Chapter 4.6, Problem 39E

(a)

To determine

To find: The rational equation to model the situation.

Answer to Problem 39E

The rational equation to model the situationis $\frac{1}{2R_2}+\frac{1}{R_2}=\frac{1}{10}-\frac{1}{20}$ .

Explanation of Solution

Given:

The general equation is $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$ and In the circuit $R_1$ is equal to twice of $R_2$ and $R_3$ is $20\text{ohms}$ and the equivalent resistance equals to $10\text{ohms}$ .

Calculation:

Substitute $10\text{ohms}$ for $R$ and $20\text{ohms}$ for $R_3$ , $2R_2$ for $R_1$ in the equation $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$ to find the model.

  $\begin{array}{c}\frac{1}{10}=\frac{1}{2R_2}+\frac{1}{R_2}+\frac{1}{20}\\ \frac{1}{2R_2}+\frac{1}{R_2}=\frac{1}{10}-\frac{1}{20}\end{array}$

Therefore, the rational equation to model the situationis $\frac{1}{2R_2}+\frac{1}{R_2}=\frac{1}{10}-\frac{1}{20}$ .

(b)

To determine

To find: The value of $R_1$ and $R_2$ .

Answer to Problem 39E

The value of $R_1$ and $R_2$ are $30\text{ohms}$ and $15\text{ohms}$ respectively.

Explanation of Solution

Given:

The general equation is $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$ and In the circuit $R_1$ is equal to twice of $R_2$ and $R_3$ is $20\text{ohms}$ and the equivalent resistance equals to $10\text{ohms}$ .

Calculation:

Solve the equation $\frac{1}{d_1}=\frac{1}{8}-\frac{1}{32}$ .

  $\begin{array}{c}\frac{1}{2R_2}+\frac{1}{R_2}=\frac{1}{10}-\frac{1}{20}\\ \frac{3}{2R_2}=\frac{1}{20}\\ 3\left(10\right)=R_2\\ R_2=30\text{ohms}\end{array}$

So, $R_1=\frac{30}{2}=15\text{ohms}$ .

Therefore, thevalue of $R_1$ and $R_2$ are $30\text{ohms}$ and $15\text{ohms}$ respectively.