Chapter 2, Problem 53P

Obtain the equivalent resistance R_ab in each of the circuits of Fig. 2.117. In (b), all resistors have a value of 30 Ω.

Figure 2.117

To determine

Calculate the equivalent resistor at terminals a-b in Figure 2.117(a).

Answer to Problem 53P

The equivalent resistor at terminals a-b in Figure 2.117(a) is $\underset{\_}{142.32\Omega }$.

Explanation of Solution

Formula used:

Consider the delta to wye conversions.

$R_1=\frac{R_b{R}_c}{R_a+R_b+R_c}$ (1)

$R_2=\frac{R_a{R}_c}{R_a+R_b+R_c}$ (2)

$R_3=\frac{R_a{R}_b}{R_a+R_b+R_c}$ (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 2.117(a) in the textbook For Prob.2.53.

Step 1:

In Figure 2.117(a), convert the delta connection into wye connection.

Consider $R_a=10\Omega ,R_b=40\Omega ,$ and, $R_c=50\Omega$.

Substitute $10\Omega$ for $R_a$, $40\Omega$ for $R_b$ and $50\Omega$ for $R_c$ in equation (1) to obtain $R_1$ value of wye connection.

$\begin{array}{c}{R}_1=\frac{\left(40\Omega \right)\left(50\Omega \right)}{10\Omega +40\Omega +50\Omega }\\ =\frac{2000}{100}\Omega \\ =20\Omega \end{array}$

Substitute $10\Omega$ for $R_a$, $40\Omega$ for $R_b$ and $50\Omega$ for $R_c$ in equation (2) to obtain $R_2$ value of wye connection.

$\begin{array}{c}{R}_2=\frac{\left(10\Omega \right)\left(50\Omega \right)}{10\Omega +40\Omega +50\Omega }\\ =\frac{500}{100}\Omega \\ =5\Omega \end{array}$

Substitute $10\Omega$ for $R_a$, $40\Omega$ for $R_b$ and $50\Omega$ for $R_c$ in equation (3) to obtain $R_3$ value of wye connection.

$\begin{array}{c}{R}_3=\frac{\left(10\Omega \right)\left(40\Omega \right)}{10\Omega +40\Omega +50\Omega }\\ =\frac{400\Omega }{100\Omega }\\ =4\Omega \end{array}$

Modify Figure 2.117(a) as shown in Figure 1.

Step 2:

In Figure 1, as $30\Omega \text{and}4\Omega$ are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq1}}=30\Omega +4\Omega \\ =34\Omega \end{array}$

Step 3:

In Figure 1, as $60\Omega \text{and}5\Omega$ are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq2}}=60\Omega +5\Omega \\ =65\Omega \end{array}$

Step 4:

In Figure 1, as $20\Omega \text{and}80\Omega$ are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq3}}=20\Omega +80\Omega \\ =100\Omega \end{array}$

Modify Figure 1 as shown in Figure 2.

Step 5:

In Figure 2, as $34\Omega \text{and}65\Omega$ are connected in parallel, therefore the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq4}}=\frac{1}{\left(\frac{1}{34\Omega }+\frac{1}{65\Omega }\right)}\\ =\frac{1}{\left[\frac{65\Omega +34\Omega }{\left(34\Omega \right)\left(65\Omega \right)}\right]}\Omega \\ =\frac{\left(34\Omega \right)\left(65\Omega \right)}{99}\Omega \\ =\frac{2210}{99}\Omega \end{array}$

$R_{\text{eq4}}=22.32\Omega$

Modify Figure 2 as shown in Figure 3.

Step 6:

In Figure 3, as $20\Omega ,22.32\Omega ,\text{and}100\Omega$ resistors are connected in series, therefore the equivalent resistance of the series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq}5}=20\Omega +22.32\Omega +100\Omega \\ =142.32\Omega \end{array}$

Conclusion:

Thus, the equivalent resistor at terminals a-b in Figure 2.117(a) is $\underset{\_}{142.32\Omega }$.

To determine

Calculate the equivalent resistor at terminals a-b in Figure 2.117(b).

Answer to Problem 53P

The equivalent resistor at terminals a-b in Figure 2.117(b) is $\underset{\_}{33.33\Omega }$.

Explanation of Solution

Given data:

All resistance have $30\Omega$.

Formula used:

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

$R_{\text{Y}}=\frac{R_{\Delta }}{3}$ (4)

Calculation:

Refer to Figure 2.117(b) in the textbook For Prob.2.53.

Step 1:

In Figure 2.117(b), at left most corner of circuit, as two resistors are connected in series, therefore the equivalent resistance for series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq}1}=30\Omega +30\Omega \\ =60\Omega \end{array}$

Step 2:

As $R_{\text{eq}1}$ and $30\Omega$ are connected in parallel, therefore the equivalent resistance for parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq2}}=\frac{1}{\left(\frac{1}{30\Omega }+\frac{1}{R_{\text{eq}1}}\right)}\\ =\frac{1}{\left(\frac{1}{30\Omega }+\frac{1}{60\Omega }\right)}\\ =\frac{1}{\left(\frac{2+1}{60\Omega }\right)}\\ =\frac{60\Omega }{3}\end{array}$

$R_{\text{eq2}}=20\Omega$

Modify Figure 2.117(b) as shown in Figure 4.

Step 3:

In Figure 4, as in upper part of the circuit all three $30\Omega$ resistors are connected in the form of delta with same branch values, therefore the delta connected resistance value can be connected into star connection as follows.

Substitute $30\Omega$ for $R_{\Delta }$ in equation (1) to obtain the branch values of wye.

$\begin{array}{c}{R}_{\text{Y}}=\frac{30\Omega }{3}\\ =10\Omega \end{array}$

Since all branch values are same in a delta connection that is $30\Omega$ , therefore all branches of delta will be same that is $10\Omega$.

Modify Figure 4 as shown in Figure 5.

Step 4:

In Figure 5, as $20\Omega \text{and}10\Omega$ are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq3}}=20\Omega +10\Omega \\ =30\Omega \end{array}$

Step 5:

In Figure 5, as in right most part of the circuit all three $30\Omega$ resistors are connected in the form of delta with same branch values, therefore the delta connected resistance value can be connected into star connection as follows.

Substitute $30\Omega$ for $R_{\Delta }$ in equation (1) to obtain the branch values of wye.

$\begin{array}{c}{R}_{\text{Y}}=\frac{30\Omega }{3}\\ =10\Omega \end{array}$

Since all branch values are same in a delta connection that is $30\Omega$ , therefore all branches of delta will be same that is $10\Omega$.

Modify Figure 5 as shown in Figure 6.

Step 6:

In Figure 6, as $10\Omega \text{and}10\Omega$ are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq4}}=10\Omega +10\Omega \\ =20\Omega \end{array}$

Step 7:

In Figure 6, as $10\Omega \text{and}30\Omega$ are connected in series, therefore the equivalent resistance of series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq5}}=10\Omega +30\Omega \\ =40\Omega \end{array}$

Step 8:

As $R_{\text{eq}4}\text{and}{R}_{\text{eq}5}$ are connected in parallel, therefore the equivalent resistance of parallel connected circuit is calculated as follows.

$\begin{array}{c}{R}_{\text{eq6}}=\frac{1}{\left(\frac{1}{R_{\text{eq4}}}+\frac{1}{R_{\text{eq5}}}\right)}\\ =\frac{1}{\left(\frac{1}{20\Omega }+\frac{1}{40\Omega }\right)}\\ =\frac{1}{\left(\frac{2+1}{40\Omega }\right)}\\ =\frac{40\Omega }{3}\end{array}$

$R_{\text{eq6}}=13.33\Omega$

Modify Figure 6 as shown in Figure 7.

Step 4:

In Figure 7, as $10\Omega ,13.33\Omega ,\text{and}10\Omega$ resistors are connected in series, therefore the equivalent resistance of the series connected circuit is calculated as follows.

$\begin{array}{c}{R}_{ab\left(b\right)}=10\Omega +13.33\Omega +10\Omega \\ =33.33\Omega \end{array}$

Conclusion:

Thus, the equivalent resistor at terminals a-b in Figure 2.117(b) is $\underset{\_}{33.33\Omega }$.