Chapter 2, Problem 52P

For the circuit shown in Fig. 2.116, find the equivalent resistance. All resistors are 3Ω.

Figure 2.116

To determine

Calculate the equivalent resistance to the circuit connection in Figure 2.116.

Answer to Problem 52P

The equivalent resistance to the circuit connection in Figure 2.116 is $\underset{\_}{9.218\Omega }$.

Explanation of Solution

Formula used:

Consider the following delta to wye conversion, when all branches in a delta consist same value.

$R_{\text{Y}}=\frac{R_{\Delta }}{3}$ (1)

Consider the expression for $N$ resistors connected in parallel.

$\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdots +\frac{1}{R_N}$

Here,

$R_1,R_2,R_3\cdots R_N$ are resistors.

Consider the expression for $N$ resistors connected in series.

$R_{\text{eq}}=R_1+R_2+R_3+\cdots +R_N$

Calculation:

Refer to Figure 2.116 in the textbook For Prob.2.52.

Step 1:

In Figure 2.116, convert the wye- sub network at right side top corner connection into delta connection.

Substitute $3\Omega$ for $R_{\text{Y}}$ in equation (1) to obtain the branch values of delta.

$\begin{array}{c}{R}_{\Delta }=3\left(3\Omega \right)\\ =9\Omega \end{array}$

Since all branches values are same in a wye connection that is $R_1=R_2=R_3=3\Omega$ , therefore all branches of delta will be same that is $R_a=R_b=R_c=9\Omega$.

Step 2:

In Figure 2.116, as two $3\Omega$ resistors at the right most bottom corner are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq1}}=3\Omega +3\Omega \\ =6\Omega \end{array}$

Modify Figure 2.116 as shown in Figure 1.

Step 3:

In Figure 1, as $9\Omega \text{and}3\Omega$ are connected in parallel, therefore the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq2}}=\frac{1}{\left(\frac{1}{9\Omega }+\frac{1}{3\Omega }\right)}\\ =\frac{1}{\left(\frac{1+3}{9\Omega }\right)}\\ =\frac{9\Omega }{4}\\ =2.25\Omega \end{array}$

Step 4:

In Figure 1, $6\Omega \text{and}3\Omega$ are connected in parallel, therefore the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq3}}=\frac{1}{\left(\frac{1}{6\Omega }+\frac{1}{3\Omega }\right)}\\ =\frac{1}{\left(\frac{1+2}{6\Omega }\right)}\\ =\frac{6\Omega }{3}\\ =2\Omega \end{array}$

Modify Figure 1 as shown in Figure 2.

Step 5:

In Figure 2, as $2.25\Omega ,3\Omega \text{and}2.25\Omega$ are connected in star connection, convert the star connection to delta connection as follows.

$\begin{array}{c}{R}_a=2.25\Omega +3\Omega +\frac{\left(2.25\Omega \right)\left(3\Omega \right)}{2.25\Omega }\\ =2.25\Omega +3\Omega +3\Omega \\ =8.25\Omega \end{array}$

$\begin{array}{c}{R}_b=2.25\Omega +2.25\Omega +\frac{\left(2.25\Omega \right)\left(2.25\Omega \right)}{3\Omega }\\ =2.25\Omega +2.25\Omega +1.6785\Omega \\ =6.1875\Omega \\ \cong 6.188\Omega \end{array}$

$\begin{array}{c}{R}_c=2.25\Omega +3\Omega +\frac{\left(2.25\Omega \right)\left(3\Omega \right)}{2.25\Omega }\\ =2.25\Omega +3\Omega +3\Omega \\ =8.25\Omega \end{array}$

Modify Figure 2 as shown in Figure 3.

Step 6:

In Figure 3, $6.188\Omega \text{and}9\Omega$ are connected in parallel, therefore the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq4}}=\frac{1}{\left(\frac{1}{6.188\Omega }+\frac{1}{9\Omega }\right)}\\ =\frac{\left(6.188\Omega \right)\left(9\Omega \right)}{6.188\Omega +9\Omega }\\ =\frac{55.692}{15.188}\Omega \\ =3.667\Omega \end{array}$

Step 7:

In Figure 3, $8.25\Omega \text{and}2\Omega$ are connected in parallel, therefore the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq5}}=\frac{1}{\left(\frac{1}{8.25\Omega }+\frac{1}{2\Omega }\right)}\\ =\frac{\left(8.25\Omega \right)\left(2\Omega \right)}{8.25\Omega +2\Omega }\\ =\frac{16.5}{10.25}\Omega \\ =1.6098\Omega \end{array}$

Modify Figure 3 as shown in Figure 4.

Step 8:

In Figure 4, as $3.667\Omega \text{and}1.6098\Omega$ are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq6}}=3.667\Omega +1.6098\Omega \\ =5.2768\Omega \\ \cong 5.277\Omega \end{array}$

Modify Figure 4 as shown in Figure 5.

Step 9:

In Figure 5, $8.25\Omega \text{and}5.277\Omega$ are connected in parallel, therefore the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq7}}=\frac{1}{\left(\frac{1}{8.25\Omega }+\frac{1}{5.277\Omega }\right)}\\ =\frac{\left(8.25\Omega \right)\left(5.277\Omega \right)}{8.25\Omega +5.277\Omega }\\ =\frac{43.535}{13.527}\Omega \\ =3.218\Omega \end{array}$

Modify Figure 5 as shown in Figure 6.

Step 10:

In Figure 6, as $3\Omega ,3.218\Omega \text{and}3\Omega$ are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq}}=3\Omega +3.218\Omega +3\Omega \\ =9.218\Omega \end{array}$

Conclusion:

Thus, the equivalent resistance to the circuit connection in Figure 2.116 is $\underset{\_}{9.218\Omega }$.