Chapter 2, Problem 2.1P

Reduce each of the networks shown in Figure P2.1 to a single equivalent resistance by combining resistances in series and parallel.
* Denotes that answers are contained in the Student Solutions flies. See Appendix E for more information about accessing the Student Solutions

FIgure P2.1

To determine

(a)

The equivalent resistance by combining resistance in series.

Answer to Problem 2.1P

The value of equivalent resistance is $20\Omega$.

Explanation of Solution

Calculation:

The required diagram is shown in Figure 1.

Mark the resistance $R_1$, $R_2$, $R_3$, $R_4$, $R_5$, $R_6$ and $R_7$ in the above figure.

The required diagram is shown in Figure 2.

The value of resistance $R_{eq1}$ is calculated as,

  $R_{eq1}=R_7\parallel R_6$

Substitute $12\Omega$ for $R_6$ and $24\Omega$ for $R_7$ in the above equation.

  $\begin{array}{c}{R}_{eq1}=12\Omega \parallel 24\Omega \\ =\frac{\left(12\Omega \right)\left(24\Omega \right)}{12\Omega +24\Omega }\\ =8\Omega \end{array}$

The required diagram is shown in Figure 3.

The value of resistance $R_{eq2}$ is calculated as,

  $R_{eq2}=R_4+R_5+R_{eq1}$

Substitute $3\Omega$ for $R_4$, $4\Omega$ for $R_5$ and $8\Omega$ for $R_{eq1}$ in the above equation.

  $\begin{array}{c}{R}_{eq2}=3\Omega +4\Omega +8\Omega \\ =15\Omega \end{array}$

The required diagram is shown in Figure 4.

The value of resistance $R_{eq3}$ is calculated as,

  $R_{eq3}=R_3\parallel R_{eq2}$

Substitute $15\Omega$ for $R_{eq2}$ and $30\Omega$ for $R_3$ in the above equation.

  $\begin{array}{c}{R}_{eq3}=15\Omega \parallel 30\Omega \\ =\frac{\left(15\Omega \right)\left(30\Omega \right)}{15\Omega +30\Omega }\\ =10\Omega \end{array}$

The required diagram is shown in Figure 5.

The value of resistance $R_{eq}$ is calculated as,

  $R_{eq}=R_1+R_2+R_{eq3}$

Substitute $3\Omega$ for $R_1$, $7\Omega$ for $R_2$ and $10\Omega$ for $R_{eq3}$ in the above equation.

  $\begin{array}{c}{R}_{eq}=3\Omega +7\Omega +10\Omega \\ =20\Omega \end{array}$

The required diagram is shown in Figure 6.

Conclusion:

Therefore, the value of equivalent resistance is $20\Omega$ .

To determine

(b)

The equivalent resistance by combining resistance in series.

Answer to Problem 2.1P

The value of equivalent resistance is $23\Omega$.

Explanation of Solution

Calculation:

The required diagram is shown in Figure 7.

The value of resistance $R_{eq1}$ is calculated as,

  $R_{eq1}=R_7\parallel R_8$

Substitute $15\Omega$ for $R_7$ and $60\Omega$ for $R_8$ in the above equation.

  $\begin{array}{c}{R}_{eq1}=15\Omega \parallel 60\Omega \\ =\frac{\left(15\Omega \right)\left(60\Omega \right)}{15\Omega +60\Omega }\\ =12\Omega \end{array}$

The required diagram is shown in Figure 8.

The value of resistance $R_{eq2}$ is calculated as,

  $R_{eq2}=R_6+R_{eq1}$

Substitute $6\Omega$ for $R_6$ and $12\Omega$ for $R_{eq1}$ in the above equation.

  $\begin{array}{c}{R}_{eq2}=6\Omega +12\Omega \\ =18\Omega \end{array}$

The equivalent resistance is shown in Figure 9.

The value of resistance $R_{eq3}$ is calculated as,

  $R_{eq3}=R_5\parallel R_{eq2}$

Substitute $18\Omega$ for $R_{eq2}$ and $9\Omega$ for $R_5$ in the above equation.

  $\begin{array}{c}{R}_{eq3}=9\Omega \parallel 18\Omega \\ =\frac{\left(9\Omega \right)\left(18\Omega \right)}{9\Omega +18\Omega }\\ =6\Omega \end{array}$

The required diagram is shown in Figure 10.

The value of resistance $R_{eq4}$ is calculated as,

  $R_{eq4}=R_4+R_{eq3}$

Substitute $6\Omega$ for $R_4$ and $6\Omega$ for $R_{eq3}$ in the above equation.

  $\begin{array}{c}{R}_{eq4}=6\Omega +6\Omega \\ =12\Omega \end{array}$

The equivalent resistance is shown in Figure 11.

The value of resistance $R_{eq5}$ is calculated as,

  $R_{eq5}=R_3\parallel R_{eq4}$

Substitute $12\Omega$ for $R_{eq4}$ and $24\Omega$ for $R_5$ in the above equation.

  $\begin{array}{c}{R}_{eq5}=12\Omega \parallel 24\Omega \\ =\frac{\left(12\Omega \right)\left(24\Omega \right)}{12\Omega +24\Omega }\\ =8\Omega \end{array}$

The required diagram is shown in Figure 12.

The value of resistance $R_{eq}$ is calculated as,

  $R_{eq}=R_1+R_2+R_{eq5}$

Substitute $10\Omega$ for $R_1$, $5\Omega$ for $R_2$ and $8\Omega$ for $R_{eq5}$ in the above equation.

  $\begin{array}{c}{R}_{eq}=10\Omega +5\Omega +8\Omega \\ =23\Omega \end{array}$

The required diagram is shown in Figure 13.

Conclusion:

Therefore, the value of equivalent resistance is $23\Omega$ .