Chapter 13, Problem 14P

Obtain the Thevenin equivalent circuit for the circuit in Fig. 13.83 at terminals a-b.

To determine

Calculate the Thevenin equivalent to the circuit at terminals a-b.

Answer to Problem 14P

TheThevenin equivalent circuit parameters are $\underset{\_}{V_{Th}=106.98\angle 34.12°\text{V}}$ and $\underset{\_}{Z_{Th}=2.332\angle 50°\Omega }$.

Explanation of Solution

Given data:

Refer to Figure 13.83 in the textbook for the circuit with coupled coils.

Consider that the value of the source voltage.

$\begin{array}{c}V=200\angle 90°\text{V}\\ =j\text{200}\text{V}\end{array}$

Calculation:

Calculate the Thevenin voltage.

Modify the Figure 13.83 by converting the current source $\left(80\text{A}\right)$ with parallel resistor $\left(2\Omega \right)$ is converted into the voltage source $\left(80\text{A}\times 2\Omega =160\text{V}\right)$ with series resistor $\left(2\Omega \right)$. The modified circuit as shown in Figure 1.

Consider that the two coils are connected series aiding.

$j\omega L=j\omega L_1+j\omega L_2-2j\omega M$

Substitute $j6$ for $j\omega L_1$, $j8$ for $j\omega L_2$, and $j2$ for $j\omega M$.

$\begin{array}{c}j\omega L=j6+j8-2\left(j2\right)\\ =j10\Omega \end{array}$

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

$-j200+\left(5+j\omega L-j3+2\right)I+160=0$

Substitute $j10\Omega$ for $j\omega L$.

$-j200+\left(5+j10-j3+2\right)I+160=0$

Re-arrange the equation.

$I=\frac{-160+j200}{5+j10-j3+2}$

$I=\frac{-160+j200}{7+j7}\text{A}$        (1)

From Figure 1, consider the following expression using Kirchhoff's voltage law.

$-j200+\left(5+j6\right)I-j2I+V_{Th}=0$

Re-arrange the Equation.

$V_{Th}=j200-\left(5+j6\right)I+j2I$

$V_{Th}=j200-\left(5+j4\right)I$        (2)

Substitute Equation (1) in (2).

$\begin{array}{c}{V}_{Th}=j200-\left(5+j4\right)\left(\frac{-160+j200}{7+j7}\right)\\ =j200-\left(6.4031\angle 38.66°\right)\left(\frac{256.12\angle 128.66°}{9.8995\angle 45°}\right)\\ =j200-165.661\angle 122.32°\\ =j200+88.57-j140\end{array}$

Simplify the equation as follows.

$\begin{array}{c}{V}_{Th}=88.57+j60\\ =106.98\angle 34.115°\text{V}\end{array}$

To obtain Thevenin impedance $\left(Z_{Th}\right)$, set all the voltage sources to zero and insert a 1-A current source at the terminals a-b as shown in Figure 2.

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

Write the Kirchhoff's voltage law expression to Figure 2 using super mesh analysis.

$\left(5+j6\right)I_1-j2I_2+\left(2+j8-j3\right)I_2-j2I_1=0$

$\left(5+j4\right)I_1+\left(2+j3\right)I_2=0$        (3)

From Figure 2, write the current expression.

$\begin{array}{l}{I}_2-I_1=1\\ I_2=I_1+1\end{array}$

Substitute $I_1+1$ for $I_2$ in Equation (3).

$\begin{array}{l}\left(5+j4\right)I_1+\left(2+j3\right)\left(I_1+1\right)=0\\ \left(5+j4\right)I_1+\left(2+j3\right)I_1+\left(2+j3\right)=0\\ \left(7+j7\right)I_1=-\left(2+j3\right)\end{array}$

$I_1=\frac{-\left(2+j3\right)}{7+j7}\text{A}$        (4)

Apply Kirchhoff's voltage law to the loop 1 contains current $I_1$ in Figure 2.

$\left(5+j6\right)I_1-j2I_1+V_o=0$

Re-arrange the equation.

$V_o=-\left(5+j6\right)I_1+j2I_1$

$V_o=-\left(5+j4\right)I_1$        (5)

Substitute Equation (4) in (5).

$\begin{array}{c}{V}_o=-\left(5+j4\right)\left(\frac{-\left(2+j3\right)}{7+j7}\right)\\ =\frac{-2+j23}{7+j7}\\ =1.5+j1.7857\\ =2.332\angle 50°\text{V}\end{array}$

Write the expression for Thevenin’s equivalent impedance.

$Z_{Th}=\frac{V_o}{1\text{A}}$

Substitute $2.332\angle 50°\text{V}$ for $V_o$.

$\begin{array}{c}{Z}_{Th}=\frac{2.332\angle 50°\text{V}}{1\text{A}}\\ =2.332\angle 50°\Omega \end{array}$

Conclusion:

Thus, the Thevenin equivalent circuit parameters are $\underset{\_}{V_{Th}=106.98\angle 34.12°\text{V}}$ and $\underset{\_}{Z_{Th}=2.332\angle 50°\Omega }$.