Chapter 16, Problem 23E

Determine the input impedance Z_in of the one-port shown in Fig. 16.51 if ω is equal to (a) 50 rad/s; (b) 1000 rad/s.

To determine

The value of the input impedance for $\omega$ equals to $50\text{rad}/\text{s}$ is $70.21\angle 90°\Omega$

Answer to Problem 23E

The value of the input impedance is $70.21\angle 90°\Omega$.

Explanation of Solution

Given data:

The value of $\omega$ is given as $50\text{rad}/\text{s}$.

The given diagram is shown in Figure 1.

Calculation:

Let the inductances be $L_1=5\text{H}$, $L_2=1{\rm H}$, $L_3=3{\rm H}$, and $L_4=2{\rm H}$ $L_5=5{\rm H}$.

Let the capacitance $C_1=0.02\text{F}$ and $C_2=50\text{mF}$.

The expression for the inductive impedance $Z_L$ is given as,

$Z_L=j\times \omega \times L$        (1)

The expression for the capacitive impedance $Z_C$ is given as,

$Z_C=\frac{1}{j\times \omega \times C}$        (2)

Substitute $Z_{L1}$ for $Z_L$, $L_1$ for $L$ and in equation (1).

$Z_{L1}=j\times \omega \times L_1$        (3)

Substitute $Z_{L2}$ for $Z_L$ and $L_2$ for $L$ in equation (2).

$Z_{L2}=j\times \omega \times L_2$        (4)

Substitute $Z_{L3}$ for $Z_L$ and $L_3$ for $L$ in equation (3).

$Z_{L3}=j\times \omega \times L_3$        (5)

Substitute $Z_{L4}$ for $Z_L$ and $L_4$ for $L$ in equation (4).

$Z_{L4}=j\times \omega \times L_4$        (6)

The value of $Z_{L4}$ and $Z_{L5}$ are equal as inductance are equal.

$Z_{L4}=Z_{L5}$        (7)

Substitute $Z_{C1}$ for $Z_C$ and $C_1$ for $C$ in equation (2).

$Z_{C1}=\frac{1}{j\times \omega \times C_1}$        (8)

Substitute $Z_{C2}$ for $Z_C$ and $C_2$ for $C$ in equation (2).

$Z_{C2}=\frac{1}{j\times \omega \times C_2}$        (9)

Substitute $50\text{rad}/\text{s}$ for $\omega$ and $5\text{H}$ for $L_1$ in equation (3).

$\begin{array}{c}{Z}_{L1}=j\times \left(50\text{rad}/\text{s}\right)\times 5\text{H}\\ =j\text{250}\Omega \end{array}$

Substitute $50\text{rad}/\text{s}$ for $\omega$ and $1\text{H}$ for $L_2$ in equation (4).

$\begin{array}{c}{Z}_{L2}=j\times \left(50\text{rad}/\text{s}\right)\times 1\text{H}\\ =j\text{50}\Omega \end{array}$

Substitute $50\text{rad}/\text{s}$ for $\omega$ and $3\text{H}$ for $L_3$ in equation (5).

$\begin{array}{c}{Z}_{L3}=j\times \left(50\text{rad}/\text{s}\right)\times 3\text{H}\\ =j1\text{50}\Omega \end{array}$

Substitute $50\text{rad}/\text{s}$ for $\omega$ and $2\text{H}$ for $L_4$ in equation (6).

$\begin{array}{c}{Z}_{L4}=j\times \left(50\text{rad}/\text{s}\right)\times 2\text{H}\\ =j10\text{0}\Omega \end{array}$

Substitute $j10\text{0}\Omega$ for $Z_{L4}$ in equation (7).

$Z_{L5}=j10\text{0}\Omega$

Substitute $50\text{rad}/\text{s}$ for $\omega$ and $0.02\text{F}$ for $C_1$ in equation (8)

$\begin{array}{c}{Z}_{C1}=\frac{1}{j\times \left(50\text{rad}/\text{s}\right)\times 0.02\text{F}}\\ =-j\Omega \end{array}$

The conversion of $50\text{mF}$ into $\text{F}$ is given by,

$50\text{mF}=50\times {10}^{-3}\text{F}$

Substitute $50\text{rad}/\text{s}$ for $\omega$ and $50\times {10}^{-3}\text{F}$ for $C_2$ in equation (9).

$\begin{array}{c}{Z}_{C2}=\frac{1}{j\times \left(50\text{rad}/\text{s}\right)\times \left(50\times {10}^{-3}\right)\text{F}}\\ =-j0.4\Omega \end{array}$

Mark the impedances and redraw the circuit.

The required diagram is shown in Figure 2

The $\Delta$ part of the circuit is converted into $\text{Y}$.

The required diagram is shown in the Figure 3.

Here,

$Z_1$, $Z_2$, and $Z_3$ are the equivalent star impedances of the circuit.

The impedance $Z_1$ is obtained as,

$\begin{array}{c}{Z}_1=\frac{\left(j150\right)\left(-j\right)}{\left(j150\Omega +j250\Omega -j\Omega \right)}\\ =\frac{150\Omega }{j399\Omega }\\ =-j0.37\Omega \end{array}$

The impedance $Z_2$ is obtained as,

$\begin{array}{c}{Z}_2=\frac{\left(j250\Omega \right)\left(j150\Omega \right)}{\left(j150\Omega +j250\Omega -j\Omega \right)}\\ =\frac{-37500\Omega }{j399\Omega }\\ =j93.98\Omega \end{array}$

The impedance $Z_3$ is obtained as,

$\begin{array}{c}{Z}_3=\frac{\left(-j\Omega \right)\left(j250\Omega \right)}{\left(j150\Omega +j250\Omega -j\Omega \right)}\\ =\frac{250\Omega }{j399\Omega }\\ =-j0.63\Omega \end{array}$

The required diagram is shown in Figure 4.

Add the impedances in series in the above network and redraw the network.

The required diagram is shown in Figure 5.

In the above circuit $j99.63\Omega$ and $j93.58$ are connected in parallel.

Thus, the parallel combination $Z_{\text{eq}}$ of the impedances is given by,

$\begin{array}{c}\frac{1}{Z_{\text{eq}}}=\frac{1}{j99.63\Omega }+\frac{1}{j93.58\Omega }\\ \frac{1}{Z_{\text{eq}}}=\frac{j193.21}{-9323.37}\\ Z_{\text{eq}}=j48.255\Omega \end{array}$

Mark the equivalent impedance and redraw the circuit.

The required figure is shown in Figure 6.

The value of the input impedance $Z_{\text{in}}$ is obtained by,

$\begin{array}{c}{Z}_{\text{in}}=\left(250\Omega \right)\left|\right|\left(j49.37\Omega +j48.25\Omega \right)\\ =\left(j250\right)\left|\right|\left(j97.62\right)\\ =\frac{\left(j250\right)\left(j97.62\right)}{\left(j250\right)+\left(j97.62\right)}\\ =\frac{-24405}{j347.62}\end{array}$

Solve it further as,

$\begin{array}{c}{Z}_{\text{in}}=j70.21\Omega \\ =70.21\angle 90°\Omega \end{array}$

Conclusion:

Therefore, the value of the input impedance for $\omega$ equals to $50\text{rad}/\text{s}$ is $70.21\angle 90°\Omega$

To determine

The input impedance of the circuit is determined for $\omega$ equals to $1000\text{rad}/\text{s}$.

Answer to Problem 23E

The value of the input impedance for $\omega$ equals to $1000\text{rad}/\text{s}$ is $140.8\angle 90°\Omega$

Explanation of Solution

Given data:

The value of $\omega$ is given as $1000\text{rad}/\text{s}$.

Calculation:

Substitute $1000\text{rad}/\text{s}$ for $\omega$ and $5\text{H}$ for $L_1$ in equation (3).

$\begin{array}{c}{Z}_{L1}=j\times \left(1000\text{rad}/\text{s}\right)\times 5\text{H}\\ =j\text{5000}\Omega \end{array}$

Substitute $1000\text{rad}/\text{s}$ for $\omega$ and $1\text{H}$ for $L_2$ in equation (4).

$\begin{array}{c}{Z}_{L2}=j\times \left(1000\text{rad}/\text{s}\right)\times 1\text{H}\\ =j100\text{0}\Omega \end{array}$

Substitute $1000\text{rad}/\text{s}$ for $\omega$ and $3\text{H}$ for $L_3$ in equation (5).

$\begin{array}{c}{Z}_{L3}=j\times \left(1000\text{rad}/\text{s}\right)\times 3\text{H}\\ =j3000\Omega \end{array}$

Substitute $1000\text{rad}/\text{s}$ for $\omega$ and $2\text{H}$ for $L_4$ in equation (6).

$\begin{array}{c}{Z}_{L4}=j\times \left(1000\text{rad}/\text{s}\right)\times 2\text{H}\\ =j200\text{0}\Omega \end{array}$

Substitute $j200\text{0}\Omega$ for $Z_{L4}$ in equation (7).

$Z_{L5}=j200\text{0}\Omega$

Substitute $1000\text{rad}/\text{s}$ for $\omega$ and $0.02\text{F}$ for $C_1$ in equation (8).

$\begin{array}{c}{Z}_{C1}=\frac{1}{j\times \left(1000\text{rad}/\text{s}\right)\times 0.02\text{F}}\\ =-j0.05\Omega \end{array}$

Substitute $1000\text{rad}/\text{s}$ for $\omega$ and $50\times {10}^{-3}\text{F}$ for $C_2$ in equation (9).

$\begin{array}{c}{Z}_{C2}=\frac{1}{j\times \left(50\text{rad}/\text{s}\right)\times \left(50\times {10}^{-3}\text{F}\right)}\\ =-j0.02\Omega \end{array}$

Mark the impedances and redraw the circuit.

The required diagram is shown in Figure 7

The $\Delta$ part of the circuit is converted into $\text{Y}$.

The required diagram is shown in the Figure 8

Here,

$Z_1$, $Z_2$, and $Z_3$ are the equivalent star impedances of the circuit.

The impedance $Z_1$ is obtained as,

$\begin{array}{c}{Z}_1=\frac{\left(j3000\right)\left(-j0.05\right)}{\left(j3000\Omega +j5000\Omega -j0.05\Omega \right)}\\ =\frac{150\Omega }{j7999.95\Omega }\\ =-j0.01875\Omega \end{array}$

The impedance $Z_2$ is obtained as,

$\begin{array}{c}{Z}_2=\frac{\left(j3000\right)\left(j5000\right)}{\left(j3000\Omega +j5000\Omega -j0.05\Omega \right)}\\ =\frac{-15000000\Omega }{j7999.95\Omega }\\ =j1875\Omega \end{array}$

The impedance $Z_3$ is obtained as,

$\begin{array}{c}{Z}_3=\frac{\left(j5000\right)\left(-j0.05\right)}{\left(j3000\Omega +j5000\Omega -j0.05\Omega \right)}\\ =\frac{550\Omega }{j7999.95\Omega }\\ =-j0.03125\Omega \end{array}$

The modified diagram is shown in Figure 9.

Add the impedances in series in the above network and redraw the network.

The required diagram is shown in Figure 10.

In the above circuit $j99.63\Omega$ and $j93.58$ are connected in parallel.

Thus, the parallel combination $Z_{\text{eq}}$ of the impedances is written as,

$\begin{array}{c}\frac{1}{Z_{\text{eq}}}=\frac{1}{j99.98\Omega }+\frac{1}{j1874.6\Omega }\\ \frac{1}{Z_{\text{eq}}}=\frac{j1974.58}{-187422.508}\\ Z_{\text{eq}}=j94.91\Omega \end{array}$

Mark the equivalent impedance and redraw the circuit.

The required figure is shown in Figure 11.

The value of the input impedance $Z_{\text{in}}$ is obtained by,

$\begin{array}{c}{Z}_{\text{in}}=\left(5000\Omega \right)\left|\right|\left(j49.97\Omega +j94.91\Omega \right)\\ =\left(j5000\right)\left|\right|\left(j144.88\right)\\ =\frac{\left(j5000\right)\left(j144.88\right)}{\left(j5000\right)+\left(j144.88\right)}\\ =\frac{-724400}{j5144.88}\end{array}$

Solve it further as,

$\begin{array}{c}{Z}_{\text{in}}=j140.8\Omega \\ =140.8\angle 90°\Omega \end{array}$

Conclusion:

Therefore the value of the input impedance for $\omega$ equals to $10000\text{rad}/\text{s}$ is $140.8\angle 90°\Omega$