Chapter 13, Problem 49P

Find current i_x in the ideal transformer circuit shown in Fig. 13.114.

To determine

Calculate the current $i_{\text{x}}$ in the ideal transformer circuit.

Answer to Problem 49P

The value of current $i_{\text{x}}$ is $\underset{\_}{0.937\cos \left(2t+51.34°\right)\text{A}}$.

Explanation of Solution

Given data:

Refer to Figure 13.114 in the textbook for the ideal transformer circuit.

From Figure 13.114, the value of $\omega$ is 2.

Calculation:

Consider the expression for the capacitive reactance.

$X_C=\frac{1}{j\omega C}$

Substitute $\frac{1}{20}\text{F}$ for C and $2$ for $\omega$.

$\begin{array}{c}{X}_C=\frac{1}{j\left(2\right)\left(\frac{1}{20}\text{F}\right)}\\ =-j10\Omega \end{array}$

Modify the Figure 13.114 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.

Apply Kirchhoff's current law at node 1 in Figure 1.

$\begin{array}{l}\frac{12-V_1}{2}=\frac{V_1-V_2}{-j10}+I_1\\ \frac{V_1-V_2}{-j10}+I_1-\frac{12-V_1}{2}=0\\ \left(\frac{1}{-j10}+\frac{1}{2}\right)V_1+\frac{V_2}{j10}+I_1=6\end{array}$

Multiply by 2 on both sides of the Equation.

$2\left(\frac{1}{-j10}+\frac{1}{2}\right)V_1+\frac{2V_2}{j10}+2I_1=12$

$2I_1+V_1\left(1+j0.2\right)-j0.2V_2=12$        (1)

Apply Kirchhoff's current law at node 2 in Figure 1.

$\begin{array}{l}{I}_2+\frac{V_1-V_2}{-j10}=\frac{V_2}{6}\\ I_2+\frac{V_1-V_2}{-j10}-\frac{V_2}{6}=0\end{array}$

Multiply by 6 on both sides of the Equation.

$6\left(I_2+\frac{V_1-V_2}{-j10}-\frac{V_2}{6}\right)=6\times 0$

$6I_2+j0.6V_1-\left(1+j0.6\right)V_2=0$        (2)

Consider the expression for the turns ratio or transformation ratio.

$\frac{V_2}{V_1}=-n$

Substitute $3$ for n.

$\begin{array}{l}\frac{V_2}{V_1}=-3\\ V_2=-3V_1\end{array}$

$3V_1+V_2=0$        (3)

Consider the expression for the turns ratio or transformation ratio.

$\frac{I_2}{I_1}=-\frac{1}{n}$

Substitute $3$ for n.

$\begin{array}{l}\frac{I_2}{I_1}=-\frac{1}{3}\\ -I_1=3I_2\end{array}$

$I_1+3I_2=0$        (4)

Write equations (1), (2), (3), and (4) in matrix form as follows.

$\left[\begin{array}{cccc}2& 0& 1+j0.2& -j0.2\\ 0& 6& j0.6& -1-j0.6\\ 0& 0& 3& 1\\ 1& 3& 0& 0\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ V_1\\ V_2\end{array}\right]=\left[\begin{array}{c}12\\ 0\\ 0\\ 0\end{array}\right]$        (5)

Write the MATLAB code to solve the equation (5).

A = [2 0 (1+j*0.2) (j*(-0.2)); 0 6 (j*0.6) (-1-j*0.6); 0 0 3 1; 1 3 0 0];

B = [12;0;0;0];

C = inv(A)*B

The output in command window:

C =

   4.5000 + 0.0000i

  -1.5000 + 0.0000i

   1.8293 - 1.4634i

  -5.4878 + 4.3902i

From the MATLAB output, the values of $I_1,I_2,I_3,V_1,\text{and}{V}_2$ are,

$\begin{array}{l}{I}_1=4.5\text{A,}\\ I_2=-1.5\text{A,}\\ V_1=\left(1.8293-j1.4634\right)\text{V,}\text{and}\\ V_2=\left(-5.4878+j4.902\right)\text{V,}\end{array}$

Write the expression for the current $I_{\text{x}}$ using Figure 1.

$I_{\text{x}}=\frac{V_1-V_2}{-j10\Omega }$

Substitute $-3V_1$ for $V_2$.

$\begin{array}{c}{I}_{\text{x}}=\frac{V_1-\left(-3V_1\right)}{-j10\Omega }\\ =\frac{4V_1}{-j10\Omega }\end{array}$

Substitute $\left(1.8293-j1.4634\right)\text{V}$ for $V_1$.

$\begin{array}{c}{I}_{\text{x}}=\frac{4\left(1.8293-j1.4634\right)\text{V}}{-j10\Omega }\\ =4\left(0.14634+j0.18293\right)\\ =4\left(0.2343\angle 51.34°\text{A}\right)\\ =0.937\angle 51.34°\text{A}\end{array}$

Convert the polar form to time-domain form.

$i_x=0.937\cos \left(2t+51.34°\right)\text{A}$

Conclusion:

Thus, the value of current $i_{\text{x}}$ is $\underset{\_}{0.937\cos \left(2t+51.34°\right)\text{A}}$.