right\}\right)=0\\ \left(V_2-V_1\right)\left(\frac{10}{j\omega }\right)+0.166V_2+V_2\left(\frac{j\omega \left(20\times {10}^{-8}\right)}{10-{\omega }^2\left(2\times {10}^{-8}\right)}\right)=0\\ \left(V_2-V_1\right)10+0.166V_{2}j\omega +V_2\left(\frac{{\omega }^2\left(20\times {10}^{-8}\right)}{10-{\omega }^2\left(2\times {10}^{-8}\right)}\right)=0\end{array}$

Substitute $100\pi$ for $\omega$ in the above expression.

$\begin{array}{l}\left(V_2-V_1\right)10+0.166V_{2}j\left(100\pi \right)+V_2\left(\frac{{\left(100\pi \right)}^2\left(20\times {10}^{-8}\right)}{10-{\left(100\pi \right)}^2\left(2\times {10}^{-8}\right)}\right)=0\\ \left(V_2-V_1\right)10+j{V}_2\left(52.15\right)-V_2\left(1.97\times {10}^{-3}\right)=0\end{array}$

$-10V_1+\left(9.99+j52.15\right)V_2=0$        (2)

Write equation (1) and (2) in matrix form.

$\left[\begin{array}{cc}10.05& -10\\ -10& 9.99+j52.15\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]=0$

The determinant of the given circuit is given by,

$\begin{array}{c}{\Delta }_Y=\left[\begin{array}{cc}10.05& -10\\ -10& 9.99+j52.15\end{array}\right]\\ =\left\{10.05\left(9.99+j52.15\right)-\left(-10\right)\left(-10\right)\right\}\\ =100.4+j524.12-100\\ =0.4+j524.12\end{array}$

Further solve as,

${\Delta }_Y=524.12\angle {89.95}^{\text{ο}}$

Conclusion:

Therefore, the determinant of the network is, ${\Delta }_Y=524.12\angle {89.95}^{\text{ο}}$, and the input admittance of the network is $Y_{\text{in}}=0.031\angle -{79.29}^{\text{ο}}\text{S}$.

 (b)

To determine

The voltage across the current source.

Answer to Problem 7E

The voltage across the current source is $3.1\angle -{79.29}^{\text{ο}}\text{V}$.

Explanation of Solution

Given data:

The value of the current source is, $I=100\text{A}$.

The frequency is $\omega =100\pi \text{rad}/\text{s}$.

Calculation:

The required diagram is shown in Figure 7.

The expression for the voltage across current source is given by,

$V=Y_{\text{in}}I$

Here,

$V$ is the voltage across current source.

$I$ is the current source.

$Y_{\text{in}}$ is the input admittance.

Substitute $100\text{A}$ for $I$ and $0.031\angle -{79.29}^{\text{ο}}\text{S}$ for $Y_{\text{in}}$ in the above expression.

$\begin{array}{c}V=100\text{A}\left(0.031\angle -{79.29}^{\text{ο}}\text{S}\right)\\ =3.1\angle -{79.29}^{\text{ο}}\text{V}\end{array}$

Conclusion:

Therefore, the voltage across the current source is $3.1\angle -{79.29}^{\text{ο}}\text{V}$.