Chapter 2, Problem 57P

Find R_eq and I in the circuit of Fig. 2.121.

Figure 2.121

To determine

Calculate the values of equivalent resistance $R_{\text{eq}}$ and current $I$ to the circuit connection in Figure 2.121.

Answer to Problem 57P

The values of equivalent resistance $R_{\text{eq}}$ and current $I$ to the circuit connection in Figure 2.121 are $\underset{\_}{32.44\Omega \text{and}1.5413\text{A}}$, respectively.

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 2.121 in the textbook For Prob.2.57.

Step 1:

In Figure 2.121, as $25\Omega \text{and}5\Omega$ resistors at the right most top corner are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq1}}=25\Omega +5\Omega \\ =30\Omega \end{array}$

Step 2:

In Figure 2.121, as $25\Omega \text{and}15\Omega$ resistors at the right most bottom corner are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq2}}=25\Omega +15\Omega \\ =40\Omega \end{array}$

Step 3:

In Figure 2.121, convert the wye- sub network into delta connection.

Substitute $10\Omega$ for $R_{\text{Y}}$ in equation (1) to obtain the branch values of delta.

$\begin{array}{c}{R}_{\Delta }=3\left(10\Omega \right)\\ =30\Omega \end{array}$

Since all branches values are same in a wye connection that is $R_1=R_2=R_3=10\Omega$ , therefore all branches of delta will be same that is $R_a=R_b=R_c=30\Omega$.

Modify Figure 2.121 as shown in Figure 1.

Step 4:

In Figure 1, as $20\Omega ,80\Omega \text{and}10\Omega$ are connected in star connection, convert the star connection to delta connection as follows.

$\begin{array}{c}{R}_a=20\Omega +40\Omega +\frac{\left(20\Omega \right)\left(40\Omega \right)}{80\Omega }\\ =20\Omega +40\Omega +10\Omega \\ =70\Omega \end{array}$

$\begin{array}{c}{R}_b=80\Omega +20\Omega +\frac{\left(80\Omega \right)\left(20\Omega \right)}{40\Omega }\\ =80\Omega +20\Omega +40\Omega \\ =140\Omega \end{array}$

$\begin{array}{c}{R}_c=80\Omega +40\Omega +\frac{\left(80\Omega \right)\left(40\Omega \right)}{20\Omega }\\ =80\Omega +40\Omega +160\Omega \\ =280\Omega \end{array}$

Modify Figure 1 as shown in Figure 2.

Step 5:

In Figure 2, as two $30\Omega$ are connected in parallel, therefore the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq3}}=\frac{30\Omega }{2}\\ =15\Omega \end{array}$

Step 6:

In Figure 2, as $30\Omega \text{and}140\Omega$ are connected in parallel, therefore the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq4}}=\frac{1}{\left(\frac{1}{30\Omega }+\frac{1}{140\Omega }\right)}\\ =\frac{\left(30\Omega \right)\left(140\Omega \right)}{30\Omega +140\Omega }\\ =\frac{4200}{170}\Omega \\ =24.706\Omega \end{array}$

Step 7:

In Figure 2, as $10\Omega \text{and}280\Omega$ are connected in parallel, therefore the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq5}}=\frac{1}{\left(\frac{1}{10\Omega }+\frac{1}{280\Omega }\right)}\\ =\frac{\left(10\Omega \right)\left(280\Omega \right)}{10\Omega +280\Omega }\\ =\frac{2800}{290}\Omega \\ =9.6552\Omega \end{array}$

Modify Figure 2 as shown in Figure 3.

Step 8:

In Figure 3, as $30\Omega ,24.706\Omega \text{and}15\Omega$ are connected in delta connection, convert the delta connection to star connection as follows.

$\begin{array}{c}{R}_1=\frac{\left(30\Omega \right)\left(15\Omega \right)}{30\Omega +15\Omega +24.706\Omega }\\ =\frac{450}{69.706}\Omega \\ =6.45568\Omega \\ \cong 6.4557\Omega \end{array}$

$\begin{array}{c}{R}_2=\frac{\left(24.706\Omega \right)\left(15\Omega \right)}{30\Omega +15\Omega +24.706\Omega }\\ =\frac{370.59}{69.706}\Omega \\ =5.31647\Omega \\ \cong 5.3165\Omega \end{array}$

$\begin{array}{c}{R}_3=\frac{\left(24.706\Omega \right)\left(30\Omega \right)}{30\Omega +15\Omega +24.706\Omega }\\ =\frac{741.18}{69.706}\Omega \\ =10.6329\Omega \end{array}$

Modify Figure 3 as shown in Figure 4.

Step 9:

In Figure 4, as $10\Omega \text{and}6.4557\Omega$ resistors are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq6}}=10\Omega +6.4557\Omega \\ =16.4557\Omega \end{array}$

Step 10:

In Figure 4, as $5.3165\Omega \text{and}70\Omega$ resistors are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq7}}=5.3165\Omega +70\Omega \\ =75.3165\Omega \end{array}$

Step 11:

In Figure 4, as $10.6329\Omega \text{and}9.6552\Omega$ resistors are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq8}}=10.6329\Omega +9.6552\Omega \\ =20.2881\Omega \end{array}$

Modify Figure 4 as shown in Figure 5.

Step 12:

In Figure 5, as $20.2881\Omega \text{and}75.3165\Omega$ are connected in parallel, therefore the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq9}}=\frac{1}{\left(\frac{1}{20.2881\Omega }+\frac{1}{75.3165\Omega }\right)}\\ =\frac{\left(20.2881\Omega \right)\left(75.3165\Omega \right)}{20.2881\Omega +75.3165\Omega }\\ =\frac{1528.028}{95.6046}\Omega \\ =15.98\Omega \end{array}$

Modify Figure 5 as shown in Figure 6.

Step 13:

In Figure 6, as $16.4557\Omega \text{and}15.98\Omega$ are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq}}=16.4557\Omega +15.98\Omega \\ =32.4357\Omega \\ \cong 32.44\Omega \end{array}$

Consider the general expression to find current $I$ in Figure 6.

$\begin{array}{c}I=\frac{50\text{V}}{32.44\Omega }\text{A}\\ =\text{1}\text{.5413}\text{A}\end{array}$

Conclusion:

Thus, the values of equivalent resistance $R_{\text{eq}}$ and current $I$ to the circuit connection in Figure 2.121 are $\underset{\_}{32.44\Omega \text{and}1.5413\text{A}}$, respectively.