Chapter 13, Problem 25P

To determine

Calculate the impedance $Z_{ab}$ and the current $I_o$ using MATLAB.

Answer to Problem 25P

The impedance $Z_{ab}$ and the current $I_o$ are $\underset{\_}{1.5085\angle 17.9°\Omega }$ and $\underset{\_}{2.2\sin \left(2t-4.88°\right)\text{A}}$ respectively.

Explanation of Solution

Given data:

Refer to Figure 13.94 in the textbook for the circuit with coupled coils.

The coupling co-efficient is 0.5.

Calculation:

Consider the expression for the mutual inductance.

$M=k\sqrt{L_1{L}_2}$

Substitute 0.5 for k, 1 H for $L_1$, and 1 H for $L_2$.

$\begin{array}{c}M=0.5\sqrt{\left(1\text{H}\right)\left(1\text{H}\right)}\\ =0.5\text{H}\end{array}$

From Figure 13.94, the value of $\omega$ is 2.

Consider the expression for the inductive reactance.

$X_L=j\omega L$        (1)

Substitute 1 H for L and $2$ for $\omega$ in equation (1) to the reactance for 1 H inductor.

$\begin{array}{c}{X}_L=j\left(2\right)\left(1\text{H}\right)\\ =j2\Omega \end{array}$

Substitute 2 H for L and $2$ for $\omega$ in equation (1) to the reactance for 2 H inductor.

$\begin{array}{c}{X}_L=j\left(2\right)\left(2\text{H}\right)\\ =j4\Omega \end{array}$

Consider the expression for the capacitive reactance.

$X_C=\frac{1}{j\omega C}$        (2)

Substitute 0.5 F for C and $2$ for $\omega$ in equation (2).

$\begin{array}{c}{X}_C=\frac{1}{j\left(2\right)\left(0.5\right)}\\ =-j\Omega \end{array}$

Modify the Figure 13.94 by transforming the time-domain circuit with coupled-coils to frequency domain of the circuit with coupled-coil. The frequency domain equivalent circuit is shown in Figure 1.

Write the expression for the impedance $Z_{ab}$ using Figure 1.

$Z_{ab}=\left(2-j\right)\parallel Z_{\text{in}}$

$Z_{ab}=\left(2-j\right)\parallel \left(1+j2+Z_R\right)$

$Z_{ab}=\frac{\left(2-j\right)\left(1+j2+Z_R\right)}{\left(2-j\right)+\left(1+j2+Z_R\right)}$        (3)

Consider the expression for the reflected impedance $Z_R$.

$Z_R=\frac{{\omega }^2{M}^2}{R_2+j\omega L_2+Z_L}$        (4)

Substitute 2 for $\omega$, 0.5 H for M, $3\Omega$ for $R_2$, $j2$ for $j\omega L_2$, and $j4$ for $Z_L$.

$\begin{array}{c}{Z}_R=\frac{{\left(2\right)}^2{\left(0.5\right)}^2}{j2+3+j4}\\ =\frac{1}{3+j6}\\ =0.0667-j0.1333\Omega \end{array}$

Substitute $0.0667-j0.1333\Omega$ for $Z_R$ in Equation (3).

$\begin{array}{c}{Z}_{ab}=\left(2-j\right)\parallel \left(1+j2+0.0667-j0.1333\right)\\ =\left(2-j\right)\parallel \left(1.0667+j1.8667\right)\end{array}$

Simplify the Equation as follows.

$\begin{array}{c}{Z}_{ab}=\frac{\left(2-j\right)\times \left(1.0667+j1.8667\right)}{\left(2-j\right)+\left(1.0667+j1.8667\right)}\\ =\frac{4.0001+j2.6667}{3.0667+j0.8667}\\ =1.4355+j0.4639\\ =1.5085\angle 17.9°\Omega \end{array}$

Write the expression for the current $I_o$ using Figure 1.

$I_o=\frac{12\angle 0°\text{V}}{Z_{ab}+4\Omega }$        (5)

Substitute $1.4355+j0.4639$ for $Z_{ab}$ in above equation to find the current $I_o$.

$\begin{array}{l}{I}_o=\frac{12\angle 0°\text{V}}{1.4355+j0.4639+4\Omega }\\ I_o=\frac{12\angle 0°\text{V}}{5.45526\angle 4.878163°\Omega }\\ I_o=2.19174-j0.18705\text{ A}\\ I_o\cong 2.1917-j0.1871\text{ A}\end{array}$

The value of current $I_o$ is,

$\begin{array}{c}{I}_o=2.1917-j0.1871\text{A}\\ =\text{2}\text{.2}\angle -\text{4}\text{.88}°\text{A}\end{array}$

Convert the current from polar form to time domain form.

$i_o=2.2\sin \left(2t-4.88°\right)\text{A}$

MATLAB code:

The MATLAB code using equations (3), (4) and (5) is,

M=0.5;

R2=3;

w=2;

L2=1;

ZL=4*j;

ZR=(w^2 * M^2)/(R2 + j*w*L2 + ZL);

Zab= (2-1*j)*(1+2*j+ZR)/(2-j+1+2*j+ZR)

Io=12/(Zab+4)

Then the MATLAB output is,

Zab = 1.4354 + 0.4639i

Io = 2.1918 - 0.1871i

Form the MATLAB output, impedance $Z_{ab}$ is,

$\begin{array}{c}{Z}_{ab}=1.4354+j0.4639\Omega \\ =1.50850\angle 17.910°\Omega \\ \cong 1.5085\angle 17.9°\Omega \end{array}$

Form the MATLAB output, current $I_o$ is,

$\begin{array}{c}{I}_o=2.1918-j0.1871\text{A}\\ =\text{2}\text{.199}\angle -\text{4}\text{.879}°\text{A}\\ \cong \text{2}\text{.2}\angle -\text{4}\text{.88}°\text{A}\end{array}$

The output is satisfied with analytical solution.

Conclusion:

Thus, the impedance $Z_{ab}$ and the current $I_o$ are $\underset{\_}{1.5085\angle 17.9°\Omega }$ and $\underset{\_}{2.2\sin \left(2t-4.88°\right)\text{A}}$ respectively.