Chapter 4, Problem 42P

For the circuit in Fig. 4.109, find the Thevenin equivalent between terminals a and b.

Figure 4.109

To determine

Find the Thevenin voltage and Thevenin resistance at terminals a-b of the circuit shown in Figure 4.109.

Answer to Problem 42P

The Thevenin voltage is $10\text{V}$ and the Thevenin resistance is $10\Omega$ at terminals a-b of the given circuit.

Explanation of Solution

Given data:

Refer to Figure 4.109 in the textbook.

The voltage source is $20\text{V}$ and $30\text{V}$.

The current source is $5\text{A}$.

Calculation:

In the given circuit, find the Thevenin resistance by turning off $20\text{V}$ and $30\text{V}$ voltage source (replacing with a short circuit) and $5\text{A}$ current source (replacing with an open circuit).

The modified circuit is shown in Figure 1.

In Figure 1, $20\Omega$ and $20\Omega$ resistance are connected in parallel. Therefore, the equivalent resistance for the parallel connection is,

$\begin{array}{l}20\Omega \left|\right|20\Omega =\frac{20\Omega \times 20\Omega }{20\Omega +20\Omega }\\ =10\Omega \end{array}$

The modified circuit is shown in Figure 2.

In Figure 2, the three $10\Omega$ resistors connected to the center of the circuit in wye connection at the point “x”. This wye connected resistors is converted to the delta connection as shown in Figure 3.

For the delta connection in Figure 3, the value of the resistor $R_1$ is calculated as follows,

$\begin{array}{c}{R}_1=\frac{\left(\left(10\times 10\right)+\left(10\times 10\right)+\left(10\times 10\right)\right)}{10}\\ =\frac{300}{10}\\ =30\Omega \end{array}$

Similarly,

$\begin{array}{c}{R}_2=\frac{\left(\left(10\times 10\right)+\left(10\times 10\right)+\left(10\times 10\right)\right)}{10}\\ =\frac{300}{10}\\ =30\Omega \end{array}$

And,

$\begin{array}{c}{R}_3=\frac{\left(\left(10\times 10\right)+\left(10\times 10\right)+\left(10\times 10\right)\right)}{10}\\ =\frac{300}{10}\\ =30\Omega \end{array}$

The modified circuit is shown in Figure 4.

In Figure 4, $10\Omega$ and $30\Omega$ resistors are connected in parallel. Therefore, the equivalent resistance for the parallel connection is,

$\begin{array}{l}10\Omega \left|\right|30\Omega =\frac{10\Omega \times 30\Omega }{10\Omega +30\Omega }\\ =7.5\Omega \end{array}$

The modified circuit is shown in Figure 5.

In Figure 5, the Thevenin resistance is,

$\begin{array}{c}{R}_{\text{Th}}=\left[\left(7.5+7.5\right)\left|\right|30\right]\\ =\left(15\left|\right|30\right)\\ =\left(\frac{15\times 30}{15+30}\right)\\ =10\Omega \end{array}$

Refer to Figure 4.109 in the textbook.

The given circuit is modified as shown in Figure 6.

In Figure 6, the voltage source with series resistance is converted into current source with parallel resistance by source transformation method.

That is,

$\begin{array}{c}{I}_1=\frac{20\text{V}}{20\Omega }\\ =1\text{A}\left\{\because 1\text{A}=\frac{1\text{V}}{1\Omega }\right\}\end{array}$

Similarly, the current source with parallel resistance is converted into voltage source with series resistance by source transformation method.

That is,

$\begin{array}{c}{V}_1=5\text{A}\times \text{10}\Omega \\ =50\text{V}\left\{\because 1\text{V}=1\text{A}\times \text{1}\Omega \right\}\end{array}$

The source transformation is shown in Figure 7.

In Figure 7, $20\Omega$ and $20\Omega$ resistance are connected in parallel. Therefore, the equivalent resistance for the parallel connection is,

$\begin{array}{l}20\Omega \left|\right|20\Omega =\frac{20\Omega \times 20\Omega }{20\Omega +20\Omega }\\ =10\Omega \end{array}$

The modified circuit is shown in Figure 8.

In Figure 8, the current source with parallel resistance is converted into voltage source with series resistance by source transformation method.

That is,

$\begin{array}{c}{V}_2=1\text{A}\times \text{10}\Omega \\ =10\text{V}\left\{\because 1\text{V}=1\text{A}\times \text{1}\Omega \right\}\end{array}$

The source transformation is shown in Figure 9.

In Figure 9, apply Kirchhoff’s voltage law to the loop $i_1$ as follows.

$\begin{array}{l}-30+50+30i_1-10i_2=0\\ 30i_1-10i_2=-20\end{array}$

$3i_1-i_2=-2$        (1)

Rearrange the equation (1) as follows,

$i_2=3i_1+2$

In Figure 9, apply Kirchhoff’s voltage law to the loop $i_2$ as follows.

$\begin{array}{l}-50-10+30i_2-10i_1=0\\ 30i_2-10i_1=60\end{array}$

$-i_1+3i_2=6$        (2)

Substitute $3i_1+2$ for $i_2$ in equation (2) as follows,

$\begin{array}{l}-i_1+3\left(3i_1+2\right)=6\\ -i_1+9i_1+6=6\\ 8i_1=6-6\\ i_1=0\text{A}\end{array}$

Substitute 0 for $i_1$ in equation (2) as follows,

$\begin{array}{l}-\left(0\right)+3i_2=6\\ 3i_2=6\\ i_2=2\text{A}\end{array}$

In Figure 9, apply Kirchhoff’s voltage law to the outer loop as follows.

$-v_{ab}-10i_1+30-10i_2=0$        (3)

Substitute 0 for $i_1$ and 2 for $i_2$ in equation (3) to find the voltage across the terminals a-b in volts.

$\begin{array}{l}-v_{ab}-10\left(0\right)+30-10\left(2\right)=0\\ -v_{ab}-0+30-20=0\\ -v_{ab}=-10\\ v_{ab}=10\text{V}\end{array}$

Since, the voltage $v_{ab}$ at terminals a-b is the Thevenin voltage. Therefore,

$V_{\text{Th}}=v_{ab}=10\text{V}$

The Thevenin equivalent is shown in Figure 10.

Conclusion:

Thus, the Thevenin voltage is $10\text{V}$ and the Thevenin resistance is $10\Omega$ at terminals a-b of the given circuit.