Chapter 28, Problem 18SP

(a)

To determine

The equivalent resistance of $8\Omega$ and $4\Omega$ resistances in series.

Answer to Problem 18SP

Solution:

$\text{12 }\Omega$

Explanation of Solution

Given data:

The resistance of the first resistor is $8\Omega$.

The resistance of the second resistor is $\text{4 }\Omega$.

Formula used:

The expression for equivalent resistance in series is written as:

$R_{eq}=R_1+R_2$

Here, $R_1$ and $R_2$ are the resistances of the resistor in series and $R_{eq}$ is the equivalent resistance.

Explanation:

Recall the expression for equivalent resistance in series.

$R_{eq}=R_1+R_2$

Substitute $8\Omega$ for $R_1$ and $\text{4 }\Omega$ for $R_2$

$\begin{array}{c} R_{eq}=8\Omega +4\Omega \\ =12\Omega \end{array}$

Conclusion:

The equivalent resistance is $\text{12 }\Omega$.

(b)

To determine

The equivalent resistance of $8\Omega$ and $4\Omega$ resistances in parallel.

Answer to Problem 18SP

Solution:

The equivalent resistance is $\text{2}\text{.7 }\Omega$.

Explanation of Solution

Given data:

The resistance of the first resistor is $8\Omega$.

The resistance of the second resistor is $\text{4 }\Omega$.

Formula used:

The expression for equivalent resistance in parallel is written as:

$\begin{array}{l}\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\ \frac{1}{R_{eq}}=\frac{R_1+R_2}{R_1{R}_2}\\ R_{eq}=\frac{R_1{R}_2}{R_1+R_2}\end{array}$

Here, $R_1$ and $R_2$ are the resistances of the resistor in parallel and $R_{eq}$ is the equivalent resistance.

Explanation:

Recall the expression for equivalent resistance in parallel.

$R_{eq}=\frac{R_1{R}_2}{R_1+R_2}$

Substitute $8\Omega$ for $R_1$ and $\text{4 }\Omega$ for $R_2$

$\begin{array}{c}{R}_{eq}=\frac{\left(8\Omega \right)\left(4\Omega \right)}{8\Omega +4\Omega }\\ =\frac{32}{12}\\ =2.7\Omega \end{array}$

Conclusion:

The equivalent resistance is $\text{2}\text{.7 }\Omega$.