Chapter 9, Problem 67P

At ω = 10³ rad/s, find the input admittance of each of the circuits in Fig. 9.74.

Figure 9.74

To determine

Find the value of input admittance $\left(Y_{\text{in}}\right)$ for the circuit given in Figure 9.74(a).

Answer to Problem 67P

The value of input admittance $\left(Y_{\text{in}}\right)$ for the circuit given in Figure 9.74(a) is $14.8\angle -20.22°\text{mS}$.

Explanation of Solution

Given data:

Refer to Figure 9.74(a) in the textbook.

The value of angular frequency $\left(\omega \right)$ is ${10}^3\frac{\text{rad}}{\text{s}}$.

Formula used:

Write a general expression to calculate the impedance of a resistor.

$Z_R=R$        (1)

Here,

$R$ is the value of resistance.

Write a general expression to calculate the impedance of an inductor.

$Z_L=j\omega L$        (2)

Here,

$L$ is the value of inductance, and

$\omega$ is the angular frequency.

Write a general expression to calculate the impedance of a capacitor.

$Z_C=\frac{1}{j\omega C}$        (3)

Here,

$C$ is the value of capacitance.

Write a general expression to calculate the input admittance.

$Y_{\text{in}}=\frac{1}{Z_{\text{in}}}$        (4)

Here,

$Z_{\text{in}}$ is the value of input impedance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Use equation (1) to find $Z_{R_1}$.

$Z_{R_1}=60\Omega$

Use equation (1) to find $Z_{R_2}$.

$Z_{R_2}=60\Omega$

Substitute $20\text{mH}$ for $L$, and ${10}^3\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (2) to find $Z_L$.

$\begin{array}{c}{Z}_L=j\left({10}^3\frac{\text{rad}}{\text{s}}\right)\left(20\text{mH}\right)\\ =j\left({10}^3\frac{\text{rad}}{\text{s}}\right)\left(20\times {10}^{-3}\Omega \cdot \text{s}\right)\left\{\begin{array}{l}\because 1\text{m}={10}^3\\ 1\text{H}=1\Omega \cdot 1\text{s}\end{array}\right\}\\ =j20\Omega \end{array}$

Substitute $12.5\mu \text{F}$ for $C$, and ${10}^3\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (3) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left({10}^3\frac{\text{rad}}{\text{s}}\right)\left(12.5\mu \text{F}\right)}\\ =\frac{1}{j\left({10}^3\frac{\text{rad}}{\text{s}}\right)\left(12.5\times {10}^{-6}\frac{\text{s}}{\Omega }\right)}\left\{\begin{array}{l}\because 1\mu ={10}^{-6}\\ 1\text{F}=\frac{\text{1}\text{s}}{1\Omega }\end{array}\right\}\\ =-j80\Omega \end{array}$

The impedance diagram of Figure 1 is drawn as shown in Figure 2.

Refer to Figure 2, the impedance $Z_L$  is connected in parallel with the series combination of the impedances $Z_{R_2}$ and $Z_C$ and this parallel combination is connected in series with the impedance $Z_{R_1}$.

The input impedance $\left(Z_{\text{in}}\right)$ is calculated as follows:

$\begin{array}{c}{Z}_{\text{in}}=60\Omega +\left(j20\Omega \parallel \left(60-j80\right)\Omega \right)\\ =60\Omega +\frac{\left(j20\Omega \right)\left(60-j80\right)\Omega }{\left(j20+60-j80\right)\Omega }\\ =60\Omega +3.33\Omega +j23.33\Omega \\ =\left(63.33+j23.33\right)\Omega \end{array}$

Substitute $63.33\Omega +j23.33\Omega$ for $Z_{\text{in}}$ in equation (4) to find $Y_{\text{in}}$.

$\begin{array}{c}{Y}_{\text{in}}=\frac{1}{\left(63.33+j23.33\right)\Omega }\\ =0.0148\angle -20.22°\text{S}\\ =14.8\times {10}^{-3}\angle -20.22°\text{S}\\ =14.8\angle -20.22°\text{mS}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Conclusion:

Thus, the value of input admittance $\left(Y_{\text{in}}\right)$ for the circuit given in Figure 9.74(a) is $14.8\angle -20.22°\text{mS}$.

To determine

Find the value of input admittance $\left(Y_{\text{in}}\right)$ for the circuit given in Figure 9.74(b).

Answer to Problem 67P

The value of input admittance $\left(Y_{\text{in}}\right)$ for the circuit given in Figure 9.74(b) is $19.704\angle 74.56°\text{mS}$.

Explanation of Solution

Calculation:

The given circuit is also redrawn as shown in Figure 3.

Use equation (1) to find $Z_{R_1}$.

 $Z_{R_1}=30\Omega$

Use equation (1) to find $Z_{R_2}$.

$Z_{R_2}=60\Omega$

Use equation (1) to find $Z_{R_3}$.

$Z_{R_3}=40\Omega$

Substitute $10\text{mH}$ for $L$, and ${10}^3\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (2) to find $Z_L$.

$\begin{array}{c}{Z}_L=j\left({10}^3\frac{\text{rad}}{\text{s}}\right)\left(10\text{mH}\right)\\ =j\left({10}^3\frac{\text{rad}}{\text{s}}\right)\left(10\times {10}^{-3}\Omega \cdot \text{s}\right)\left\{\begin{array}{l}\because 1\text{m}={10}^3\\ 1\text{H}=1\Omega \cdot 1\text{s}\end{array}\right\}\\ =j10\Omega \end{array}$

Substitute $20\mu \text{F}$ for $C$, and ${10}^3\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (3) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left({10}^3\frac{\text{rad}}{\text{s}}\right)\left(20\mu \text{F}\right)}\\ =\frac{1}{j\left({10}^3\frac{\text{rad}}{\text{s}}\right)\left(20\times {10}^{-6}\frac{\text{s}}{\Omega }\right)}\left\{\begin{array}{l}\because 1\mu ={10}^{-6}\\ 1\text{F}=\frac{\text{1}\text{s}}{1\Omega }\end{array}\right\}\\ =-j50\Omega \end{array}$

The impedance diagram of Figure 3 is drawn as shown in Figure 4.

Refer to Figure 4, the impedances $Z_{R_1}$ and $Z_{R_2}$ are connected in parallel.

The equivalent impedance $\left(Z_1\right)$ is calculated as follows:

$\begin{array}{c}{Z}_R=\frac{\left(30\Omega \right)\left(60\Omega \right)}{30\Omega +60\Omega }\\ =\frac{180{\Omega }^2}{90\Omega }\\ =20\Omega \end{array}$

Now, the Figure 4 is reduced as shown in Figure 5.

Refer to Figure 5, the impedance $Z_R$ is connected in parallel with the series combination of the impedances $Z_{R_3}$ and $Z_L$ and this parallel combination is connected in series with the impedance $Z_C$.

The input impedance $\left(Z_{\text{in}}\right)$ is calculated as follows:

$\begin{array}{c}{Z}_{\text{in}}=-j50\Omega +\left(20\Omega \parallel \left(40+j10\right)\Omega \right)\\ =-j50\Omega +\frac{\left(20\Omega \right)\left(40+j10\right)\Omega }{\left(20+40+j10\right)\Omega }\\ =-j50\Omega +13.51342\Omega +j1.08109\Omega \\ =\left(13.51342\Omega -j48.9189\right)\Omega \end{array}$

Substitute $\left(13.51342\Omega -j48.9189\right)\Omega$ for $Z_{\text{in}}$ in equation (4) to find $Y_{\text{in}}$.

$\begin{array}{c}{Y}_{\text{in}}=\frac{1}{\left(13.51342\Omega -j48.9189\right)\Omega }\\ =0.019704\angle 74.56°\text{S}\\ =19.704\times {10}^{-3}\angle 74.56°\text{S}\\ =19.704\angle 74.56°\text{mS}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Conclusion:

Thus, the value of input admittance $\left(Y_{\text{in}}\right)$ for the circuit given in Figure 9.74(b) is $19.704\angle 74.56°\text{mS}$.