Chapter 13, Problem 18P

Find the Thevenin equivalent to the left of the load Z in the circuit of Fig. 13.87.

To determine

Calculate the Thevenin equivalent to the left side of load Z in the coupled coils circuit.

Answer to Problem 18P

The Thevenin equivalent circuit parameters are $\underset{\_}{V_{Th}=638.78\angle 9.76°\text{V}}$ and $\underset{\_}{Z_{Th}=11.32\angle 85.01°\Omega }$.

Explanation of Solution

Given data:

Refer to Figure 13.87 in the textbook for the circuit with coupled coils.

Calculation:

To the Thevenin’s voltage, open-circuit the impedance Z. Then, the current $I_2$ is zero.

Modify the Figure 13.87 by convert the circuit into the frequency domain and convert the coupled inductors into their dependent source equivalent. The modified circuit as shown in Figure 1.

Apply Kirchhoff's voltage law to the loop 1 contains current $I_1$ in Figure 1.

$-440+4\left(I_1-I_2\right)-j4I_1+j5I_1+j6\left(I_1-I_2\right)+j5I_2=0$

Substitute 0 for $I_2$.

$\begin{array}{l}-440+4\left(I_1-0\right)-j4I_1+j5I_1+j6\left(I_1-0\right)+j5\left(0\right)=0\\ -440+\left(4+j\left(-4+5+6\right)\right)I_1=0\\ -440+\left(4+j7\right)I_1=0\end{array}$

Modify the Equation as follows.

$\begin{array}{c}{I}_1=\frac{-440}{-\left(4+j7\right)}\\ =\frac{440}{4+j7}\\ =\frac{440}{8.06226\angle 60.255°}\\ =54.575\angle -60.255°\text{A}\end{array}$

Write the expression for the open-circuit voltage.

$\begin{array}{c}{V}_{oc}=\left(j5\right)I_1+\left(4+j6\right)I_1\\ =\left(4+j11\right)I_1\end{array}$

Substitute $54.575\angle -60.255°\text{A}$ for $I_1$.

$\begin{array}{c}{V}_{oc}=\left(4+j11\right)\left(54.575\angle -60.255°\text{A}\right)\\ =\left(11.7047\angle 70.017°\right)\left(54.575\angle -60.255°\text{A}\right)\\ =638.78\angle 9.76°\text{V}\end{array}$

Consider the expression for the Thevenin voltage.

$V_{Th}=V_{oc}$

Substitute $638.78\angle 9.76°\text{V}$ for $V_{oc}$.

$V_{Th}=638.78\angle 9.76°\text{V}$

Modify the Figure 1 by short-circuiting the load side contains current $I_2$. The modified circuit as shown in Figure 2.

From Figure 2, consider that the loops 1 and 2 contain the currents $I_1$ and $I_2$ respectively.

Apply Kirchhoff's voltage law to the loop 1 in Figure 1.

$-440+\left(-j4+j5\right)I_1-j5I_2+\left(4+j6\right)\left(I_1-I_2\right)=0$

$\left(4+j7\right)I_1-\left(4+j11\right)I_2=440$        (1)

Apply Kirchhoff's voltage law to the loop 2 in Figure 1.

$\left(4+j6\right)\left(I_2-I_1\right)+j22I_2-j5I_1=0$

$\left(-4-j11\right)I_1+\left(4+j28\right)I_2=0$        (2)

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

$\left[\begin{array}{cc}\left(4+j7\right)& \left(-4-j11\right)\\ \left(-4-j11\right)& \left(4+j28\right)\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{c}440\\ 0\end{array}\right]$        (3)

Write the MATLAB code to solve the equation (3).

A = [(4+j*7) -(4+j*11);-(4+j*11) (4+j*28)];

B = [440; 0];

I = inv(A)*B

The output in command window:

I =

  61.069 - 121.926i

  14.369 - 54.571i

From the MATLAB output, the currents $I_1$ and $I_2$ are,

$\begin{array}{c}{I}_1=\left(61.069-j121.926\right)\text{A}\\ =136.3648\angle -63.39°\text{A}\end{array}$

And

$\begin{array}{c}{I}_2=\left(14.369-j54.571\right)\text{A}\\ =56.43\angle -75.248°\text{A}\end{array}$

The short-circuit current $I_{sc}$ is,

$I_{sc}=I_2$

Substitute $56.43\angle -75.248°\text{A}$ for $I_2$.

$I_{sc}=56.43\angle -75.248°\text{A}$

Write the expression for Thevenin’s equivalent impedance.

$Z_{Th}=\frac{V_{Th}}{I_{sc}}$

Substitute $638.78\angle 9.76°\text{V}$ for $V_{Th}$ and $56.43\angle -75.248°\text{A}$ for $I_{sc}$.

$\begin{array}{c}{Z}_{Th}=\frac{638.78\angle 9.76°\text{V}}{56.43\angle -75.248°\text{A}}\\ =11.32\angle 85.01°\Omega \end{array}$

Conclusion:

Thus, the Thevenin equivalent circuit parameters are $\underset{\_}{V_{Th}=638.78\angle 9.76°\text{V}}$ and $\underset{\_}{Z_{Th}=11.32\angle 85.01°\Omega }$.