Chapter 5, Problem 62E

Employ Δ–Y conversion techniques as appropriate to determine R_in as labeled in Fig. 5.101.

FIGURE 5.101

To determine

Employ Δ–Y conversion techniques as appropriate to determine $R_{\text{in}}$.

Answer to Problem 62E

The value of $R_{\text{in}}$ is $3.57\Omega$.

Explanation of Solution

Formula used:

The expression for the equivalent resistor when resistors are connected in series is as follows:

$R_{eq}=R_1+R_2+\mathrm{......}+R_N$ (1)

Here,

$R_1$, $R_2$,…, $R_N$ are the resistances.

The expression for the equivalent resistor when resistors are connected in parallel is as follows:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\mathrm{......}+\frac{1}{R_N}$ (2)

Here,

$R_1$, $R_2$,…, $R_N$ are the resistances.

The $Y$-network and $\Delta$-network circuit diagram is given in Figure 1.

Refer to the redrawn Figure 1:

The expression for the conversion of $Y$-network into $\Delta$-network is as follows:

$R_a=\frac{R_1{R}_2+R_2{R}_3+R_3{R}_1}{R_1}$ (3)

$R_b=\frac{R_1{R}_2+R_2{R}_3+R_3{R}_1}{R_2}$ (4)

$R_c=\frac{R_1{R}_2+R_2{R}_3+R_3{R}_1}{R_3}$ (5)

Here,

$R_1$, $R_2$ and $R_3$ are the resistors connected in $Y$-network and

$R_a$, $R_b$ and $R_c$ are the resistors connected in $\Delta$-network.

The expression for the conversion of $\Delta$-network into $Y$-network is as follows:

$R_1=\frac{R_b{R}_c}{R_a+R_b+R_c}$ (6)

$R_2=\frac{R_a{R}_c}{R_a+R_b+R_c}$ (7)

$R_3=\frac{R_a{R}_b}{R_a+R_b+R_c}$ (8)

Calculation:

The redrawn circuit diagram is given in Figure 2:

Refer to the redrawn Figure 2:

$6\Omega$ and $12\Omega$ resistor are connected in series.

Substitute $6\Omega$ for $R_1$ and $12\Omega$ for $R_2$ equation (1):

$\begin{array}{c}{R}_{eq}=6\Omega +12\Omega \\ =18\Omega \end{array}$

The simplified circuit diagram is given in Figure 3.

Refer to the redrawn Figure 3:

$\text{18 }\Omega$ and $\text{20 }\Omega$ resistor are connected in parallel.

Substitute $\text{18 }\Omega$ for $R_1$ and $\text{20 }\Omega$ for $R_2$ equation (2):

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{\text{18 }\Omega }+\frac{1}{\text{20 }\Omega }\\ =\frac{19}{\text{180 }\Omega }\end{array}$

Rearrange the equation for $R_{eq}$:

$\begin{array}{c}{R}_{eq}=\frac{\text{180}}{19}\Omega \\ =9.474\Omega \end{array}$

The simplified circuit diagram is given in Figure 4.

Refer to the redrawn Figure 4:

The $9\Omega$, $5\Omega$ and $\text{6 }\Omega$ resistors are connected in $Y$-network.

Substitute $9\Omega$ for $R_1$, $5\Omega$ for $R_2$ and $\text{6 }\Omega$ for $R_3$ in equation (3):

$\begin{array}{c}{R}_a=\frac{\left(9\Omega \right)\left(\text{5 }\Omega \right)+\left(\text{5 }\Omega \right)\left(\text{6 }\Omega \right)+\left(\text{6 }\Omega \right)\left(9\Omega \right)}{9\Omega }\\ =\frac{129}{9}\Omega \\ =14.3\Omega \end{array}$

Substitute $9\Omega$ for $R_1$, $5\Omega$ for $R_2$ and $\text{6 }\Omega$ for $R_3$ in equation (4):

$\begin{array}{c}{R}_b=\frac{\left(9\Omega \right)\left(\text{5 }\Omega \right)+\left(\text{5 }\Omega \right)\left(\text{6 }\Omega \right)+\left(\text{6 }\Omega \right)\left(9\Omega \right)}{\text{5 }\Omega }\\ =\frac{129}{5}\Omega \\ =25.8\Omega \end{array}$

Substitute $9\Omega$ for $R_1$, $5\Omega$ for $R_2$ and $\text{6 }\Omega$ for $R_3$ in equation (5):

$\begin{array}{c}{R}_c=\frac{\left(9\Omega \right)\left(\text{5 }\Omega \right)+\left(\text{5 }\Omega \right)\left(\text{6 }\Omega \right)+\left(\text{6 }\Omega \right)\left(9\Omega \right)}{\text{6 }\Omega }\\ =\frac{129}{6}\Omega \\ =21.5\Omega \end{array}$

The simplified circuit diagram is given in Figure 5:

Refer to the redrawn Figure 5:

$\text{4 }\Omega$ and $\text{14}\text{.3 }\Omega$ resistor are connected in parallel:

Substitute $\text{4 }\Omega$ for $R_1$ and $\text{14}\text{.3 }\Omega$ for $R_2$ in equation (2):

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{\text{4 }\Omega }+\frac{1}{\text{14}\text{.3 }\Omega }\\ =\frac{183}{\text{572 }\Omega }\end{array}$

Rearrange the equation for $R_{eq}$:

$\begin{array}{c}{R}_{eq}=\frac{\text{572}}{183}\Omega \\ =3.126\Omega \end{array}$

$\text{3 }\Omega$ and $\text{25}\text{.8 }\Omega$ resistor are connected in parallel.

Substitute $\text{3 }\Omega$ for $R_1$ and $\text{25}\text{.8 }\Omega$ for $R_2$ equation (2):

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{\text{3 }\Omega }+\frac{1}{\text{25}\text{.8 }\Omega }\\ =0.37209{\Omega }^{-}\end{array}$

Rearrange the equation for $R_{eq}$:

$\begin{array}{c}{R}_{eq}=\frac{1}{0.37209}\Omega \\ =2.688\Omega \end{array}$

The simplified circuit diagram is given in Figure 6.

Refer to the redrawn Figure 6:

$21.5\Omega$, $3.126\Omega$ and $\text{2}\text{.688 }\Omega$ resistors are connected in $\Delta$-network.

Substitute $21.5\Omega$ for $R_a$, $3.126\Omega$ for $R_b$ and $\text{2}\text{.688 }\Omega$ for $R_c$ in equation (6):

$\begin{array}{c}{R}_1=\frac{\left(3.126\Omega \right)\left(\text{2}\text{.688 }\Omega \right)}{21.5\Omega +3.126\Omega +\text{2}\text{.688 }\Omega }\\ =\frac{8.403}{27.314}\Omega \\ =0.3076\Omega \end{array}$

Substitute $21.5\Omega$ for $R_a$, $3.126\Omega$ for $R_b$ and $\text{2}\text{.688 }\Omega$ for $R_c$ in equation (7):

$\begin{array}{c}{R}_2=\frac{\left(\text{2}\text{.688 }\Omega \right)\left(21.5\Omega \right)}{21.5\Omega +3.126\Omega +\text{2}\text{.688 }\Omega }\\ =\frac{57.792}{27.314}\Omega \\ =2.11\Omega \end{array}$

Substitute $21.5\Omega$ for $R_a$, $3.126\Omega$ for $R_b$ and $\text{2}\text{.688 }\Omega$ for $R_c$ in equation (8):

$\begin{array}{c}{R}_3=\frac{\left(21.5\Omega \right)\left(3.126\Omega \right)}{21.5\Omega +3.126\Omega +\text{2}\text{.688 }\Omega }\\ =\frac{67.209}{27.314}\Omega \\ =2.461\Omega \end{array}$

The simplified circuit diagram is given in Figure 7.

Refer to the redrawn Figure 7:

$10\Omega$ and $0.3076\Omega$ resistor are connected in series.

Substitute $10\Omega$ for $R_1$ and $0.3076\Omega$ for $R_2$ equation (1):

$\begin{array}{c}{R}_{eq}=10\Omega +0.3076\Omega \\ =10.3076\Omega \end{array}$

The simplified circuit diagram is given in Figure 8:

Refer to the redrawn Figure 8:

The $10.3076\Omega$, $\text{2}\text{.116 }\Omega$ and $\text{2}\text{.461 }\Omega$ resistors are connected in $Y$-network.

Substitute $10.3076\Omega$ for $R_1$, $\text{2}\text{.116 }\Omega$ for $R_2$ and $\text{2}\text{.461 }\Omega$ for $R_3$ in equation (3):

$\begin{array}{c}{R}_a=\frac{\left(10.3076\Omega \right)\left(\text{2}\text{.116 }\Omega \right)+\left(\text{2}\text{.116 }\Omega \right)\left(\text{2}\text{.461 }\Omega \right)+\left(\text{2}\text{.461 }\Omega \right)\left(10.3076\Omega \right)}{10.3076\Omega }\\ =\frac{52.4}{10.3076}\Omega \\ =5.4\Omega \end{array}$

Substitute $10.3076\Omega$ for $R_1$, $\text{2}\text{.116 }\Omega$ for $R_2$ and $\text{2}\text{.461 }\Omega$ for $R_3$ in equation (4):

$\begin{array}{c}{R}_b=\frac{\left(10.3076\Omega \right)\left(\text{2}\text{.116 }\Omega \right)+\left(\text{2}\text{.116 }\Omega \right)\left(\text{2}\text{.461 }\Omega \right)+\left(\text{2}\text{.461 }\Omega \right)\left(10.3076\Omega \right)}{\text{2}\text{.116 }\Omega }\\ =\frac{52.4}{\text{2}\text{.116}}\Omega \\ =24.76\Omega \end{array}$

Substitute $10.3076\Omega$ for $R_1$, $\text{2}\text{.116 }\Omega$ for $R_2$ and $\text{2}\text{.461 }\Omega$ for $R_3$ in equation (5):

$\begin{array}{c}{R}_c=\frac{\left(10.3076\Omega \right)\left(\text{2}\text{.116 }\Omega \right)+\left(\text{2}\text{.116 }\Omega \right)\left(\text{2}\text{.461 }\Omega \right)+\left(\text{2}\text{.461 }\Omega \right)\left(10.3076\Omega \right)}{\text{2}\text{.116 }\Omega }\\ =\frac{52.4}{\text{2}\text{.461}}\Omega \\ =21.29\Omega \end{array}$

The simplified circuit diagram is given in Figure 9:

Refer to the redrawn Figure 9:

$\text{7 }\Omega$ and $\text{24}\text{.76 }\Omega$ resistor are connected in parallel.

Substitute $\text{7 }\Omega$ for $R_1$ and $\text{24}\text{.76 }\Omega$ for $R_2$ in equation (2):

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{\text{7 }\Omega }+\frac{1}{\text{24}\text{.76 }\Omega }\\ =\frac{794}{4333\Omega }\end{array}$

Rearrange the equation for $R_{eq}$:

$\begin{array}{c}{R}_{eq}=\frac{4333}{794}\Omega \\ =5.457\Omega \end{array}$

$\text{21}\text{.29 }\Omega$ and $\text{9}\text{.474 }\Omega$ resistor are connected in parallel.

Substitute $\text{21}\text{.29 }\Omega$ for $R_1$ and $\text{24}\text{.76 }\Omega$ for $R_2$ in equation (2):

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{21.29}+\frac{1}{\text{9}\text{.474 }\Omega }\\ =0.152522{\Omega }^{-}\end{array}$

Rearrange the equation for $R_{eq}$:

$\begin{array}{c}{R}_{eq}=\frac{1}{0.152522}\Omega \\ =6.56\Omega \end{array}$

The simplified circuit diagram is given in Figure 10.

Refer to the redrawn Figure 10:

$5.457\Omega$ and $6.56\Omega$ resistor are connected in series.

Substitute $5.457\Omega$ for $R_1$ and $6.56\Omega$ for $R_2$ in equation (1):

$\begin{array}{c}{R}_{eq}=5.457\Omega +6.56\Omega \\ =12.017\Omega \end{array}$

The simplified circuit diagram is given in Figure 11.

Refer to the redrawn Figure 11:

$\text{12}\text{.017 }\Omega$ and $\text{5}\text{.082 }\Omega$ resistor are connected in parallel.

Substitute $\text{12}\text{.017 }\Omega$ for $R_1$ and $\text{5}\text{.082 }\Omega$ for $R_2$ equation (2):

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{\text{12}\text{.017 }\Omega }+\frac{1}{\text{5}\text{.082 }\Omega }\\ =0.2799{\Omega }^{-}\end{array}$

Rearrange the equation for $R_{eq}$:

$\begin{array}{c}{R}_{eq}=\frac{1}{0.2799}\Omega \\ =3.57\Omega \end{array}$

The simplified circuit diagram is given in Figure 12.

Conclusion:

Thus, the value of $R_{\text{in}}$ is $3.57\Omega$.