Chapter 14, Problem 80P

1. (a) For the circuit in Fig. 14.97, draw the new circuit after it has been scaled by K_m = 200 and K_f = 10⁴.
2. (b) Obtain the Thevenin equivalent impedance at terminals a-b of the scaled circuit at ω = 10⁴ rad/s.

Figure 14.97

To determine

Draw the new circuit for the circuit in Figure 14.97 after it has been magnitude scaled by a factor of $200$ and frequency scaled by a factor of ${10}^4$.

Explanation of Solution

Given data:

Refer to Figure 14.97 in the textbook.

The value of the magnitude scaling factor $\left(K_m\right)$ is $200$.

The value of the frequency scaling factor $\left(K_f\right)$ is $1{\text{0}}^4$.

Formula used:

Consider the equations used in magnitude and frequency scaling.

Write the expression to calculate the scaled resistor.

$R^{\prime }=K_{m}R$        (1)

Here,

$K_m$ is the magnitude scaling factor, and

$R$ is the value of the resistor.

Write the expression to calculate the scaled inductor.

$L^{\prime }=\frac{K_m}{K_f}L$        (2)

Here,

$K_f$ is the frequency scaling factor, and

$L$ is the value of the inductor.

Write the expression to calculate the scaled capacitor.

$C^{\prime }=\frac{1}{K_m{K}_f}C$        (3)

Here,

$C$ is the value of the capacitor.

Calculation:

The given circuit is redrawn as Figure 1.

Refer to Figure 1, the value of the resistor $\left(R\right)$ is $2\Omega$, the inductor $\left(L\right)$ is $\text{1}\text{H}$ and the capacitor $\left(C\right)$ is $0.5\text{F}$.

Substitute $2\Omega$ for $R$ and $200$ for $K_m$ in equation (1) to find $R^{\prime }$.

$\begin{array}{c}{R}^{\prime }=\left(200\right)\left(2\Omega \right)\\ =400\Omega \end{array}$

Substitute $\text{1}\text{H}$ for $L$, $200$ for $K_m$ and $1{\text{0}}^4$ for $K_f$ in equation (2) to find $L^{\prime }$.

$\begin{array}{c}{L}^{\prime }=\frac{200}{{10}^4}\left(\text{1}\text{H}\right)\\ =\text{20}\times {\text{10}}^{-3}\text{H}\\ =\text{20}\text{mH}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Substitute $\text{0}\text{.5}\text{F}$ for $C$, $200$ for $K_m$ and $1{\text{0}}^4$ for $K_f$ in equation (3) to find $C^{\prime }$.

$\begin{array}{c}{C}^{\prime }=\frac{1}{\left(200\right)\left(1{\text{0}}^4\right)}\left(\text{0}\text{.5}\text{F}\right)\\ =0.25\times {10}^{-6}\text{F}\\ =0.25\mu \text{F}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

The redesigned circuit is drawn as Figure 2 which is obtained by using the magnitude and frequency scaling on the circuit in Figure 1.

Conclusion:

Thus, the new circuit for the circuit in Figure 14.97 is drawn by using the magnitude and frequency scaling.

To determine

Find the value of the Thevenin equivalent impedance at terminals a-b of the scaled circuit.

Answer to Problem 80P

The value of the Thevenin equivalent impedance $\left(Z_{\text{Th}}\right)$ at a-b terminals of the scaled circuit is $\left(632.5\angle -18.435°\right)\Omega$.

Explanation of Solution

Given data:

The value of the angular frequency $\left(\omega \right)$ is $1{\text{0}}^4\frac{\text{rad}}{\text{s}}$.

Formula used:

Write the expression to calculate the impedance of the passive elements resistor, inductor and capacitor in s-domain.

$Z_R\left(s\right)=R$        (4)

$Z_L\left(s\right)=sL$        (5)

$Z_C\left(s\right)=\frac{1}{sC}$        (6)

Here,

$R$ is the value of the resistor,

$L$ is the value of the inductor, and

$C$ is the value of the capacitor.

Calculation:

Use equation (4) to find $Z_{R^{\prime }}\left(s\right)$.

$Z_{R^{\prime }}\left(s\right)=R^{\prime }$

Use equation (5) to find $Z_{L^{\prime }}\left(s\right)$.

$Z_{L^{\prime }}\left(s\right)=s{L}^{\prime }$

Use equation (6) to find $Z_{C^{\prime }}\left(s\right)$.

$Z_{C^{\prime }}\left(s\right)=\frac{1}{s{C}^{\prime }}$

Insert a $1\text{A}$ current source at a-b terminals and redraw the scaled circuit in Figure 2 using s-domain as Figure 3.

Apply Kirchhoff’s current law on Figure 3 to find $V_1$.

$\begin{array}{l}\frac{V_1}{\left(\frac{1}{s{C}^{\prime }}\right)}+\frac{V_1-V_2}{s{L}^{\prime }}=1\\ V_{1}s{C}^{\prime }+\frac{V_1-V_2}{s{L}^{\prime }}=1\end{array}$

Rearrange the above equation.

$\left(\frac{V_1-V_2}{s{L}^{\prime }}\right)=1-V_{1}s{C}^{\prime }$        (7)

Apply Kirchhoff’s current law on Figure 3 to find $V_2$.

$\begin{array}{l}\frac{V_2-V_1}{s{L}^{\prime }}+\frac{V_2}{R^{\prime }}=0.5I_x\\ -\left(\frac{V_1-V_2}{s{L}^{\prime }}\right)+\frac{V_2}{R^{\prime }}=0.5I_x\end{array}$

$\left(\frac{V_1-V_2}{s{L}^{\prime }}\right)=\frac{V_2}{R^{\prime }}-0.5I_x$        (8)

Refer to Figure 3, the current $I_x$ can be expressed as,

$\begin{array}{c}{I}_x=\frac{V_1}{\left(\frac{1}{s{C}^{\prime }}\right)}\\ =s{C}^{\prime }{V}_1\end{array}$

Substitute $s{C}^{\prime }{V}_1$ for $I_x$ in equation (8).

$\left(\frac{V_1-V_2}{s{L}^{\prime }}\right)=\frac{V_2}{R^{\prime }}-0.5s{C}^{\prime }{V}_1$        (9)

By comparing the equations (7) and (9), the following equation is obtained.

$\begin{array}{l}\frac{V_2}{R^{\prime }}-0.5s{C}^{\prime }{V}_1=1-s{C}^{\prime }{V}_1\\ \frac{V_2}{R^{\prime }}=1-s{C}^{\prime }{V}_1+0.5s{C}^{\prime }{V}_1\\ \frac{V_2}{R^{\prime }}=1-0.5s{C}^{\prime }{V}_1\end{array}$

Rearrange the above equation to find $V_2$.

$V_2=R^{\prime }\left(1-0.5s{C}^{\prime }{V}_1\right)$

Substitute $R^{\prime }\left(1-0.5s{C}^{\prime }{V}_1\right)$ for $V_2$ in equation (7).

$\begin{array}{l}\left(\frac{V_1-R^{\prime }\left(1-0.5s{C}^{\prime }{V}_1\right)}{s{L}^{\prime }}\right)=1-V_{1}s{C}^{\prime }\\ V_1-R^{\prime }+0.5s{R}^{\prime }{C}^{\prime }{V}_1=s{L}^{\prime }\left(1-V_{1}s{C}^{\prime }\right)\\ V_1-R^{\prime }+0.5s{R}^{\prime }{C}^{\prime }{V}_1=s{L}^{\prime }-V_1{s}^2{L}^{\prime }{C}^{\prime }\\ V_1+V_1{s}^2{L}^{\prime }{C}^{\prime }+0.5s{R}^{\prime }{C}^{\prime }{V}_1=s{L}^{\prime }+R^{\prime }\end{array}$

Simplify the above equation.

$V_1\left(s^2{L}^{\prime }{C}^{\prime }+0.5s{R}^{\prime }{C}^{\prime }+1\right)=s{L}^{\prime }+R^{\prime }$

Simplify the above equation to find $V_1$.

$V_1=\frac{s{L}^{\prime }+R^{\prime }}{s^2{L}^{\prime }{C}^{\prime }+0.5s{R}^{\prime }{C}^{\prime }+1}$

Refer to Figure 3, the Thevenin equivalent impedance across the a-b terminals are calculated as follows.

$Z_{\text{Th}}=\frac{V_1}{1\text{A}}$

Substitute $\frac{s{L}^{\prime }+R^{\prime }}{s^2{L}^{\prime }{C}^{\prime }+0.5s{R}^{\prime }{C}^{\prime }+1}$ for $V_1$ in above equation to find $Z_{\text{Th}}$.

$\begin{array}{c}{Z}_{\text{Th}}=\frac{\left(\frac{s{L}^{\prime }+R^{\prime }}{s^2{L}^{\prime }{C}^{\prime }+0.5s{R}^{\prime }{C}^{\prime }+1}\right)}{1}\\ =\frac{s{L}^{\prime }+R^{\prime }}{s^2{L}^{\prime }{C}^{\prime }+0.5s{R}^{\prime }{C}^{\prime }+1}\end{array}$

Substitute $j\omega$ for $s$ in above equation to find $Z_{\text{Th}}$.

$\begin{array}{c}{Z}_{\text{Th}}=\frac{\left(j\omega \right)L^{\prime }+R^{\prime }}{{\left(j\omega \right)}^2{L}^{\prime }{C}^{\prime }+0.5\left(j\omega \right)R^{\prime }{C}^{\prime }+1}\\ =\frac{j\omega L^{\prime }+R^{\prime }}{j^2{\omega }^2{L}^{\prime }{C}^{\prime }+j0.5\omega R^{\prime }{C}^{\prime }+1}\\ =\frac{j\omega L^{\prime }+R^{\prime }}{-{\omega }^2{L}^{\prime }{C}^{\prime }+j0.5\omega R^{\prime }{C}^{\prime }+1}\left\{\because j^2=-1\right\}\\ =\frac{j\omega L^{\prime }+R^{\prime }}{1-{\omega }^2{L}^{\prime }{C}^{\prime }+j0.5\omega R^{\prime }{C}^{\prime }}\end{array}$

Substitute ${10}^4\frac{\text{rad}}{\text{s}}$ for $\omega$, $400\Omega$ for $R^{\prime }$, $\text{20}\text{mH}$ for $L^{\prime }$ and $0.25\mu \text{F}$ for $C^{\prime }$ in above equation to find $Z_{\text{Th}}$.

$\begin{array}{c}{Z}_{\text{Th}}=\frac{j\left({10}^4\frac{\text{rad}}{\text{s}}\right)\left(\text{20}\text{mH}\right)+\left(400\Omega \right)}{1-{\left({10}^4\frac{\text{rad}}{\text{s}}\right)}^2\left(\text{20}\text{mH}\right)\left(0.25\mu \text{F}\right)+j0.5\left({10}^4\frac{\text{rad}}{\text{s}}\right)\left(400\Omega \right)\left(0.25\mu \text{F}\right)}\\ =\frac{j\left({10}^4\right)\left(\text{20}\times {\text{10}}^{-3}\right)\left(\frac{\text{H}}{\text{s}}\right)+400\Omega }{\left(\begin{array}{l}1-\left({\left({10}^4\right)}^2\left(\text{20}\times {\text{10}}^{-3}\right)\left(0.25\times {\text{10}}^{-6}\right)\frac{\text{H}\cdot \text{F}}{{\text{s}}^{\text{2}}}\right)\\ +j\left(0.5\left({10}^4\right)\left(400\right)\left(0.25\times {\text{10}}^{-6}\right)\frac{\Omega \text{F}}{\text{s}}\right)\end{array}\right)}\left\{\begin{array}{l}\because 1\text{m}={10}^{-3}\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\frac{j200\left(\frac{\Omega \cdot \text{s}}{\text{s}}\right)+400\Omega }{\left(1-\left(0.5\frac{\left(\frac{{\text{s}}^2}{\text{F}}\right)\cdot \text{F}}{{\text{s}}^{\text{2}}}\right)+j\left(0.5\frac{\Omega \left(\frac{\text{s}}{\Omega }\right)}{\text{s}}\right)\right)}\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s}\\ 1\text{H}=\frac{1{\text{s}}^2}{1\text{F}}\\ 1\text{F}=\frac{1\text{s}}{1\Omega }\end{array}\right\}\\ =\frac{400\Omega +j200\Omega }{\left(1-0.5+j0.5\right)}\end{array}$

Simplify the above equation to find $Z_{\text{Th}}$.

$\begin{array}{c}{Z}_{\text{Th}}=\frac{\left(400+j200\right)\Omega }{\left(0.5+j0.5\right)}\\ =\left(600-j200\right)\Omega \\ =\left(632.5\angle -18.435°\right)\Omega \end{array}$

Conclusion:

Thus, the value of the Thevenin equivalent impedance $\left(Z_{\text{Th}}\right)$ at a-b terminals of the scaled circuit is $\left(632.5\angle -18.435°\right)\Omega$.