Chapter 2, Problem 56P

Determine V in the circuit of Fig. 2.120.

Figure 2.120

To determine

Calculate the value of voltage $V$ in Figure 2.120.

Answer to Problem 56P

The value of voltage $V$ in Figure 2.120 is $\underset{\_}{42.15\text{V}}$.

Explanation of Solution

Formula used:

Consider the wye to delta conversions.

$R_a=\frac{R_1{R}_2+R_2{R}_3+R_3{R}_1}{R_1}$ (1)

$R_b=\frac{R_1{R}_2+R_2{R}_3+R_3{R}_1}{R_2}$ (2)

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

Here,

$R_1,R_2,R_3,R_a,R_b,R_c$ are resistors.

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 2.120 in the textbook For Prob.2.56.

Step 1:

From Figure 2.120, consider $R_1=15\Omega ,R_2=10\Omega ,\text{and}{R}_3=12\Omega$.

Substitute $15\Omega$ for $R_1$, $10\Omega$ for $R_2$ and $12\Omega$ for $R_3$ in equation (1) to obtain $R_a$ value of delta connection.

$\begin{array}{c}{R}_a=\frac{\left(15\Omega \right)\left(10\Omega \right)+\left(10\Omega \right)\left(12\Omega \right)+\left(12\Omega \right)\left(15\Omega \right)}{15\Omega }\\ =\frac{150+120+180}{15}\Omega \\ =\frac{450}{15}\Omega \\ =30\Omega \end{array}$

Substitute $15\Omega$ for $R_1$, $10\Omega$ for $R_2$ and $12\Omega$ for $R_3$ in equation (2) to obtain $R_b$ value of delta connection.

$\begin{array}{c}{R}_b=\frac{\left(15\Omega \right)\left(10\Omega \right)+\left(10\Omega \right)\left(12\Omega \right)+\left(12\Omega \right)\left(15\Omega \right)}{10\Omega }\\ =\frac{150+120+180}{10}\Omega \\ =\frac{450}{10}\Omega \\ =45\Omega \end{array}$

Substitute $15\Omega$ for $R_1$, $10\Omega$ for $R_2$ and $12\Omega$ for $R_3$ in equation (3) to obtain $R_c$ value of delta connection.

$\begin{array}{c}{R}_c=\frac{\left(15\Omega \right)\left(10\Omega \right)+\left(10\Omega \right)\left(12\Omega \right)+\left(12\Omega \right)\left(15\Omega \right)}{12\Omega }\\ =\frac{150+120+180}{12}\Omega \\ =\frac{450}{12}\Omega \\ =37.5\Omega \end{array}$

Modify Figure 2.120 as shown in Figure 1.

Step 2:

In Figure 1, as $30\Omega \text{and}20\Omega$ are connected in parallel, the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq1}}=\frac{1}{\left(\frac{1}{30\Omega }+\frac{1}{20\Omega }\right)}\\ =\frac{1}{\left(\frac{30\Omega +20\Omega }{\left(30\Omega \right)\left(20\Omega \right)}\right)}\\ =\frac{\left(30\Omega \right)\left(20\Omega \right)}{50\Omega }\\ =\frac{600}{50}\Omega \end{array}$

$R_{\text{eq1}}=12\Omega$

Step 3:

In Figure 1, as $35\Omega \text{and}45\Omega$ are connected in parallel, the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq2}}=\frac{1}{\left(\frac{1}{35\Omega }+\frac{1}{45\Omega }\right)}\\ =\frac{1}{\left(\frac{35\Omega +45\Omega }{\left(35\Omega \right)\left(45\Omega \right)}\right)}\\ =\frac{\left(35\Omega \right)\left(45\Omega \right)}{80\Omega }\\ =\frac{1575}{80}\Omega \end{array}$

$R_{\text{eq2}}=19.68\Omega$

Step 4:

In Figure 1, as $30\Omega \text{and}37.5\Omega$  resistors are connected in parallel, the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq3}}=\frac{1}{\left(\frac{1}{30\Omega }+\frac{1}{37.5\Omega }\right)}\\ =\frac{1}{\left(\frac{30\Omega +37.5\Omega }{\left(30\Omega \right)\left(37.5\Omega \right)}\right)}\\ =\frac{\left(30\Omega \right)\left(37.5\Omega \right)}{67.5\Omega }\\ =\frac{1125}{67.5}\Omega \end{array}$

$R_{\text{eq3}}=16.667\Omega$

Modify Figure 1 as shown in Figure 2.

Step 5:

In Figure 2, as $15\Omega$ and $13.846\Omega$ are connected in series, the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq}4}=15\Omega +13.846\Omega \\ =28.846\Omega \end{array}$

Modify Figure 2 as shown in Figure 3.

Step 6:

In Figure 3, as $19.68\Omega \text{and}28.667\Omega$ are connected in parallel, the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq5}}=\frac{1}{\left(\frac{1}{19.68\Omega }+\frac{1}{28.667\Omega }\right)}\\ =\frac{1}{\left(\frac{19.68\Omega +28.667\Omega }{\left(19.68\Omega \right)\left(28.667\Omega \right)}\right)}\\ =\frac{\left(19.68\Omega \right)\left(28.667\Omega \right)}{48.347\Omega }\\ =\frac{564.16}{48.347}\Omega \end{array}$

$R_{\text{eq5}}=11.66\Omega$

Modify Figure 3 as shown in Figure 4.

Step 7:

Apply voltage division rule to Figure 4.

$\begin{array}{c}V=100\text{V}\left(\frac{11.66\Omega }{11.66\Omega +16\Omega }\right)\\ =100\text{V}\left(\frac{11.66\Omega }{27.66\Omega }\right)\\ =100\text{V}\left(0.4215\right)\\ =42.15\text{V}\end{array}$

Conclusion:

Thus, the value of voltage $V$ in Figure 2.120 is $\underset{\_}{42.15\text{V}}$.